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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