23.11 - More Solved Examples on Standard Integrals

In the previous section, we saw some solved examples related to the standard integrals. In this section, we will see a few more solved examples.

Solved example 23.14
Find the following integrals:
$\small{(i)~\int{\left[\frac{dx}{\sqrt{x^2\,+\,2x\,+\,2}} \right]}~~~(ii)~\int{\left[\frac{dx}{\sqrt{(x-m)(x-n)}} \right]}}$

$\small{(iii)~\int{\left[\frac{dx}{\sqrt{(x-1)(x-2)}} \right]}~~~(iv)~\int{\left[\frac{dx}{\sqrt{8 + 3x - x^2}} \right]}}$
Solution:
Part (i):
1. We have the following data:
a = 1, b = 2 and c = 2

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

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

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

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

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

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

(ii) Here we need to use formula VI.

(iii) Let $\small{t = x+1}$

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

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

$\small{~=~\log \left|x+1\,+\,\sqrt{(x+1)^2\,+\,1} \right|+ \rm{C}}$

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

Part (ii): $\small{\int{\left[\frac{1}{\sqrt{(x-m)(x-n)}} \right]dx}}$

1. Let us rearrange the given expression:

$\small{\frac{1}{\sqrt{(x-m)(x-n)}}~=~\frac{1}{\sqrt{x^2 -mx -nx + mn}}~=~\frac{1}{\sqrt{x^2 -(m+n)x+mn}}}$

So we have the following data:
a = 1, b = -(m+n) and c = mn

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

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

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,mn~-~\left(\frac{m+n}{2} \right)^2}$

$\small{\,=\,mn~-~\left(\frac{m^2 + 2mn +n^2}{4} \right)\,=\frac{4mn -m^2 - 2mn -n^2}{4}}$

$\small{\,=\frac{2mn -m^2 -n^2}{4}\,=\frac{-(m-n)^2}{4}}$

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

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

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

(ii) Here we need to use formula IV.

(iii) Let $\small{t = x - \frac{m+n}{2} }$

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

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

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

$\small{~=~\log \left|x - \frac{m+n}{2}\,+\,\sqrt{\frac{[2x -(m+n)]^2}{4}\,-\,\frac{[m-n]^2}{4}} \right|+ \rm{C}}$

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

$\small{~=~\log \left|x - \frac{m+n}{2}\,+\,\frac{1}{2} \sqrt{4x^2 - 4x(m+n) + (m+n)^2 -(m-n)^2} \right|+ \rm{C}}$

$\small{~=~\log \left|x - \frac{m+n}{2}\,+\,\frac{1}{2} \sqrt{4x^2 - 4xm -4xn +m^2 + 2mn + n^2 -[m^2 -2mn + n^2]} \right|+ \rm{C}}$

$\small{~=~\log \left|x - \frac{m+n}{2}\,+\,\frac{1}{2} \sqrt{4x^2 - 4xm -4xn +m^2 + 2mn + n^2 -m^2 +2mn - n^2]} \right|+ \rm{C}}$

$\small{~=~\log \left|x - \frac{m+n}{2}\,+\,\frac{1}{2} \sqrt{4x^2 - 4xm -4xn  + 4mn ]} \right|+ \rm{C}}$

$\small{~=~\log \left|x - \frac{m+n}{2}\,+\, \sqrt{x^2 - xm -xn  + mn ]} \right|+ \rm{C}}$

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

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

Part (iii): $\small{\int{\left[\frac{dx}{\sqrt{(x-1)(x-2)}} \right]}}$

Easy method:
• From the previous solved example, we have the formula:
$\small{\int{\left[\frac{dx}{\sqrt{(x-m)(x-n)}} \right]}~=~\log \left|x - \frac{m+n}{2}\,+\, \sqrt{(x-m)(x-n) } \right|+ \rm{C}}$

• So we get:
$\small{\int{\left[\frac{dx}{\sqrt{(x-1)(x-2)}} \right]}~=~\log \left|x - \frac{1+2}{2}\,+\, \sqrt{(x-1)(x-2) } \right|+ \rm{C}}$ 

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

• The reader may try the usual method also.

Part (iv): $\small{\int{\left[\frac{dx}{\sqrt{8 + 3x - x^2}} \right]}}$

1. We have the following data:
a = -1, b = 3 and c = 8

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

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

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,-\frac{41}{4}}$

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

$\small{\left(\frac{1}{\sqrt{-1}} \right)  \int{\left[\frac{dx}{\sqrt{\left( x - \frac{3}{2} \right)^2~-~\left(\frac{41}{4} \right)}}\right]}}$

$\small{~=~ \int{\left[\frac{dx}{\sqrt{(-1)\left[\left( x - \frac{3}{2} \right)^2~-~\left(\frac{\sqrt{41}}{2} \right)^2 \right]}}\right]}~=~ \int{\left[\frac{dx}{\sqrt{\left[\left(\frac{\sqrt{41}}{2} \right)^2~-~\left( x - \frac{3}{2} \right)^2 \right]}}\right]}}$

(ii) Here we need to use formula V.

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

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

$\small{ \int{\left[\frac{dt}{\sqrt{\left(\frac{\sqrt{41}}{2} \right)^2 ~-~t^2 }}\right]}}~=~ \sin^{-1}\frac{t}{\frac{\sqrt{41}}{2}}+ \rm{C}$

$\small{~=~ \sin^{-1}\frac{ x - \frac{3}{2}}{\frac{\sqrt{41}}{2}}+ \rm{C}~=~\sin^{-1}\frac{ 2x-3}{\sqrt{41}}+ \rm{C}}$

Solved example 23.15
Find the following integrals:
$\small{(i)~\int{\left[\frac{x+3}{x^2 - 2x - 5} \right]dx}~~~(ii)~\int{\left[\frac{5x-2}{1+2x+3x^2} \right]dx}}$

$\small{(iii)~\int{\left[\frac{dx}{9x^2 + 6x + 5} \right]}~~~(iv)~\int{\left[\frac{dx}{x^2\,+\,2x\,+\,2} \right]}}$
Solution:
Part (i):
1. We have the following data:
a = 1, b = -2, c = -5, p = 1 and q = 3

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

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

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

(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:
$\small{\int{\left[\frac{px + q}{{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

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

• The R.H.S has two terms

5. Calculation of first term in (4):

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

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

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

$\small{~=~ \frac{1}{2} \log \left|{x^2 - 2x - 5} \right|\,+\,\rm{C_1}}$

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 $\small{\left(\frac{4}{{1}} \right)  \int{\left[\frac{dx}{{\left( x - 1\right)^2~-~\left(\sqrt 6 \right)^2}}\right]}}$

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

• Here we need to use formula I.

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

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

$\small{4\int{\left[\frac{dt}{{t^2 ~-~(\sqrt{6})^2 }}\right]}}~=~\frac{4}{2(\sqrt{6})} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + \rm{C_2}$

$\small{~=~\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + \rm{C_2}}$

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

$\small{\frac{1}{2} \log \left|{x^2 - 2x - 5} \right|\,+\,\rm{C_1}~+~\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right|+ \rm{C_2}}$

$\small{\frac{1}{2} \log \left|{x^2 - 2x - 5} \right|~+~\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right|+ \rm{C}}$

Part (ii): $\small{\int{\left[\frac{5x-2}{1+2x+3x^2} \right]dx}}$

1. We have the following data:
a = 3, b = 2, c = 1, p = 5 and q = -2

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

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x +\frac{1}{3}}$

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

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,\frac{5}{6}}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,\frac{-11}{3}}$

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

$\small{\int{\left[\frac{\frac{5}{6}(6x \,+\, 2)\,-\frac{11}{3}}{{3x^2 + 2x + 1}} \right]dx}~=~\int{\left[\frac{\frac{5}{6}(6x \,+\, 2)}{{3x^2 + 2x + 1}} \right]dx}~-~\int{\left[\frac{\frac{11}{3}}{{3x^2 + 2x + 1}} \right]dx}}$

• The R.H.S has two terms

5. Calculation of first term in (4):

(i) Let $\small{t=3x^2 + 2x + 1}$

(ii) Then $\small{\frac{dt}{dx}=6x +2~⇒~(6x+2)dx\,=\,dt }$

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

$\small{~=~ \frac{5}{6} \log \left|{3x^2 + 2x + 1} \right|\,+\,\rm{\frac{5 C_1}{6}}}$

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 $\small{\left(\frac{11}{{3}} \right)\left(\frac{1}{3} \right)  \int{\left[\frac{dx}{{\left( x +\frac{1}{3} \right)^2~+~\left(\frac{\sqrt 2}{3} \right)^2}}\right]}}$

• Here we need to use formula III.

(iii) Let $\small{t = x+\frac{1}{3}}$

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

$\small{\frac{11}{9} \int{\left[\frac{dt}{{t^2 ~+~(\frac{\sqrt 2}{3})^2 }}\right]}}~=~\frac{11}{9}   \left[\left( \frac{1}{\frac{\sqrt{2}}{3}} \right) \tan^{-1}\left(\frac{t}{\frac{\sqrt{2}}{3}} \right) + \rm{C_2} \right]$

$\small{~=~\frac{11}{9}   \left[\left( \frac{3}{\sqrt 2} \right) \tan^{-1}\left(\frac{3t}{\sqrt 2} \right) + \rm{C_2} \right]}$

$\small{~=~ \left( \frac{11}{3 \sqrt 2} \right) \tan^{-1}\left(\frac{3\left(x + \frac{1}{3} \right)}{\sqrt 2} \right) + \rm{\frac{11 C_2}{9}} }$

$\small{~=~ \left( \frac{11}{3 \sqrt 2} \right) \tan^{-1}\left(\frac{3x + 3 }{\sqrt 2} \right) + \rm{\frac{11 C_2}{9}} }$

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

$\small{\frac{5}{6} \log \left|{3x^2 + 2x + 1} \right|\,+\,\rm{\frac{5 C_1}{6}}~-~\left( \frac{11}{3 \sqrt 2} \right) \tan^{-1}\left(\frac{3x + 3 }{\sqrt 2} \right) + \rm{\frac{11 C_2}{9}}}$

$\small{\frac{5}{6} \log \left|{3x^2 + 2x + 1} \right|~-~\left( \frac{11}{3 \sqrt 2} \right) \tan^{-1}\left(\frac{3x + 3 }{\sqrt 2} \right) + \rm{C}}$

Part (iii):
$\small{\int{\left[\frac{dx}{9x^2 + 6x + 5} \right]}}$

1. We have the following data:
a = 9, b = 6 and c = 5

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

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x +\frac{1}{3}}$

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

3. For this problem, 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]

4. So we need to calculate

$\small{\frac{1}{9} \int{\left[\frac{du}{{\left(x + \frac{1}{3} \right)^2~+~\left(\frac{2}{3} \right)^2}} \right]}}$

• Here we need to use formula III.

(iii) Let $\small{t = x+\frac{1}{3}}$

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

$\small{\frac{1}{9} \int{\left[\frac{dt}{{t^2 ~+~\left(\frac{ 2}{3}\right)^2 }}\right]}}~=~\frac{1}{9}   \left[\left( \frac{1}{\frac{{2}}{3}} \right) \tan^{-1}\left(\frac{t}{\frac{{2}}{3}} \right) + \rm{C_1} \right]$

$\small{~=~\frac{1}{9}   \left[\left( \frac{3}{ 2} \right) \tan^{-1}\left(\frac{3t}{ 2} \right) + \rm{C_1} \right]}$

$\small{~=~ \left( \frac{1}{6} \right) \tan^{-1}\left(\frac{3\left(x + \frac{1}{3} \right)}{ 2} \right) + \rm{\frac{C_1}{9}} }$

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

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

1. We have the following data:
a = 1, b = 2 and c = 2

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

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

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

3. For this problem, 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]

4. So we need to calculate

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

• Here we need to use formula III.

(iii) Let $\small{t = x+1}$

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

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

$\small{~=~ \tan^{-1}\left(t\right) + \rm{C} }$

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

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The link below gives a few more solved examples:

Exercise 23.4

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

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