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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{dx}{\sqrt{a x^2\,+\,b x \,+\,c}} \right]}} & {~=~} &{\int{\left[\frac{dx}{\sqrt{a\left[\left(x\,+\,\frac{b}{2a} \right)^2 ~+~\left(\frac{c}{a}\,-\,\frac{b^2}{4a^2} \right) \right]}} \right]}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{1}{a} \int{\left[\frac{du}{\sqrt{u^2 ~\pm~ k^2 }} \right]}} \\
\end{array}}$
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:
$\small{\frac{1}{\sqrt 5}\int{\left[\frac{dt}{t^2\,-\,\left(\frac{1}{5} \right)^2} \right]}\,=\, \frac{1}{\sqrt 5} \left[\log \left| t\,+\,\sqrt{t^2\,-\,\frac{1}{25}} \right| \right]\,+\,\rm{C}}$
$\small{\,=\, \frac{1}{\sqrt 5} \left[\log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{\left(x - \frac{1}{5} \right)^2\,-\,\frac{1}{25}} \right| \right]\,+\,\rm{C}}$
$\small{\,=\, \frac{1}{\sqrt 5} \left[\log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{x^2 - \frac{2x}{5} + \frac{1}{25} - \frac{1}{25}} \right| \right]\,+\,\rm{C}}$
$\small{\,=\, \frac{1}{\sqrt 5} \left[\log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{x^2 - \frac{2x}{5} + } \right| \right]\,+\,\rm{C}}$
$\small{\,=\, \frac{1}{\sqrt 5} \log \left| x\,-\,\frac{1}{5}\,+\,\sqrt{x^2 - \frac{2x}{5} + } \right| \,+\,\rm{C}}$
In the next section, we will see a few more solved examples.
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