23.31 - Miscellaneous Examples (3) on Integrals

In the previous section, we saw some miscellaneous examples on integrals. In this section, we will see a few more miscellaneous examples.

Solved Example 23.148
Integrate $\small{\frac{e^x}{(1+e^x)(2+e^x)}}$    
Solution:
1. Let us put $\small{u = 1+e^x}$

• Then $\small{\frac{du}{dx}~=~e^x \Rightarrow e^x\,dx~=~du}$

• Also, since $\small{u = 1+e^x}$, we can write:

$\small{2+e^x = u+1}$

2. So we want:

$\small{I = \int{\left[\frac{e^x}{(1+e^x)(2+e^x)}\right]dx}= \int{\left[\frac{1}{(u)(u+1)}\right]du}= \int{\left[\frac{1}{u^2 + u}\right]du}}$

3. This is a standard integral of the form:

$\small{ \int{\left[\frac{px + q}{ax^2 + bx + c}\right]du}}$, where p = 0, q = 1, a = 1, b = 1 and c = 0

We get: $\small{I~=~\int{\left[\frac{1}{u^2 + u}\right]du}~=~(-1)\log \left| \frac{1}{u}~+~1 \right|~+~\rm{C}}$

$\small{~=~(-1)\log \left| \frac{1+u}{u} \right|~+~\rm{C}~=~\log \left| \frac{u}{1+u} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{I~=~\log \left| \frac{1+e^x}{2+e^x} \right|~+~\rm{C}}$

Solved Example 23.149
Integrate $\small{f'(ax+b) \left[f(ax+b) \right]^n}$
Solution:
1. Let us put $\small{u = f(ax+b)}$

• Then $\small{\frac{du}{dx}~=~a\,f'(ax+b) \Rightarrow a\,f'(ax+b)\,dx~=~du}$

2. So we want:

$\small{I = \int{\left[f'(ax+b) \left[f(ax+b) \right]^n\right]dx}= \int{\left[\frac{(a)f'(ax+b) \left[f(ax+b) \right]^n}{a}\right]dx}}$

$\small{= \int{\left[\frac{u^n}{a}\right]du}}$

3. This integration gives: $\small{I = \frac{u^{n+1}}{a(n+1)}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{I = \frac{[f(ax+b]^{n+1}}{a(n+1)}~+~\rm{C}}$

Solved Example 23.150
Integrate $\small{\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x}$
Solution:
1. 1. First we will rearrange the given expression:

$\small{\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x~=~\left(\frac{2 + 2\sin(x) \cos (x)}{2 \cos^2 x} \right)e^x}$

$\small{~=~\left(\frac{1}{\cos^2 x}~+~\tan x \right)e^x~=~\left(\sec^2 x~+~\tan x \right)e^x}$

2. So we want:

$\small{I = \int{\left[\left(\sec^2 x~+~\tan x \right)e^x\right]dx}}$

• $\small{\sec^2 x}$ is the derivative of $\small{\tan x}$. So this is of the form:

$\small{\int{\left[\left(f(x)~+~f'(x) \right)e^x\right]dx}~=~e^x\,f(x)~+~\rm{C}}$

3. Thus we get:

$\small{I = \int{\left[\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x\right]dx}= \int{\left[\left(\sec^2 x~+~\tan x \right)e^x\right]dx}~=~e^x\,\tan(x)~+~\rm{C}}$

Solved Example 23.151
Integrate $\small{\frac{\sqrt{x^2 + 1}\left[\log(x^2 + 1)~-~2 \log x \right]}{x^4}}$
Solution:
1. First we will rearrange the given expression:

Put $\small{x~=~\tan u}$

$\small{\Rightarrow 1+x^2~=~1+\tan^2 u~=~\sec^2 u}$

• Also, $\small{\frac{dx}{du}~=~\sec^2 u \Rightarrow dx~=~\sec^2 u \, du}$

• So we can write:

$\small{I~=~\int{\left[\frac{\sqrt{x^2 + 1}\left[\log(x^2 + 1)~-~2 \log x \right]}{x^4} \right]dx}~=~\int{\left[\frac{\sqrt{\sec^2 u}\left[\log(\sec^2 u)~-~ \log (\tan^2 u) \right]}{\tan^4 u} \right]\sec^2 u\,du}}$

$\small{~=~\int{\left[\frac{\left[\log\left(\frac{\sec^2 u}{\tan^2 u} \right) \right]}{\tan^4 u} \right]\sec^3 u\,du}~=~\int{\left[\frac{\left[\log\left(\frac{1}{\sin^2 u} \right) \right]}{\tan^4 u\left(\cos^3 u \right)} \right]du}}$

$\small{~=~\int{\left[\frac{\cos u\left[\log\left(\frac{1}{\sin^2 u} \right) \right]}{\sin^4 u} \right]du}}$

2. We will use the method of integration by parts.

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(u)=\log\left(\frac{1}{\sin^2 u} \right)}$

   ♦ Let second function be: $\small{g(u)=\frac{\cos u}{\sin^4 u}}$

(b) Finding A:

$\small{A~=~\int{\left[g(u) \right]dx}~=~\frac{- 1}{3\sin^3 u}}$

(c) $\small{\big[f(u) \left(A \right) \big]~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]}$

• This is the first term.

(d) $\small{f'(u)~=~\frac{-2 \cos(u)}{\sin(u)}}$

(e) $\small{\int{\big[f'(u)\,\left(A \right)  \big]du}~=~\int{\big[\frac{-2 \cos(u)}{\sin(u)}\,\left(\frac{- 1}{3 \sin^3 (u)} \right)\big]du}}$

$\small{~=~\frac{2}{3}\int{\big[\frac{\cos(u)}{\sin^4(u)}\big]du}~=~\frac{-2}{9\sin^3 u}}$

• This is the second term.

(f) So we get:

$\small{I~=~\text{First term - Second term}}$

$\small{~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]~-~\big[\frac{-2}{9 \sin^3 u} \big]}$

$\small{~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]~+~\big[\frac{2}{9 \sin^3 u} \big]}$

$\small{~=~\big[\frac{2}{9 \sin^3 u} \big]~-~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{1}{3 \sin^3 u} \big]}$

$\small{~=~\big[\frac{1}{3\sin^3 u} \big]\bigg[\frac{2}{3}~-~ \log\left(\frac{1}{\sin^2 u} \right)\bigg]}$

3. Now we will substitute for u:

(a) We wrote $\small{x~=~\tan u}$

$\small{\Rightarrow 1+x^2~=~1+\tan^2 u~=~\sec^2 u}$

$\small{\Rightarrow \cos^2 u~=~\frac{1}{1+x^2}}$

$\small{\Rightarrow \sin^2 u~=~1~-~\frac{1}{1+x^2}~=~\frac{x^2}{1+x^2}}$

$\small{\Rightarrow \frac{1}{\sin^2 u}~=~1~+~\frac{1}{x^2}}$

$\small{\Rightarrow \frac{1}{\sin u}~=~\left(1~+~\frac{1}{x^2} \right)^{1/2}}$

$\small{\Rightarrow \frac{1}{\sin^3 u}~=~\left(1~+~\frac{1}{x^2} \right)^{3/2}}$

(b) So we get:

$\small{I~=~\big[\frac{1}{3\sin^3 u} \big]\bigg[\frac{2}{3}~-~ \log\left(\frac{1}{\sin^2 u} \right)\bigg]}$

$\small{~=~\big[\frac{1}{3} \left(1~+~\frac{1}{x^2} \right)^{3/2} \big]\bigg[\frac{2}{3}~-~ \log\left(1~+~\frac{1}{x^2} \right)\bigg]~+~\rm{C}}$

Solved Example 23.152
Integrate $\small{\frac{1}{e^x~+~e^{-x}}}$
Solution:
1. First we will rearrange the given expression:

$\small{\frac{1}{e^x~+~e^{-x}}~=~\frac{1}{e^x~+~\frac{1}{e^x}}~=~\frac{e^x}{e^{2x}~+~1}}$

Put $\small{u~=~e^x}$

• Then, $\small{\frac{du}{dx}~=~e^x \Rightarrow e^x\,dx ~=~du}$

2. So we can write:

$\small{I~=~\int{\left[\frac{1}{e^x~+~e^{-x}} \right]dx}~=~\int{\left[\frac{e^x}{e^{2x}~+~1} \right]du}~=~\int{\left[\frac{1}{u^{2}~+~1} \right]du}}$

3. This is a standard integral. We get:

$\small{I~=~\tan^{-1}u~+~\rm{C}}$

4. Substituting for u, we get:

$\small{I~=~\tan^{-1}(e^x)~+~\rm{C}}$

Solved Example 23.153
Evaluate $\small{\int_0^1{\left[e^{2 - 3x} \right]dx}}$ as a limit of a sum.
Solution:
1. 1. $\small{\int_{0}^{1}{\left[e^{2-3x} \right]dx}}$ is the area bounded by the four items:
   ♦ The curve $\small{y = f(x) = e^{2-3x}}$
   ♦ The vertical line x = 0 (y-axis)
   ♦ The vertical line x = 1
   ♦ The horizontal line y = 0 (x-axis)
   
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = 0 and b = 1

So $\small{h~=~\frac{b-a}{n}~=~\frac{1-0}{n}~=~\frac{1}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{1}{n} \left[f(a+\frac{1}{n})~+~f(a+\frac{2}{n})~+~f(a+\frac{3}{n})~+~~.~.~.~+f(a+(n)\frac{1}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{1}{n} \bigg[\big[e^{2-3(a+\frac{1}{n})}\big]~+~\big[e^{2-3(a+\frac{2}{n})}\big]~+~\big[e^{2-3(a+\frac{3}{n})}\big]~+~~.~.~.~}$

$\small{~~~~~~~~~.~.~.~+~\big[e^{2-3(a+(n)\frac{1}{n})}\big]\bigg]}$

4. Let us determine the quantity inside the large square brackets. We have to do a summation:

$\small{e^{2-3(a+\frac{1}{n})}~+~e^{2-3(a+\frac{2}{n})}~+~e^{2-3(a+\frac{3}{n})}~+~.~.~.~ \text{n terms}}$

$\small{~=~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-3}{n}} \right)~+~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-6}{n}} \right)~+~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-9}{n}} \right)~+~.~.~.~ \text{n terms}}$

$\small{~=~e^{2-3a}\left(e^{\frac{-3}{n}}~+~e^{\frac{-6}{n}}~+~e^{\frac{-9}{n}}~+~.~.~.~ \text{n terms} \right)}$

$\small{~=~e^{2}\left(e^{\frac{-3}{n}}~+~e^{\frac{-6}{n}}~+~e^{\frac{-9}{n}}~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~~\because {3a}~=~{3(0)}~=~0}$

• So inside the brackets, we have a geometric progression.

• First term = $\small{e^{\frac{-3}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{-6}{n}}}{e^{\frac{-3}{n}}}~=~\frac{e^{\frac{-9}{n}}}{e^{\frac{-6}{n}}}~=~e^{\frac{-3}{n}}}$

• Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^{n-1}~-~1 \right)}{r~-~1}}$

$\small{r^{n-1}~=~\left(e^{\frac{-3}{n}}\right)^{n-1}~=~e^{\frac{-3n+3}{n}}~=~e^{-3+\frac{3}{n}}}$

• So we get:

Sum of all terms of the G.P

$\small{~=~\frac{e^{\frac{-3}{n}}~\times~\left({e^{-3+\frac{3}{n}}}~-~1 \right)}{e^{\frac{-3}{n}}~-~1}~=~\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1}}$

• So the summation is:

$\small{e^2 \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

5. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{1}{n} \Bigg[e^2 \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right] \Bigg]}$

$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{1}{n}  \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{1}{n}  \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~e^2 \lim_{n\rightarrow \infty} \frac{1}{n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{(-3)}{(-3)n}  \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~e^2 \lim_{n\rightarrow \infty} \frac{(-3)}{(-3)n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~\frac{e^{-1}}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{{1} }{e^{\frac{-3}{n}}~-~1} \right]~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

6. In the above expression, the limits can be evaluated as follows:

$\small{\frac{e^{-1}}{(-3)} \lim_{n\rightarrow \infty}   \left[\frac{{1} }{\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}} \right]
~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty}   \left[\frac{{e^{\frac{-3}{n}}} }{\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}}  \right]}$

$\small{~=~\frac{e^{-1}}{(-3)}  \bigg[\frac{1}{1} \bigg]~-~\frac{e^2}{(-3)}\big[\frac{1}{1} \big]~=~\frac{e^{-1}~-~e^2}{(-3)}}$

$\small{~=~\frac{e^{2}~-~e^{-1}}{3}}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{-3}{n}} \right]~=~e^{\frac{-3}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{-3}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$   

Solved Example 23.154
Evaluate $\small{\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{\sin x \cos x}{\cos^4 x + \sin^4 x}~=~\frac{\sin x \cos x}{\cos^4 x + \sin^4 x~+~2\sin^2x\,\cos^2x~-~2\sin^2x\,\cos^2x}}$

$\small{~=~\frac{\sin x \cos x}{\left(\cos^2 x + \sin^2 x \right)^2~-~2\sin^2x\,\cos^2x}~=~\frac{\sin x \cos x}{1~-~2\sin^2x\,\cos^2x}}$

$\small{~=~\frac{\sin x \cos x}{1~-~(4\sin^2x\,\cos^2x)/2}~=~\frac{2\sin x \cos x}{2~-~4\sin^2x\,\cos^2x}}$

$\small{~=~\frac{\sin (2x)}{2~-~\sin^2(2x)}~=~\frac{\sin (2x)}{1+[1~-~\sin^2(2x)]}~=~\frac{\sin (2x)}{1+[\cos^2(2x)]}}$

2. So we want: $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \right]dx}~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$

3. First we will find the indefinite integral:

$\small{F~=~\int{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$

• Put $\small{u = \cos(2x)}$

Then $\small{\frac{du}{dx}~=~-2\sin(2x) \Rightarrow -2\sin(2x)\,dx~=~du}$

• We want:

$\small{F~=~\int{\left[\frac{(-2)\sin (2x)}{(-2)\left[1+\cos^2(2x) \right]} \right]dx}~=~\int{\left[\frac{1}{(-2)\left[1+u^2 \right]} \right]dx}~=~\frac{1}{(-2)}\int{\left[\frac{1}{1+u^2 } \right]dx}}$

• This is a standard integral. We get:

$\small{F~=~\left(\frac{-1}{2} \right)\tan^{-1}u}$

• Subsituting for u, we get:

$\small{F~=~\left(\frac{-1}{2} \right)\tan^{-1}\left[\cos(2x) \right]}$

4. Now we can evaluate the definite integral. We get:

$\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$

$\small{~=~\big[\left(\frac{-1}{2} \right)\tan^{-1}\left[\cos(2x) \right]\big]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[\cos(2x) \right]\big]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[\cos\left(\frac{\pi}{2} \right) \right]~-~\tan^{-1}\left[\cos\left(0 \right) \right]\big]}$

$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[0 \right]~-~\tan^{-1}\left[1 \right]\big]}$

$\small{~=~\left(\frac{-1}{2} \right) \big[0~-~\frac{\pi}{4}\big]~=~\frac{\pi}{8}}$

Solved Example 23.155
Evaluate $\small{\int_{\frac{\pi}{2}}^{\pi}{\left[e^x\left(\frac{1-\sin x}{1 - \cos x} \right) \right]dx}}$
Solution:
1. Let us rearrange the portion inside braces:

$\small{\frac{1-\sin x}{1 - \cos x}~=~\frac{1-\sin x}{2 \sin^2\left(\frac{x}{2} \right)}~=~\frac{1}{2 \sin^2\left(\frac{x}{2} \right)}~-~\frac{\sin x}{2 \sin^2\left(\frac{x}{2} \right)}}$

$\small{~=~\frac{1}{2 \sin^2\left(\frac{x}{2} \right)}~-~\frac{2 \sin\left(\frac{x}{2} \right) \cos\left(\frac{x}{2} \right)}{2 \sin^2\left(\frac{x}{2} \right)}}$

$\small{~=~\frac{\csc^2\left(\frac{x}{2} \right)}{2}~-~\cot\left(\frac{x}{2} \right)~=~(-1)\left[\cot\left(\frac{x}{2} \right)~-~\frac{\csc^2\left(\frac{x}{2} \right)}{2} \right]}$

$\small{~=~(-1)\left[\cot\left(\frac{x}{2} \right)~+~\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$

2. So the given expression can be written as:

$\small{(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)~+~\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$

• In the above result, $\small{\left[\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$ is the derivative of $\small{\left[\cot\left(\frac{x}{2} \right) \right]}$

• So the given expression can be written as:

$\small{(-1)\,e^x\left[f(x)~+~f'(x) \right]}$

3. So the indefinite integral can be written as:

$\small{F~=~(-1)\,e^x\left[f(x)\right]~=~(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)\right]}$

4. Now we can evaluate the definite integral. We get:

$\small{I~=~\int_{\frac{\pi}{2}}^{\pi}{\left[e^x\left(\frac{1-\sin x}{1 - \cos x} \right) \right]dx}}$

$\small{~=~\big[(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)\right]\big]_{\frac{\pi}{2}}^{\pi}}$

$\small{~=~\big[(-1)\,e^{\pi}\left[\cot\left(\frac{\pi}{2} \right)\right]\big]~-~\big[(-1)\,e^{\frac{\pi}{2}}\left[\cot\left(\frac{\pi}{4} \right)\right]\big]}$

$\small{~=~\big[(-1)\,e^{\pi}\left[0\right]\big]~-~\big[(-1)\,e^{\frac{\pi}{2}}\left[1\right]\big]}$

$\small{~=~e^{\frac{\pi}{2}}}$

Solved Example 23.156
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x}~=~\frac{\cos^2 x}{\cos^2 x ~+~ 4(1-\cos^2 x)}~=~\frac{\cos^2 x}{4~-~3 \cos^2 x}}$

$\small{~=~\left(\frac{-1}{3} \right)\left[\frac{(-3)\cos^2 x}{4~-~3 \cos^2 x} \right]~=~\left(\frac{-1}{3} \right)\left[\frac{4~-~3\cos^2 x~-~4}{4~-~3 \cos^2 x} \right]}$

$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4}{4~-~3 \cos^2 x} \right]~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{\frac{4}{\cos^2 x}}{\frac{4}{\cos^2 x}~-~\frac{3\cos^2 x}{\cos^2 x}} \right]}$

$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{4 \sec^2 x~-~3} \right]~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{4 (1+\tan^2 x)~-~3} \right]}$

$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \right]}$ 

2. Now we can find the indefinite integral:

$\small{F~=~\int{\left[\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x} \right]dx}~=~\int{\bigg[\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \right] \bigg]dx}}$


$\small{~=~\int{\bigg[\left(\frac{-1}{3} \right) \bigg]dx}~+~\left(\frac{1}{3} \right)\int{\bigg[\frac{4 \sec^2 x}{1~+~4 \tan^2 x}  \bigg]dx}}$

$\small{~=~\frac{-x}{3}~+~\left(\frac{1}{3} \right)\int{\bigg[\frac{4 \sec^2 x}{1~+~4 \tan^2 x}  \bigg]dx}}$

$\small{~=~F_1~+~F_2}$

• $\small{F_2}$ can be calculated as follows:

• Put $\small{u~=~\tan x}$. Then $\small{\frac{du}{dx}~=~\sec^2 x \Rightarrow \sec^2 x\,dx~=~du}$

• So we get:

$\small{F_2~=~\left(\frac{1}{3} \right)\int{\bigg[\frac{4}{1~+~4(u)^2}  \bigg]dx}~=~\left(\frac{4}{3} \right)\int{\bigg[\frac{1}{1~+~(2u)^2}  \bigg]dx}}$

$\small{~=~\left(\frac{4}{3} \right)\left(\frac{1}{2} \right)\bigg[\tan^{-1}(2u)  \bigg]~=~\left(\frac{2}{3} \right)\bigg[\tan^{-1}(2 \tan x)  \bigg]}$

• Therefore, $\small{F~=~\frac{-x}{3}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan x)  \right]}$

3. Now we can evaluate the definite integral. We get:

$\small{I~=~\big[\frac{-x}{3}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan x)  \right]\big]_{0}^{\frac{\pi}{2}}}$

$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan \frac{\pi}{2})  \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan 0)  \right]\big]}$

$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(\infty)  \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(0)  \right]\big]}$

$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\frac{\pi}{2}  \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[0  \right]\big]}$

$\small{~=~\big[\frac{-\pi}{6}~+~\frac{2\pi}{6}\big]~=~\frac{\pi}{6}}$

Solved Example 23.157
Evaluate $\small{\int_{0}^{1}{\left[\frac{1}{\sqrt{1+x}~-~\sqrt{x}} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{1}{\sqrt{1+x}~-~\sqrt{x}}~=~\frac{1(\sqrt{1+x}~+~\sqrt{x})}{(\sqrt{1+x}~-~\sqrt{x})(\sqrt{1+x}~+~\sqrt{x})}}$

$\small{~=~\frac{\sqrt{1+x}~+~\sqrt{x}}{1+x~-~x}~=~\sqrt{1+x}~+~\sqrt{x}}$

2. So we can write:

$\small{\int_{0}^{1}{\left[\frac{1}{\sqrt{1+x}~-~\sqrt{x}} \right]dx}~=~\int_{0}^{1}{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}}$

3. Now we can write the indefinite integral:

$\small{F~=~\int{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}~=~\frac{(1+x)^{3/2}}{3/2}~+~\frac{(x)^{3/2}}{3/2}}$

$\small{~=~\frac{(1+x)^{3/2}~+~(x)^{3/2}}{3/2}}$

4. Now we can evaluate the definite integral. We get:

$\small{I~=~\int_{0}^{1}{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}}$

$\small{~=~\big[\frac{(1+x)^{3/2}~+~(x)^{3/2}}{3/2}\big]_{0}^{1}}$

$\small{~=~\big[\frac{(1+1)^{3/2}~+~(1)^{3/2}}{3/2}\big]~-~\big[\frac{(1+0)^{3/2}~+~(0)^{3/2}}{3/2}\big]}$

$\small{~=~\big[\frac{2^{3/2}~+~1}{3/2}\big]~-~\big[\frac{1~+~0}{3/2}\big]~=~\frac{2^{3/2}}{3/2}~=~\frac{2^{5/2}}{3}~=~\frac{4 \sqrt2}{3}}$

Solved Example 23.158
Evaluate $\small{\int_{0}^{\pi}{\left[\frac{x \tan x}{\sec x~+~\tan x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{x \tan x}{\sec x~+~\tan x}~=~\frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x}~+~\frac{\sin x}{\cos x}}~=~\frac{x \sin x}{1~+~\sin x}}$

2. So we want:

$\small{I~=~\int_{0}^{\pi}{\left[\frac{x \tan x}{\sec x~+~\tan x} \right]dx}~=~\int_{0}^{\pi}{\left[\frac{x \sin x}{1~+~\sin x} \right]dx}}$

3. Applying P₄, we get:

$\small{I~=~\int_{0}^{\pi}{\left[\frac{(\pi-x) \sin (\pi-x)}{1~+~\sin (\pi-x)} \right]dx}~=~\int_{0}^{\pi}{\left[\frac{(\pi-x) \sin x}{1~+~\sin x} \right]dx}}$

$\small{~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}~-~\int_{0}^{\pi}{\left[\frac{ x \sin x}{1~+~\sin x} \right]dx}}$

$\small{\Rightarrow I~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}~-~I}$

$\small{\Rightarrow 2I~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}}$

$\small{\Rightarrow I~=~\frac{\pi}{2}\int_{0}^{\pi}{\left[\frac{ \sin x}{1~+~\sin x} \right]dx}}$

$\small{\Rightarrow I~=~\frac{\pi}{2}\int_{0}^{\pi}{\left[\frac{\sin x (1~-~\sin x)}{(1~+~\sin x)(1~-~\sin x)} \right]dx}}$

$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin x~-~\sin^2 x}{1~-~\sin^2 x} \right]dx}~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin x~-~\sin^2 x}{\cos^2 x} \right]dx}}$

$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x~-~\tan^2 x \right]dx}}$

$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x~-~\left(\sec^2 x - 1 \right) \right]dx}}$

$\small{\Rightarrow I~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x \right]dx}~-~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec^2 x \right]dx}~+~\frac{\pi}{2} \int_{0}^{\pi}{\left[1 \right]dx}}$

$\small{\Rightarrow I~=~I_1~-~I_2~+~I_3}$

4. First we will calculate I₁:

$\small{F_1~=~\frac{\pi}{2} \int{\left[\sec x\, \tan x \right]dx}~=~\frac{\pi}{2}\,\sec x}$

$\small{\Rightarrow I_1~=~\left[\frac{\pi}{2}\,\sec x \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\,\sec \left(\pi \right) \right]~-~\left[\frac{\pi}{2}\,\sec \left(0 \right) \right]}$

$\small{\Rightarrow I_1~=~\left[\frac{-\pi}{2} \right]~-~\left[\frac{\pi}{2} \right]~=~-\pi}$

5. Next we will calculate I₂:

$\small{F_2~=~\frac{\pi}{2} \int{\left[\sec^2 x \right]dx}~=~\frac{\pi}{2}\,\tan x}$

$\small{\Rightarrow I_2~=~\left[\frac{\pi}{2}\,\tan x \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\,\tan \left(\pi \right) \right]~-~\left[\frac{\pi}{2}\,\tan \left(0 \right) \right]}$

$\small{\Rightarrow I_2~=~\left[0 \right]~-~\left[0 \right]~=~0}$

6. Finally we will calculate I₃:

$\small{F_3~=~\frac{\pi}{2} \int{\left[1\right]dx}~=~\frac{\pi}{2}\,x}$

$\small{\Rightarrow I_3~=~\left[\frac{\pi}{2}\,(x) \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\, \left(\pi \right) \right]~-~\left[\frac{\pi}{2} \left(0 \right) \right]~=~\frac{\pi^2}{2}}$

7. From (3), (4), (5) and (6), we get:

$\small{I~=~I_1~-~I_2~+~I_3}$

$\small{~=~-\pi~-~0~+~\frac{\pi^2}{2}}$

$\small{~=~\left(\frac{\pi^2}{2}~-~\pi \right)~=~\left(\frac{\pi^2}{2}~-~\frac{2 \pi}{2} \right)}$

$\small{~=~\frac{\pi}{2}\left(\pi~-~2 \right)}$

Solved Example 23.159
Evaluate $\small{\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x~+~\cos x}{9~+~16 \sin(2x)} \right]dx}}$
Solution:
1. Put $\small{u = \sin x ~-~\cos x}$

• Then $\small{\frac{du}{dx}~=~\cos x ~+~\sin x}$

$\small{\Rightarrow dx(\cos x ~+~\sin x)~=~du}$

• Also, $\small{u^2~=~\sin^2 x~+~\cos^2 x~-~2 \sin x \cos x~=~1~-~2 \sin x \cos x}$

$\small{\Rightarrow u^2~=~1~-~\sin(2x)}$

$\small{\Rightarrow \sin(2x)~=~1~-~u^2}$

2. So we want:

$\small{F ~=~\int{\left[\frac{\sin x~+~\cos x}{9~+~16 \sin(2x)} \right]dx}~=~\int{\left[\frac{1}{9~+~16(1-u^2)} \right]du}}$

$\small{~=~\int{\left[\frac{1}{9~+~16-16u^2} \right]du}~=~\int{\left[\frac{1}{25-16u^2} \right]du}~=~\int{\left[\frac{1}{5^2-(4u)^2} \right]du}}$

3. This is a standard integral. We have:

$\small{\int{\left[\frac{1}{a^2~-~t^2} \right]dt}~=~\frac{1}{2a} \log \left| \frac{a+t}{a-t}  \right|~+~\rm{C}}$

In our present case, a = 5 and t = 4u

• So we get:

$\small{F~=~\int{\left[\frac{1}{5^2-(4u)^2} \right]du}~=~\frac{1}{2(5)(4)} \log \left| \frac{5+4u}{5-4u}  \right|~=~\frac{1}{40} \log \left| \frac{5+4u}{5-4u}  \right|}$

4. Substituting for u, we get:

$\small{F~=~\frac{1}{40} \log \left|\frac{5~+~4 \sin x~-~4 \cos x}{5~-~4 \sin x~+~4 \cos x}  \right|}$

5. Now we can find the definite integral:

$\small{I~=~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin x~-~4 \cos x}{5~-~4 \sin x~+~4 \cos x}  \right| \right]_0^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin \left(\frac{\pi}{4} \right)~-~4 \cos \left(\frac{\pi}{4} \right)}{5~-~4 \sin \left(\frac{\pi}{4} \right)~+~4 \cos \left(\frac{\pi}{4} \right)}  \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin \left(0 \right)~-~4 \cos \left(0 \right)}{5~-~4 \sin \left(0 \right)~+~4 \cos \left(0 \right)}  \right| \right]}$

$\small{~=~\left[\frac{1}{40} \log \left|\frac{5~+~0}{5~-~0}  \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{5~+~0~-~4}{5~-~0~+~4}  \right| \right]}$

$\small{~=~\left[\frac{1}{40} \log \left|1  \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{1}{9}  \right| \right]}$

$\small{~=~\left[\frac{1}{40} (0) \right]~-~\left[\frac{1}{40} \log \left|\frac{1}{9}  \right| \right]~=~~-~\left[\frac{1}{40} \log \left|\frac{1}{9}  \right| \right]}$

$\small{~=~\frac{1}{40} \log \left(9 \right)}$

Solved Example 23.160
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\sin(2x) \tan^{-1}(\sin x) \right]dx}}$
Solution:
1. Put $\small{u = \sin x}$

• Then $\small{\frac{du}{dx}~=~\cos x \Rightarrow dx~=~\frac{du}{\cos x}}$

• We wrote: $\small{u = \sin x}$

    ♦ When x approaches 0, u approaches 0
    ♦ When x approaches $\small{\frac{\pi}{2}}$, u approaches 1

2. We want:

$\small{I = \int_{0}^{\frac{\pi}{2}}{\left[\sin(2x) \tan^{-1}(\sin x) \right]dx}}$

$\small{= \int_{0}^{\frac{\pi}{2}}{\left[2 \sin(x)\cos(x) \tan^{-1}(\sin x) \right]dx}}$

$\small{= \int_{0}^{\frac{\pi}{2}}{\left[2 u\cos(x) \tan^{-1}(u) \right]\frac{du}{\cos x}}}$

$\small{= \int_{0}^{1}{\left[2 u \tan^{-1}(u) \right]du}= 2 \int_{0}^{1}{\left[u \tan^{-1}(u) \right]du}}$

3. We will apply the method of integration by parts

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(u)=\tan^{-1}u}$

   ♦ Let second function be: $\small{g(u)=u}$

(b) Finding A:

$\small{A~=~\int{\left[g(u) \right]dx}~=~\frac{u^2}{2}}$

(c) $\small{\big[f(u) \left(A \right) \big]~=~\big[ \tan^{-1}u\big] \,\big[\frac{u^2}{2} \big]~=~\big[\frac{u^2\,\tan^{-1}u}{2} \big]}$

• This is the first term.

(d) $\small{f'(u)~=~\frac{1}{1 + u^2}}$

(e) $\small{\int{\big[f'(u)\,\left(A \right)  \big]du}~=~\int{\big[\frac{1}{1 + u^2}\,\left(\frac{u^2}{2} \right)\big]du}}$

$\small{~=~\int{\big[\frac{u^2}{2(1 + u^2)}\big]du}~=~\frac{1}{2}\int{\big[\frac{u^2}{1 + u^2}\big]du}}$

$\small{~=~\frac{1}{2}\int{\big[\frac{u^2 + 1 - 1}{1 + u^2}\big]du}~=~\frac{1}{2}\int{\big[1~-~\frac{1}{1 + u^2}\big]du}}$

$\small{~=~\frac{1}{2}\int{\big[1\big]du}~-~\frac{1}{2}\int{\big[\frac{1}{1 + u^2}\big]du}}$

$\small{~=~\frac{u}{2}~-~\frac{\tan^{-1}u}{2}}$

• This is the second term.

(f) So we get:

$\small{\int_{0}^{\frac{1}{2}}{\left[u \tan^{-1}(u) \right]du}~=~\text{First term - Second term}}$

$\small{~=~\big[\frac{u^2\,\tan^{-1}u}{2}\big]~-~\big[\frac{u}{2}~-~\frac{\tan^{-1}u}{2} \big]}$

$\small{~=~\frac{u^2\,\tan^{-1}u}{2}~-~\frac{u}{2}~+~\frac{\tan^{-1}u}{2}}$

$\small{~=~\frac{(u^2 + 1)(\tan^{-1}u)~-~u}{2}}$

4. So from (2), we get:

$\small{I= 2 \int_{0}^{1}{\left[u \tan^{-1}(u) \right]du}~=~2 \big[\frac{(u^2 + 1)(\tan^{-1}u)~-~u}{2} \big]_0^1}$

$\small{~=~ \big[(u^2 + 1)(\tan^{-1}u)~-~u \big]_0^1}$

$\small{~=~ \big[(1^2 + 1)(\tan^{-1}1)~-~1 \big]~-~\big[(0 + 1)(\tan^{-1}0)~-~0 \big]}$

$\small{~=~ \big[(2)(\pi/4)~-~1 \big]~-~\big[(1)(0)~-~0 \big]}$

$\small{~=~ \frac{\pi}{2}~-~1}$

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The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

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In the next chapter, we will see applications of integrals.

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23.20 - Solved Examples on Standard Integrals

In the previous section, we saw the derivation of three standard integrals. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.63
Find $\small{\int{\left[\sqrt{4 - x^2} \right]dx}}$
Solution:
1. The given integral can be written as: $\small{\int{\left[\sqrt{2^2 - x^2} \right]dx}}$

2. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = 2

• So we get:

$\small{\int{\left[\sqrt{2^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{2^2 - x^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{x}{2}~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{4 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{4 - x^2} ~+~2 \sin^{-1}\frac{x}{2}~+~\rm{C}}$

Solved Example 23.64
Find $\small{\int{\left[\sqrt{1 - 4 x^2} \right]dx}}$
Solution:
1. The given integral can be written as:

$\small{\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}}$

2. Put u = 2x Then du/dx = 2 ⇒ du = 2 dx

• So the given integral becomes:

$\small{\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}~=~\int{\left[\frac{2 \sqrt{1^2 - (2x)^2}}{2} \right]dx}}$

$\small{~=~\int{\left[\frac{ \sqrt{1^2 - u^2}}{2} \right]du}~=~\frac{1}{2} \int{\left[\sqrt{1^2 - u^2} \right]dx}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = 1

• So we get:

$\small{\frac{1}{2}\int{\left[\sqrt{1^2 - u^2} \right]du}~=~~\frac{u}{4}\sqrt{1^2 - u^2} ~+~\frac{1^2}{4} \sin^{-1}\frac{u}{1}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\frac{1}{2}\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}~=~~\frac{2x}{4}\sqrt{1^2 - (2x)^2} ~+~\frac{1}{4} \sin^{-1}(2x)~+~\rm{C}}$

$\small{\Rightarrow \frac{1}{2}\int{\left[\sqrt{1 - 4x^2} \right]dx}~=~~\frac{x}{2}\sqrt{1 - 4x^2} ~+~\frac{1}{4} \sin^{-1}(2x)~+~\rm{C}}$

Solved Example 23.65
Find $\small{\int{\left[\sqrt{x^2 + 4x + 1} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4 - 3} \right]dx}~=~\int{\left[\sqrt{(x+2)^2 - (\sqrt{3})^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - (\sqrt{3})^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

Here x = u and a = √3

• So we get:

$\small{\int{\left[\sqrt{u^2 - (\sqrt{3})^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 - (\sqrt{3})^2} ~-~\frac{(\sqrt{3})^2}{2} \log \left|u+\sqrt{u^2 - (\sqrt{3})^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 - (\sqrt{3})^2} \right]dx}~=~\frac{x+2}{2}\sqrt{(x+2)^2 - (\sqrt{3})^2} ~-~\frac{(\sqrt{3})^2}{2} \log \left|(x+2)-\sqrt{(x+2)^2 + (\sqrt{3})^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 1} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 1} ~-~ \frac{3}{2} \log \left|x+2-\sqrt{x^2 + 4x + 1} \right|~+~\rm{C}}$

Solved Example 23.66
Find $\small{\int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4 - 9} \right]dx}~=~\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

Here x = u and a = 3

• So we get:

$\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 - 3^2} ~-~\frac{3^2}{2} \log \left|u+\sqrt{u^2 - 3^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{(x+2)^2 - 3^2} ~-~\frac{3^2}{2} \log \left|(x+2)-\sqrt{(x+2)^2 + 3^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{x^2 + 4x - 5} ~-~ \frac{9}{2} \log \left|x+2-\sqrt{x^2 + 4x - 5} \right|~+~\rm{C}}$

Solved Example 23.67
Find $\small{\int{\left[\sqrt{3 - 2x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{-(x^2 +2x- 3)} \right]dx}~=~\int{\left[\sqrt{-[x^2 +2x+1-4]}\right]dx}}$

$\small{~=~\int{\left[\sqrt{-\left[(x+1)^2-2^2\right]}\right]dx}~=~\int{\left[\sqrt{2^2 - (x+1)^2}\right]dx}}$

2. Put u = x+1 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{2^2 - u^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = 2 and x = u

• So we get:

$\small{\int{\left[\sqrt{2^2 - u^2} \right]dx}~=~~\frac{u}{2}\sqrt{2^2 - u^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{u}{2}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{2^2 - (x+1)^2} \right]dx}}$

$\small{~=~~\frac{x+1}{2}\sqrt{2^2 - (x+1)^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{x+1}{2}~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{3-2x-x^2} \right]dx}}$

$\small{~=~\frac{x+1}{2}\sqrt{3-2x-x^2}~+~2 \sin^{-1}\frac{x+1}{2}~+~\rm{C}}$

Solved Example 23.68
Find $\small{\int{\left[\sqrt{1 - 4x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{-(x^2 +4x- 1)} \right]dx}~=~\int{\left[\sqrt{-[x^2 +4x+4-5]}\right]dx}}$

$\small{~=~\int{\left[\sqrt{-\left[(x+2)^2-(\sqrt{5})^2\right]}\right]dx}~=~\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2}\right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = √5 and x = u

• So we get:

$\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]dx}~=~~\frac{u}{2}\sqrt{(\sqrt{5})^2 - u^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{u}{\sqrt{5}}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2} \right]dx}}$

$\small{~=~~\frac{x+2}{2}\sqrt{(\sqrt{5})^2 - (x+2)^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{1-4x-x^2} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{1-4x-x^2}~+~\frac{5}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$

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The link below gives a few more solved examples:

Exercise 23.7

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We have completed a discussion on the three standard integrals. In the next section, we will see definite integrals.

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23.19 - Derivation of Some Standard Integrals

In the previous section, we completed a discussion on the method of integration by parts. In this section, we will see three standard integrals.

I. The integral $\bf{\int{\left[\sqrt{x^2 - a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 - a^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 - a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2-a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2-a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2 + a^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2}{\sqrt{x^2-a^2}}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 - a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 - a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\sqrt{x^2-a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2-a^2} \big]dx}~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2-a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula IV that we saw in section 23.8. Thus we get:

$\small{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

II. The integral $\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 + a^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 + a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2+a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2+a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2 - a^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2}{\sqrt{x^2-a^2}}~-~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2}~-~\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 + a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 + a^2} ~-~\int{\big[\sqrt{x^2+a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2+a^2} \big]dx}~=~x \sqrt{x^2 + a^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2+a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula VI that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

III. The integral $\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{a^2 - x^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{a^2 - x^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-x}{\sqrt{a^2-x^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-x}{\sqrt{a^2-x^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2 - a^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2}{\sqrt{a^2-x^2}}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x\sqrt{a^2 - x^2} \big]~-~\bigg[\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} \bigg]}$

$\small{~=~x\sqrt{a^2 - x^2} ~-~\int{\big[\sqrt{a^2-x^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{a^2-x^2} \big]dx}~=~x\sqrt{a^2 - x^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{a^2-x^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula V that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

---

Now we will see some solved examples.

Solved example 23.61
Find $\small{\int{\left[\sqrt{x^2 +2x+ 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +2x+ 1+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 2^2} \right]dx}}$

2. Put u = x+1 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ 4} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = 2

• So we get:

$\small{\int{\left[\sqrt{u^2 + 2^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + 2^2} ~+~\frac{2^2}{2} \log \left|u+\sqrt{u^2 + 2^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+1)^2 + 2^2} \right]dx}~=~\frac{x+1}{2}\sqrt{(x+1)^2 + 2^2} ~+~\frac{2^2}{2} \log \left|(x+1)+\sqrt{(x+1)^2 + 2^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 2x + 5} \right]dx}~=~\frac{x+1}{2}\sqrt{x^2 + 2x + 5} ~+~2 \log \left|(x+1)+\sqrt{x^2 + 2x + 5} \right|~+~\rm{C}}$

Solved Example 23.62
Find $\small{\int{\left[\sqrt{x^2 +4x+ 6} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4+ 2} \right]dx}~=~\int{\left[\sqrt{(x+2)^2+ 2} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ (\sqrt{2})^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ (\sqrt{2})^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = √2

• So we get:

$\small{\int{\left[\sqrt{u^2 + (\sqrt{2})^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|u+\sqrt{u^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 + (\sqrt{2})^2} \right]dx}~=~\frac{x+2}{2}\sqrt{(x+1)^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|(x+2)+\sqrt{(x+2)^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 6} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 6} ~+~ \log \left|x+2+\sqrt{x^2 + 4x + 6} \right|~+~\rm{C}}$

---

We have seen the three standard integrals. In the next section, we will see a few more solved examples.

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23.10 - Solved Examples on Standard Integrals

In the previous two sections, we saw a total of ten basic integrals. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 23.12
Find the following integrals:
$\small{(i)~\int{\left[\frac{x+2}{2x^2\,+\,6x\,+\,5} \right]dx}~~~(ii)~\int{\left[\frac{x+3}{\sqrt{5\,-\,4x\,+\,x^2}} \right]dx}}$
Solution:
Part (i):
1. We have the following data:
a = 2, b = 6, c = 5, p = 1 and q = 2

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x + \frac{3}{2}}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,\frac{1}{4}}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,\frac{1}{4}}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,\frac{1}{2}}$

3. For this problem, we need to use formula IX:
$\small{\int{\left[\frac{px + q}{ax^2 + bx + c} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{ax^2 + bx + c} \right]dx}}$

4. So we need to calculate

$\small{\int{\left[\frac{\frac{1}{4}(4x + 6)\,+\,\frac{1}{2}}{2x^2 + 6x + 5} \right]dx}~=~\int{\left[\frac{\frac{1}{4}(4x + 6)}{2x^2 + 6x + 5} \right]dx}~+~\int{\left[\frac{\frac{1}{2}}{2x^2 + 6x + 5} \right]dx}}$

• The R.H.S has two terms

5. Calculation of first term in (4):

(i) Let $\small{t=2x^2 + 6x + 5}$

(ii) Then $\small{\frac{dt}{dx}=4x + 6~⇒~(4x+6)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{\frac{1}{4}}{t} \right]dt}\,=\,\frac{1}{4} \int{\left[\frac{1}{t} \right]dt}=\frac{1}{4} \log |t|=\frac{1}{4} \log \left|2x^2 + 6x + 5 \right |\,+\,\rm{C_1}}$

6. Calculation of second term in (4):

(i) Here we need to use formula VII:

$\small{\int{\left[\frac{dx}{ax^2 + bx + c} \right]~=~\frac{1}{a} \int{\left[\frac{du}{u^2~\pm~k^2}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(ii) So we want $\small{\left(\frac{1}{2} \right)\left(\frac{1}{2} \right)  \int{\left[\frac{du}{\left( x + \frac{3}{2}\right)^2~+~\left(\frac{1}{2} \right)^2}\right]}}$

• Here we need to use formula III.

(iii) Let $\small{t = x+\frac{3}{2}}$

(iv) Then $\small{\frac{dt}{dx}=1~⇒~dx\,=\,dt }$

$\small{\left(\frac{1}{2} \right)\left(\frac{1}{2} \right)  \int{\left[\frac{dt}{t^2~+~\left(\frac{1}{2} \right)^2}\right]}}~=~\left(\frac{1}{2} \right)\left(\frac{1}{2} \right) \left(\frac{1}{1/2} \right) \tan^{-1} \frac{t}{1/2}~+~\rm{C_2}$

$\small{~=~ \frac{1}{2} \tan^{-1} 2t\,+\,\rm{C_2}~=~\frac{1}{2} \tan^{-1} 2\left(x+\frac{3}{2} \right)\,+\ \rm{C_2}~=~\frac{1}{2} \tan^{-1} (2x+3)+ \rm{C_2}}$

7. Now we can write the R.H.S of (4):

$\small{\frac{1}{4} \log \left|2x^2 + 6x + 5 \right |\,+\,\rm{C_1}~+~\frac{1}{2} \tan^{-1} (2x+3)+ \rm{C_2}}$

$\small{~=~\frac{1}{4} \log \left|2x^2 + 6x + 5 \right |\,+\,\frac{1}{2} \tan^{-1} (2x+3)+ \rm{C}}$

Part (ii): $\small{\int{\left[\frac{x+3}{\sqrt{5\,-\,4x\,+\,x^2}} \right]dx}}$

1. We have the following data:
a = -1, b = -4, c = 5, p = 1 and q = 3

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x +2}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,-9}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,\frac{-1}{2}}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,1}$

3. For this problem, we need to use formula X:
$\small{\int{\left[\frac{px + q}{\sqrt{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{\sqrt{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

$\small{\int{\left[\frac{\frac{-1}{2}(2x \,-\, 4)\,+1}{\sqrt{-x^2 - 4x + 5}} \right]dx}~=~\int{\left[\frac{\frac{-1}{2}(2x \,-\, 4)}{\sqrt{-x^2 - 4x + 5}} \right]dx}~+~\int{\left[\frac{1}{\sqrt{-x^2 - 4x + 5}} \right]dx}}$

• The R.H.S of (4) has two terms.

5. Calculation of first term in the R.H.S of (4):

(i) Let $\small{t=-x^2 - 4x + 5}$

(ii) Then $\small{\frac{dt}{dx}=-2x - 4~⇒~(-2x-4)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{\frac{-1}{2}}{\sqrt{t}} \right]dt}\,=\,\frac{-1}{2} \int{\left[\frac{1}{\sqrt{t}} \right]dt}=\frac{-1}{2} \frac{t^{1/2}}{1/2}\,+\,\rm{C_1}}$

$\small{~=~(-1)\,t^{1/2}\,+\,\rm{C_1}~=~(-1)\,\sqrt{-x^2\,-\,4x\,+\,5 }\,+\,\rm{C_1}}$

6. Calculation of second term in the R.H.S of (4):

(i) Here we need to use formula VIII:

$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(ii) So we want $\small{\left(\frac{1}{\sqrt{-1}} \right)  \int{\left[\frac{dx}{\sqrt{\left( x + 2\right)^2~-~\left(3 \right)^2}}\right]}}$

$\small{~=~ \int{\left[\frac{dx}{\sqrt{(-1)\left[\left( x + 2\right)^2~-~\left(3 \right)^2 \right]}}\right]}~=~ \int{\left[\frac{dx}{\sqrt{3^2 ~-~\left( x + 2\right)^2 }}\right]}}$

• Here we need to use formula V.

(iii) Let $\small{t = x+2}$

(iv) Then $\small{\frac{dt}{dx}=1~⇒~dx\,=\,dt }$

$\small{\int{\left[\frac{dt}{\sqrt{3^2 ~-~t^2 }}\right]}}~=~\sin^{-1}\frac{t}{3}+ \rm{C_2}$

$\small{~=~\sin^{-1}\frac{x+2}{3}+ \rm{C_2}}$

7. Now we can write the R.H.S of (4):

$\small{(-1)\,\sqrt{-x^2\,-\,4x\,+\,5 }\,+\,\rm{C_1}~+~\sin^{-1}\frac{x+2}{3}+ \rm{C_2}}$

$\small{~=~-\,\sqrt{-x^2\,-\,4x\,+\,5 }\,+\,\sin^{-1}\frac{x+2}{3}\,+\, \rm{C}}$

Solved example 23.13
Find the following integrals:
$\small{(i)~\int{\left[\frac{5x+3}{\sqrt{x^2\,+\,4x\,+\,10}} \right]dx}~~~(ii)~\int{\left[\frac{6x+7}{\sqrt{(x-5)(x-4)}} \right]dx}}$

$\small{(iii)~\int{\left[\frac{x+2}{\sqrt{4x\,-\,x^2}} \right]}~~~(iv)~\int{\left[\frac{dx}{\sqrt{9x\,-\,4x^2}} \right]}}$
Solution:
Part (i):
1. We have the following data:
a = 1, b = 4, c = 10, p = 5 and q = 3

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x +2}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,6}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,\frac{5}{2}}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,-7}$

3. For this problem, we need to use formula X:
$\small{\int{\left[\frac{px + q}{\sqrt{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{\sqrt{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

$\small{\int{\left[\frac{\frac{5}{2}(2x \,+\, 4)\,-7}{\sqrt{x^2 + 4x + 10}} \right]dx}~=~\int{\left[\frac{\frac{5}{2}(2x \,+\, 4)}{\sqrt{x^2 + 4x + 10}} \right]dx}~-~\int{\left[\frac{7}{\sqrt{x^2 + 4x + 10}} \right]dx}}$

• The R.H.S has two terms

5. Calculation of first term in (4):

(i) Let $\small{t=x^2 + 4x + 10}$

(ii) Then $\small{\frac{dt}{dx}=2x + 4~⇒~(2x+4)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{\frac{5}{2}}{\sqrt{t}} \right]dt}\,=\,\frac{5}{2} \int{\left[\frac{1}{\sqrt{t}} \right]dt}=\frac{5}{2} \frac{t^{1/2}}{1/2}\,+\,\rm{C_1}}$

$\small{~=~(5)\,t^{1/2}\,+\,\rm{C_1}~=~5\,\sqrt{x^2\,+\,4x\,+\,10 }\,+\,\rm{C_1}}$

6. Calculation of second term in (4):

(i) Here we need to use formula VIII:

$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(ii) So we want $\small{\left(\frac{7}{\sqrt{1}} \right)  \int{\left[\frac{dx}{\sqrt{\left( x + 2\right)^2~+~\left(\sqrt 6 \right)^2}}\right]}}$

$\small{~=~ 7\int{\left[\frac{dx}{\sqrt{\left[\left( x + 2\right)^2~+~\left(\sqrt 6 \right)^2 \right]}}\right]}}$ 

• Here we need to use formula VI.

(iii) Let $\small{t = x+2}$

(iv) Then $\small{\frac{dt}{dx}=1~⇒~dx\,=\,dt }$

$\small{7\int{\left[\frac{dt}{\sqrt{t^2 ~+~(\sqrt{6})^2 }}\right]}}~=~7\log \left|t + \sqrt{t^2 + (\sqrt{6})^2} \right| + \rm{C_2}$

$\small{~=~7\log \left|x+2 + \sqrt{(x+2)^2 + (\sqrt{6})^2} \right| + \rm{C_2}}$

$\small{~=~7\log \left|x+2 + \sqrt{x^2 + 4x + 4 + 6} \right| + \rm{C_2}}$

$\small{~=~7\log \left|x+2 + \sqrt{x^2 + 4x +10} \right| + \rm{C_2}}$

7. Now we can write the R.H.S of (4):

$\small{5\,\sqrt{x^2\,+\,4x\,+\,10 }\,-\,\rm{C_1}~+~7\log \left|x+2 + \sqrt{x^2 + 4x +10} \right|+ \rm{C_2}}$

$\small{~=~5\,\sqrt{x^2\,+\,4x\,+\,10 }\,-\,7\log \left|x+2 + \sqrt{x^2 + 4x +10} \right|\,+\, \rm{C}}$

Part (ii): $\small{\int{\left[\frac{6x+7}{\sqrt{(x-5)(x-4)}} \right]dx}}$

1. The given expression can be rearranged as:

$\small{\int{\left[\frac{6x+7}{x^2\,-\,9x\,+\,20} \right]dx}}$

• So we have the following data:
a = 1, b = -9, c = 20, p = 6 and q = 7

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x \,-\,\frac{9}{2}}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,\frac{-1}{4}}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,3}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,34}$

3. For this problem, we need to use formula X:
$\small{\int{\left[\frac{px + q}{\sqrt{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{\sqrt{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

$\small{\int{\left[\frac{3(2x \,-\, 9)\,+\,34}{\sqrt{x^2 - 9x + 20}} \right]dx}~=~\int{\left[\frac{3(2x \,-\, 9)}{\sqrt{x^2 - 9x + 20}} \right]dx}~+~\int{\left[\frac{34}{\sqrt{x^2 - 9x + 20}} \right]dx}}$

• The R.H.S has two terms

5. Calculation of first term in (4):

(i) Let $\small{t=x^2 - 9x + 20}$

(ii) Then $\small{\frac{dt}{dx}=2x - 9~⇒~(2x-9)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{3}{\sqrt{t}} \right]dt}\,=\,3 \int{\left[\frac{1}{\sqrt{t}} \right]dt}=3 \frac{t^{1/2}}{1/2}\,+\,\rm{C_1}}$

$\small{~=~(6)\,t^{1/2}\,+\,\rm{C_1}~=~6\,\sqrt{x^2\,-\,9x\,+\,20 }\,+\,\rm{C_1}}$

6. Calculation of second term in (4):

(i) Here we need to use formula VIII:

$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(ii) So we want $\small{\left(\frac{34}{\sqrt{1}} \right)  \int{\left[\frac{dx}{\sqrt{\left( x + 2\right)^2~+~\left(\sqrt 6 \right)^2}}\right]}}$

$\small{~=~ 7\int{\left[\frac{dx}{\sqrt{\left[\left( x - \frac{9}{2} \right)^2~-~\left(\frac{1}{2} \right)^2 \right]}}\right]}}$

• Here we need to use formula IV.

(iii) Let $\small{t = x-\frac{9}{2}}$

(iv) Then $\small{\frac{dt}{dx}=1~⇒~dx\,=\,dt }$

$\small{34 \int{\left[\frac{dt}{\sqrt{t^2 ~-~\left(\frac{1}{2}\right)^2 }}\right]}}~=~34 \log \left|t + \sqrt{t^2 - \left(\frac{1}{2}\right)^2 } \right| + \rm{C_2}$

$\small{~=~34\log \left|x-\frac{9}{2} + \sqrt{\left(x-\frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \right| + \rm{C_2}}$

$\small{~=~34\log \left|x-\frac{9}{2} + \sqrt{x^2 - 9x + \frac{81}{4}~-~\frac{1}{4}} \right| + \rm{C_2}}$

$\small{~=~34\log \left|x-\frac{9}{2} + \sqrt{x^2 - 9x + \frac{80}{4}} \right| + \rm{C_2}}$

$\small{~=~34\log \left|x-\frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + \rm{C_2}}$

7. Now we can write the R.H.S of (4):

$\small{6\,\sqrt{x^2\,-\,9x\,+\,20 }\,+\,\rm{C_1}~+~34\log \left|x-\frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + \rm{C_2}}$

$\small{~=~6\,\sqrt{x^2\,-\,9x\,+\,20 }\,+\,34\log \left|x-\frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| \,+\, \rm{C}}$

Part (iii):
$\small{\int{\left[\frac{x+2}{\sqrt{4x\,-\,x^2}} \right]}}$

1. We have the following data:
a = -1, b = 4, c = 0, p = 1 and q = 2

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x -2}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,-4}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,\frac{-1}{2}}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,4}$

3. For this problem, we need to use formula X:
$\small{\int{\left[\frac{px + q}{\sqrt{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{\sqrt{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

$\small{\int{\left[\frac{\frac{-1}{2}(-2x \,+\, 4)\,+4}{\sqrt{-x^2 + 4x}} \right]dx}~=~\int{\left[\frac{\frac{-1}{2}(-2x \,+\, 4)}{\sqrt{-x^2 + 4x}} \right]dx}~+~\int{\left[\frac{4}{\sqrt{-x^2 + 4x}} \right]dx}}$

• The R.H.S of (4) has two terms.

5. Calculation of first term in the R.H.S of (4):

(i) Let $\small{t=-x^2 + 4x}$

(ii) Then $\small{\frac{dt}{dx}=-2x + 4~⇒~(-2x+4)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{\frac{-1}{2}}{\sqrt{t}} \right]dt}\,=\,\frac{-1}{2} \int{\left[\frac{1}{\sqrt{t}} \right]dt}=\frac{-1}{2} \frac{t^{1/2}}{1/2}\,+\,\rm{C_1}}$

$\small{~=~(-1)\,t^{1/2}\,+\,\rm{C_1}~=~(-1)\,\sqrt{-x^2\,+\,4x}\,+\,\rm{C_1}}$

6. Calculation of second term in the R.H.S of (4):

(i) Here we need to use formula VIII:

$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(ii) So we want $\small{\left(\frac{1}{\sqrt{-1}} \right)  \int{\left[\frac{dx}{\sqrt{\left( x - 2\right)^2~-~\left(2 \right)^2}}\right]}}$

$\small{~=~ \int{\left[\frac{dx}{\sqrt{(-1)\left[\left( x - 2\right)^2~-~\left(2 \right)^2 \right]}}\right]}~=~ \int{\left[\frac{dx}{\sqrt{2^2 ~-~\left(x - 2\right)^2 }}\right]}}$

• Here we need to use formula V.

(iii) Let $\small{t = x-2}$

(iv) Then $\small{\frac{dt}{dx}=1~⇒~dx\,=\,dt }$

$\small{\int{\left[\frac{dt}{\sqrt{2^2 ~-~t^2 }}\right]}}~=~\sin^{-1}\frac{t}{2}+ \rm{C_2}$

$\small{~=~\sin^{-1}\frac{x-2}{2}+ \rm{C_2}}$

7. Now we can write the R.H.S of (4):

$\small{(-1)\,\sqrt{-x^2\,+\,4x}\,+\,\rm{C_1}~+~\sin^{-1}\frac{x-2}{2}+ \rm{C_2}}$

$\small{~=~-\,\sqrt{-x^2\,+\,4x}\,+\,\sin^{-1}\frac{x-2}{2}\,+\, \rm{C}}$

Part (iv): $\small{\int{\left[\frac{dx}{\sqrt{9x\,-\,4x^2}} \right]}}$

1. We have the following data:
a = -4, b = 9 and c = 0

2. Based on this data, we can calculate two items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x -\frac{9}{8}}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,-\frac{81}{64}}$

3. For this problem, we need to use formula VIII:
$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(i) So we want

$\small{\left(\frac{1}{\sqrt{-4}} \right)  \int{\left[\frac{dx}{\sqrt{\left( x - \frac{9}{8} \right)^2~-~\left(\frac{9}{8} \right)^2}}\right]}}$

$\small{~=~ \int{\left[\frac{dx}{\sqrt{(-4)\left[\left( x - \frac{9}{8} \right)^2~-~\left(\frac{9}{8} \right)^2 \right]}}\right]}~=~ \int{\left[\frac{dx}{\sqrt{(4)\left[\left(\frac{9}{8} \right)^2~-~\left( x - \frac{9}{8} \right)^2 \right]}}\right]}}$

$\small{~=~ \frac{1}{2} \int{\left[\frac{dx}{\sqrt{\left(\frac{9}{8} \right)^2~-~\left( x - \frac{9}{8} \right)^2 }}\right]}}$

(ii) Here we need to use formula V.

(iii) Let $\small{t = x-\frac{9}{8}}$

(iv) Then $\small{\frac{dt}{dx}=1~⇒~dx\,=\,dt }$

$\small{\frac{1}{2} \int{\left[\frac{dt}{\sqrt{\left(\frac{9}{8} \right)^2 ~-~t^2 }}\right]}}~=~\frac{1}{2} \sin^{-1}\frac{t}{9/8}+ \rm{\frac{C_1}{2}}$

$\small{~=~\frac{1}{2} \sin^{-1}\frac{ x-\frac{9}{8}}{\frac{9}{8}}+ \rm{\frac{C_1}{2}}~=~\frac{1}{2} \sin^{-1}\frac{ x-\frac{9}{8}}{\frac{9}{8}}+ \rm{\frac{C_1}{2}}}$

$\small{~=~\frac{1}{2} \sin^{-1}\frac{ 8x-9}{9}+ \rm{C}}$

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In the next section, we will see a few more solved examples.

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23.9 - More Standard Integrals

In the previous section, we saw six basic integrals. For convenience, they are shown again below:
Formula I
$\small{\int{\left[\frac{dx}{x^2\,-\,m^2} \right]}\,=\,\frac{1}{2m}\log \left | \frac{x-m}{x+m} \right |\,+\,C}$

Formula II
$\small{\int{\left[\frac{dx}{m^2\,-\,x^2} \right]}\,=\,\frac{1}{2m}\log \left | \frac{m+x}{m-x} \right |\,+\,C}$

Formula III
$\small{\int{\left[\frac{dx}{x^2\,+\,m^2} \right]}\,=\,\frac{1}{m} \tan^{-1}\frac{x}{m}\,+\,\rm{C}}$

Formula IV
$\small{\int{\left[\frac{dx}{\sqrt{x^2\,-\,m^2}} \right]}\,=\, \log \left|x\,+\,\sqrt{x^2\,-\,m^2} \right| \,+\,\rm{C}}$

Formula V
$\small{\int{\left[\frac{dx}{\sqrt{m^2\,-\,x^2}} \right]}\,=\, \sin^{-1} \frac{x}{m} \,+\,\rm{C}}$

Formula VI
$\small{\int{\left[\frac{dx}{\sqrt{x^2\,+\,m^2}} \right]}\,=\, \log \left|x\,+\,\sqrt{x^2\,+\,m^2} \right| \,+\,\rm{C}}$

• Using the above, we can derive four more formulas. We will see it in this section.

Formula VII
To find $\small{\int{\left[\frac{dx}{a x^2\,+\,b x \,+\,c} \right]}}$

Calculations can be written in 4 steps:
1. First we rearrange the denominator:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{ax^2 \,+\,bx\,+\,c}    & {~=~}    &{a\left[x^2\,+\,\frac{b}{a} x\,+\,\frac{c}{a} \right]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{a\left[\left(x\,+\,\frac{b}{2a} \right)^2 ~+~\left(\frac{c}{a}\,-\,\frac{b^2}{4a^2} \right) \right]}    \\
\end{array}}$                           

2. Analyzing the terms of the R.H.S:
Inside the square brackets of the R.H.S, we have two terms.
• The second term is a constant. It sign depends on the values and signs of the constants a, b and c.
   ♦ So we will write this term as a new constant $\small{\pm k^2}$.
   ♦ That is., $\small{\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\pm k^2}$
• The first term is a variable.
   ♦ So we will write this as a new variable $\small{u}$.
   ♦ That is., $\small{x\,+\,\frac{b}{2a}~=~u}$
   ♦ Then we get: $\small{\frac{du}{dx}\,=\,1}$
   ♦ Which gives $\small{du\,=\,dx}$

3. So we want:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{dx}{a x^2\,+\,b x \,+\,c} \right]}}    & {~=~}    &{\int{\left[\frac{dx}{a\left[\left(x\,+\,\frac{b}{2a} \right)^2 ~+~\left(\frac{c}{a}\,-\,\frac{b^2}{4a^2} \right) \right]} \right]}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{a} \int{\left[\frac{du}{u^2 ~\pm~ k^2 } \right]}}    \\
\end{array}}$

4. For this integration, depending upon the sign of $\small{k^2}$, we can use either formula I or formula III.

Formula VIII
To find $\small{\int{\left[\frac{dx}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]}}$

Calculations can be written in 4 steps:
(Note that, the only difference from formula VII is that, the denominator is inside the square root)
1. Just as we did for formula VII, ax² + bx + c can be written as:
$\small{a\left[\left(x\,+\,\frac{b}{2a} \right)^2 ~+~\left(\frac{c}{a}\,-\,\frac{b^2}{4a^2} \right) \right]}$

2. Second step is also the same as for formula VII.
◼ We put:
• $\small{\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\pm k^2}$
• $\small{x\,+\,\frac{b}{2a}~=~u}$
  ⇒ $\small{\frac{du}{dx}\,=\,1}$
  ⇒ $\small{du\,=\,dx}$

3. So we want:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{dx}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]}}    & {~=~}    &{\int{\left[\frac{dx}{\sqrt{a\left[\left(x\,+\,\frac{b}{2a} \right)^2 ~+~\left(\frac{c}{a}\,-\,\frac{b^2}{4a^2} \right) \right]}} \right]}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{a} \int{\left[\frac{du}{\sqrt{u^2 ~\pm~ k^2 }} \right]}}    \\
\end{array}}$                           

4. For this integration, depending upon the sign of $\small{k^2}$, we can use either formula IV or formula VI.

Formula IX
To find $\small{\int{\left[\frac{px\,+\,q}{a x^2\,+\,b x \,+\,c} \right]dx}}$

Calculations can be written in 6 steps:
1. Consider the derivative of ax² + bx + c. It is 2ax + b
• So the derivative has an x-term and a constant term.
2. In the given problem, the numerator (px + q) also has an x-term and a constant.
• But (px + q) need not be same as (2ax + b). So we cannot make a direct substitution.
3. However, we can make (px + q) equivalent to (2ax + b), using a simple trick.
• We write: px + q = A(2ax + b) + B
• Then by comparing the terms, we can write two equations:
(i) px = 2Aax   ⇒  p = 2Aa    ⇒  A = p/2a
(ii) q = Ab + B   ⇒ B = q − Ab
• Thus from the two equations, we get A and B
4. So we want:
$\small{\int{\left[\frac{px\,+\,q}{a x^2\,+\,b x \,+\,c} \right]dx}~=~\int{\left[\frac{A(2ax \,+\, b) + B}{a x^2\,+\,b x \,+\,c} \right]dx}}$        
5.This integration can be done as follows:
• Put u = ax² + bx + c
• Then du/dx = 2ax + b
⇒ du = (2ax + b) dx
• So the problem becomes:
$\small{\int{\left[\frac{A(2ax \,+\, b) + B}{a x^2\,+\,b x \,+\,c} \right]dx}~=~\int{\left[\frac{A(2ax \,+\, b)}{a x^2\,+\,b x \,+\,c} \right]dx}~+~\int{\left[\frac{B}{a x^2\,+\,b x \,+\,c} \right]dx}}$

= $\small{\int{\left[\frac{A}{u} \right]du}~+~\int{\left[\frac{B}{a x^2\,+\,b x \,+\,c} \right]dx}}$

6. In the above result,
• The first integral can be calculated using the technique that we learned earlier. It will give $\small{\log \left |u \right|\,+\,\rm{C_1}}$
• The second integral can be calculated using formula VII that we saw at the beginning of this section.

Formula X
To find $\small{\int{\left[\frac{px\,+\,q}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]dx}}$

Calculations can be written in 4 steps:
(Note that, the only difference from formula VII is that, the denominator is inside the square root)
1. Just as we did for formula IX, we can find A and B, and then rewrite the numerator.

2. So we want:
$\small{\int{\left[\frac{px\,+\,q}{a x^2\,+\,b x \,+\,c} \right]dx}~=~\int{\left[\frac{A(2ax \,+\, b) + B}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]dx}}$        
3.This integration can be done as follows:
• Put u = ax² + bx + c
• Then du/dx = 2ax + b
⇒ du = (2ax + b) dx
• So the problem becomes:
$\small{\int{\left[\frac{A(2ax \,+\, b) + B}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]dx}~=~\int{\left[\frac{A(2ax \,+\, b)}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]dx}~+~\int{\left[\frac{B}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]dx}}$

= $\small{\int{\left[\frac{A}{\sqrt{u}} \right]du}~+~\int{\left[\frac{B}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]dx}}$

4. In the above result,
• The first integral can be calculated using the technique that we learned earlier. It will give $\small{2 \sqrt{u}\,+\,\rm{C_1}}$
• The second integral can be calculated using formula VIII that we saw at the beginning of this section.

---

Now we will see some solved examples.

Solved example 23.10
Find the following integrals:
$\small{(i)~\int{\left[\frac{dx}{x^2\,-\,16} \right]}~~~(ii)~\int{\left[\frac{dx}{\sqrt{2x\,-\,x^2}} \right]}}$
Solution:
Part (i):
1. The given function can be rearranged as:
$\small{\frac{1}{x^2\,-\,16}~=~\frac{1}{x^2\,-\,4^2}}$
2. We have formula I: $\small{\int{\left[\frac{dx}{x^2\,-\,m^2} \right]}\,=\,\frac{1}{2m}\log \left | \frac{x-m}{x+m} \right |\,+\,C}$
• In our present case, m = 4
3. So we get:
$\small{\int{\left[\frac{dx}{x^2\,-\,4^2} \right]}\,=\,\frac{1}{2(4)}\log \left | \frac{x-4}{x+4} \right |\,+\,C\,=\,\frac{1}{8}\log \left | \frac{x-4}{x+4} \right |\,+\,C}$

Part (ii): $\small{\int{\left[\frac{dx}{\sqrt{2x\,-\,x^2}} \right]}}$
1. Recall how we analyzed formula VIII $\small{\int{\left[\frac{dx}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]}~=~\frac{1}{\sqrt a}\int{\left[\frac{dx}{\sqrt{u^2~\pm~ k^2}} \right]}}$

• The given function can be written as:$\frac{1}{\sqrt{- x^2\,+\,2 x \,+\,0}}$

• So in our present case, a = -1, b = 2 and c = 0

2. So we can calculate u and k²:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,\frac{2}{2(-1)}~=~(x-1)}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{0}{(-1)}\,-\,\frac{(2)^2}{4(-1)^2}~=~0\,-\,\frac{4}{4}~=~-1}$

3. So we want:

$\small{\int{\left[\frac{dx}{\sqrt{- x^2\,+\,2 x}} \right]}~=~\frac{1}{\sqrt{(-1)}}\int{\left[\frac{du}{\sqrt{u^2~\pm~ k^2}} \right]}}$

$\small{~=~\frac{1}{\sqrt{(-1)}}\int{\left[\frac{dx}{\sqrt{(x-1)^2~-~1}} \right]}~=~\int{\left[\frac{dx}{\sqrt{(-1)\left[(x-1)^2~-~1 \right]}} \right]}}$

$\small{~=~\int{\left[\frac{dx}{\sqrt{1\,-\,(x-1)^2}} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. This integration can be done as shown below:

(i) Put t = (x−1). Then dt/dx = 1, which gives dt = dx

• So we want:

$\small{\int{\left[\frac{dx}{\sqrt{1\,-\,(x-1)^2}} \right]}~=~\int{\left[\frac{dt}{\sqrt{1^2\,-\,t^2}} \right]}}$

(ii) We have formula V: $\small{\int{\left[\frac{dt}{\sqrt{m^2\,-\,t^2}} \right]}\,=\, \sin^{-1}\frac{t}{m}\,+\,\rm{C}}$

• In our present case, m = 1

5. So we get:
$\small{\int{\left[\frac{dt}{\sqrt{m^2\,-\,t^2}} \right]}\,=\, \sin^{-1}\frac{t}{1}\,+\,C\,=\, \sin^{-1}(x-1)\,+\,C}$

Solved example 23.11
Find the following integrals:
$\small{(i)~\int{\left[\frac{dx}{x^2\,-\,6x\,+\,13} \right]}~~~(ii)~\int{\left[\frac{dx}{3x^2\,+\,13x\,-\,10} \right]}~~~(iii)~\int{\left[\frac{dx}{\sqrt{5x^2\,-\,2x}} \right]}}$
Solution:
Part (i):
1. Recall how we analyzed formula VII: $\small{\int{\left[\frac{dx}{a x^2\,+\,b x \,+\,c} \right]}~=~\frac{1}{a}\int{\left[\frac{dx}{u^2~\pm~ k^2} \right]}}$

• In our present case, a = 1, b = −6 and c = 13

2. So we can calculate u and k²:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,\frac{-6}{2(1)}~=~(x-3)}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{13}{1}\,-\,\frac{(-6)^2}{4(1)^2}~=~13\,-\,\frac{36}{4}~=~4}$

3. So we want:

$\small{\int{\left[\frac{dx}{x^2\,-\,6x\,+\,13} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{1}\int{\left[\frac{dx}{(x-3)^2~+~4} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. This integration can be done as shown below:

(i) Put t = (x−3). Then dt/dx = 1, which gives dt = dx

• So we want:

$\small{\int{\left[\frac{dx}{(x-3)^2~+~4} \right]}~=~\int{\left[\frac{dt}{t^2\,+\,2^2} \right]}}$

(ii) We have formula III: $\small{\int{\left[\frac{dt}{t^2\,+\,m^2} \right]}\,=\, \frac{1}{m} \tan^{-1}\frac{t}{m}\,+\,\rm{C}}$

• In our present case, m = 2

5. So we get:
$\small{\int{\left[\frac{dt}{t^2\,+\,2^2} \right]}\,=\, \frac{1}{2} \tan^{-1}\frac{t}{2}\,+\,C\,=\,\frac{1}{2} \tan^{-1}\frac{x-3}{2}\,+\,C}$

Part (ii): $\small{\int{\left[\frac{dx}{3x^2\,+\,13x\,-\,10} \right]}}$

1. Recall how we analyzed formula VII: $\small{\int{\left[\frac{dx}{a x^2\,+\,b x \,+\,c} \right]}~=~\frac{1}{a}\int{\left[\frac{dx}{u^2~\pm~ k^2} \right]}}$

• In our present case, a = 3, b = 13 and c = −10

2. So we can calculate u and k²:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,\frac{13}{2(3)}~=~(x+\frac{13}{6})}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{-10}{3}\,-\,\frac{13^2}{4(3)^2}~=~\frac{-10}{3}\,-\,\frac{169}{36}~=~-\frac{289}{36}}$

3. So we want:

$\small{\int{\left[\frac{dx}{3x^2\,+\,13x\,-\,10} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{3}\int{\left[\frac{dx}{(x+\frac{13}{6})^2~-~\frac{289}{36}} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. This integration can be done as shown below:

(i) Put t = $x\,+\,\frac{13}{6}$. Then dt/dx = 1, which gives dt = dx

• So we want:

$\small{\frac{1}{3}\int{\left[\frac{dx}{(x+\frac{13}{6})^2~-~\frac{289}{36}} \right]}~=~\frac{1}{3}\int{\left[\frac{dt}{t^2\,-\,\left(\frac{17}{6} \right)^2} \right]}}$

(ii) We have formula I: $\small{\int{\left[\frac{dt}{t^2\,-\,m^2} \right]}\,=\, \frac{1}{2m} \log \left| \frac{x-m}{x+m} \right|\,+\,\rm{C}}$

• In our present case, m = $\frac{17}{6}$

5. So we get:
$\small{\frac{1}{3}\int{\left[\frac{dt}{t^2\,-\,\left(\frac{17}{6} \right)^2} \right]}\,=\, \frac{1}{3} \left[\frac{1}{2(17/6)} \log \left| \frac{t-\frac{17}{6}}{t+\frac{17}{6}} \right| \right]\,+\,\rm{C_1}}$

$\small{\,=\, \frac{1}{3} \left[\frac{3}{17} \log \left| \frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}} \right| \right]\,+\,\rm{C_1}\,=\, \frac{1}{17} \log \left| \frac{6x-4}{6x+30} \right| \,+\,\rm{C_1}}$

$\small{\,=\,  \frac{1}{17} \log \left| \left(\frac{3x-2}{x+5} \right)\left(\frac{2}{6} \right) \right| \,+\,\rm{C_1}\,=\,\frac{1}{17} \log \left| \frac{3x-2}{x+5}  \right|\,+\,\frac{1}{17} \log \left|\frac{2}{6}  \right| \,+\,\rm{C_1}}$

$\small{\,=\,\frac{1}{17} \log \left| \frac{3x-2}{x+5}  \right| \,+\,\rm{C}}$

Part (iii): $\small{\int{\left[\frac{dx}{\sqrt{5x^2\,-\,2x}} \right]}}$

1. Recall how we analyzed formula VIII: $\small{\int{\left[\frac{dx}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]}~=~\frac{1}{\sqrt a}\int{\left[\frac{dx}{\sqrt{u^2~\pm~ k^2}} \right]}}$

• In our present case, a = 5, b = −2 and c = 0

2. So we can calculate u and k²:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,\frac{-2}{2(5)}~=~(x-\frac{1}{5})}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{0}{5}\,-\,\frac{(-2)^2}{4(5)^2}~=~0\,-\,\frac{1}{25}~=~-\frac{1}{25}}$

3. So we want:

$\small{\int{\left[\frac{dx}{\sqrt{5x^2\,-\,2x}} \right]}~=~\frac{1}{\sqrt{a}}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{\sqrt{5}}\int{\left[\frac{dx}{(x-\frac{1}{5})^2~-~\frac{1}{25}} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. This integration can be done as shown below:

(i) Put t = $x\,-\,\frac{1}{5}$. Then dt/dx = 1, which gives dt = dx

• So we want:

$\small{\frac{1}{\sqrt{5}}\int{\left[\frac{dx}{\sqrt{(x-\frac{1}{5})^2~-~\frac{1}{25}}} \right]}~=~\frac{1}{\sqrt 5}\int{\left[\frac{dt}{\sqrt{t^2\,-\,\left(\frac{1}{5} \right)^2}} \right]}}$

(ii) We have formula IV: $\small{\int{\left[\frac{dt}{\sqrt{t^2\,-\,m^2}} \right]}\,=\, \log \left|t\,+\,\sqrt{t^2\,-\,m^2} \right|\,+\,\rm{C}}$

• In our present case, m = $\frac{1}{5}$

5. So we get:

$\small{\frac{1}{\sqrt 5}\int{\left[\frac{dt}{t^2\,-\,\left(\frac{1}{5} \right)^2} \right]}\,=\, \frac{1}{\sqrt 5} \left[\log \left| t\,+\,\sqrt{t^2\,-\,\frac{1}{25}} \right| \right]\,+\,\rm{C}}$

$\small{\,=\, \frac{1}{\sqrt 5} \left[\log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{\left(x - \frac{1}{5} \right)^2\,-\,\frac{1}{25}} \right| \right]\,+\,\rm{C}}$

$\small{\,=\, \frac{1}{\sqrt 5} \left[\log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{x^2 - \frac{2x}{5} + \frac{1}{25} - \frac{1}{25}} \right| \right]\,+\,\rm{C}}$

$\small{\,=\, \frac{1}{\sqrt 5} \left[\log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{x^2 - \frac{2x}{5} + } \right| \right]\,+\,\rm{C}}$

$\small{\,=\, \frac{1}{\sqrt 5} \log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{x^2 - \frac{2x}{5} + } \right| \,+\,\rm{C}}$

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In the next section, we will see a few more solved examples.

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