12th NCERT INTEGRALS Exercise 7.5 Questions 23
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Integrate the rational functions in Exercise 1 to 21

Question (1)

$\frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}$

Solution

$I = \int {\frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}dx}$
As denominator has factors. We will use partial fraction method
$\text{let}\frac{A}{{x + 1}} + \frac{B}{{x + 2}} = \frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}$
∴ A(x+2) + B(x+1) = x
If x = -2 ⇒ -B = -2 ⇒ B = 2
if x=-1 ⇒ A = -1 $\therefore \frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{{ - 1}}{{x + 1}} + \frac{2}{{x + 2}}$
$\therefore I = \int {\frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}dx}$
$I = \int {\left( {\frac{{ - 1}}{{x + 1}} + \frac{2}{{x + 2}}} \right)} dx$
$I = - \int {\frac{1}{{x + 1}}} dx + 2\int {\frac{1}{{x + 2}}} dx$
$I = - \log \left| {x + 1} \right| + 2\log \left| {x + 2} \right| + c$
$I = - \log \left| {x + 1} \right| + \log {\left( {x + 2} \right)^2} + c$
$I = \log \left| {\frac{{{{\left( {x + 2} \right)}^2}}}{{x + 1}}} \right| + c$

Question (2)

$\frac{1}{{{x^2} - 9}}$

Solution

$I = \int {\frac{1}{{{x^2} - 9}}dx}$
$I = \int {\frac{1}{{{x^2} - {{\left( 3 \right)}^2}}}dx}$
Use formula $\int {\frac{1}{{{x^2} - {a^2}}}dx = \frac{1}{{2\left( a \right)}}} \log \left| {\frac{{x - a}}{{x + a}}} \right| + c,x > a$
$I = \frac{1}{{2\left( 3 \right)}}\log \left| {\frac{{x - 3}}{{x + 3}}} \right| + c$
$I = \frac{1}{6}\log \left| {\frac{{x - 3}}{{x + 3}}} \right| + c$

Question (3)

$\frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}$

Solution

$I = \int {\frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx}$
As denominator has factors. We will use partial fraction method
$\text{let}\frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{c}{{x - 3}} = \frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}$
LCM = (x-1)(x-2)(x-3)
A(x-2)(x-3) + B(x-1) (x-3) + c(x-1) (x-2) = 3x-1
if x= 1 ⇒ A(-1)(-2) = 3-1
∴ 2A=2 ⇒ A = 1
x=2 ⇒ B(1)(-1) = 6-1
or (-1)B = 5 ⇒ B = -5
If x = 3 ⇒ c(2)(1) = 9-1
⇒ 2C= 8 ⇒ C = 4 $\therefore \frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{1}{{x - 1}} + \frac{{ - 5}}{{x - 2}} + \frac{4}{{x - 3}}$ $I = \int {\frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx}$ $I = \int {\left( {\frac{1}{{x - 1}} + \frac{{ - 5}}{{x - 2}} + \frac{4}{{x - 3}}} \right)dx}$ $I = \int {\frac{1}{{x - 1}}dx} - 5\int {\frac{1}{{x - 2}}dx + 4\int {\frac{1}{{x - 3}}} } dx$ $I = \log \left| {x - 1} \right| - 5\log \left| {x - 2} \right| + 4\log \left| {x - 3} \right| + c$

Question (4)

$\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}$

Solution

$$I = \int {\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx}$$
$$\text{Let}\frac{A}{{\left( {x - 1} \right)}} + \frac{B}{{\left( {x - 2} \right)}} + \frac{C}{{\left( {x - 3} \right)}} = \frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}$$ ∴ A(x-2)(x-3) + B(x-1)(x-3) + C (x-1)(x-2) = x
If x = 1 ⇒ A(-1)(-2) = 1 ⇒ A= ½
If x = 2 ⇒ B(1)(-1)=2 ⇒ B = -2
If x = 3 ⇒ C(2)(1) = 3 ⇒ C = 3/2 $\therefore \frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{{\frac{1}{2}}}{{\left( {x - 1} \right)}} + \frac{{ - 2}}{{\left( {x - 2} \right)}} + \frac{{\frac{3}{2}}}{{\left( {x - 3} \right)}}$ $I = \int {\left( {\frac{{\frac{1}{2}}}{{\left( {x - 1} \right)}} + \frac{{ - 2}}{{\left( {x - 2} \right)}} + \frac{{\frac{3}{2}}}{{\left( {x - 3} \right)}}} \right)dx}$ $I = \frac{1}{2}\int {\frac{1}{{\left( {x - 1} \right)}}dx - 2\int {\frac{1}{{\left( {x - 2} \right)}}} } + \frac{3}{2}\int {\frac{1}{{\left( {x - 3} \right)}}}$
$I = \frac{1}{2}\log \left| {x - 1} \right| - 2\log \left| {x - 2} \right| + \frac{3}{2}\log \left| {x - 3} \right| + c$

Question (5)

$\frac{{2x}}{{{x^2} + 3x + 2}}$

Solution

$I = \int {\frac{{2x}}{{{x^2} + 3x + 2}}dx}$ $I = 2\int {\frac{x}{{\left( {x + 2} \right)\left( {x + 1} \right)}}dx}$ $\text{Let} \frac{A}{{x + 2}} + \frac{B}{{x + 1}} = \frac{x}{{\left( {x + 2} \right)\left( {x + 1} \right)}}$ A(x+1) + B(x+2) = x
If x = -2 ⇒ -A = -2 ⇒ A = +2
If x = -1 ⇒ B = -1
$\therefore \frac{{x}}{{\left( {x + 2} \right)\left( {x + 1} \right)}} = \frac{2}{{x + 2}} - \frac{1}{{x + 1}}$ $I = 2\int {\frac{x}{{\left( {x + 2} \right)\left( {x + 1} \right)}}dx}$ $I = 2\int {\left( {\frac{2}{{x + 2}} - \frac{1}{{x + 1}}} \right)dx}$ $I = 4\log \left| {x + 2} \right| - 2\log \left| {x + 1} \right| + c$

Question (6)

$\frac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}$

Solution

$I = \int {\frac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx}$ $I = \int {\frac{{\require{cancel} \cancel{-}+ \left( {{x^2} - 1} \right)}}{{\cancel{-}+ \left( {2{x^2} - x} \right)}}dx}$ $I = \int {\frac{{{x^2} - 1}}{{2{x^2} - x}}dx}$

As power of numerator and denominator is same, we will do long division to reduce the numerator

$$\require{enclose} \begin{array}{r1} \frac{1}{2} \quad\; \\ 2{x^2} - x \enclose{longdiv}{\; {x^2} -\;1}\ \\ \underline {\;-{x^2} \mp \ \frac{1}{2}x}\ \\ \frac{1}{2}x - 1\phantom{00} \\ \end{array}$$
$\therefore \frac{{{x^2} - 1}}{{2{x^2} - x}} = \frac{1}{2} + \frac{{\frac{1}{2}x - 1}}{{2{x^2} - x}}$
$I = \int {\left( {\frac{1}{2} + \frac{{\frac{1}{2}x - 1}}{{2{x^2} - x}}} \right)} dx$ $\therefore I = \int {\left( {\frac{1}{2} + \frac{{\frac{1}{2}x - 1}}{{2{x^2} - x}}} \right)} dx$ $I = \frac{1}{2}\int {dx + \frac{1}{2}\int {\frac{{x - 2}}{{x\left( {2x - 1} \right)}}dx} }$ $I = \frac{x}{2} + \frac{1}{2}{I_2} \qquad \text{equation I}$
$\text{here}\quad {I_2} = \int {\frac{{x - 2}}{{x\left( {2x - 1} \right)}}dx}$ $\text{Let}\quad\frac{A}{x} + \frac{B}{{2x - 1}} = \frac{{x - 2}}{{x\left( {2x - 1} \right)}}$ ∴ A(2x-1) + Bx = x-2
If x = 0 ⇒ -A = -2 ⇒ A = 2
If x = ½ ⇒ ½B = ½ - 2 ⇒ ½ = -3/2
B= -3
${I_2} = \int {\frac{{x - 2}}{{x\left( {2x - 1} \right)}}dx}$ $\therefore {I_2} = \int {\left( {\frac{2}{x} - \frac{3}{{2x - 1}}} \right)} dx$ ${I_2} = 2\int {\frac{1}{x}dx - 3\int {\frac{1}{{2x - 1}}dx} }$
${I_2} = 2\log \left| x \right| - \frac{{3\log \left| {2x - 1} \right|}}{2} + c$ Now from equation I
$I = \frac{x}{2} + \log \left| x \right| - \frac{3}{4}\log \left| {2x - 1} \right| + c$

Question (7)

$\frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}$

Solution

$I = \int {\frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}dx}$
$\text{Let}\quad \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 1}} = \frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}$ ∴ A (x2 + 1) + (Bx+C)(x-1) = x
If x = 1 ⇒ 2A = 1 ⇒ A = 1/2
If x = 0 ⇒ A - C = 0 ⇒ A = C = ½
If x = -1 ⇒ 2A - 2( - B + C) = - 1
∴ 2A+2B - 2C = -1
1+2B-1 = -1 ⇒ B = -½
$I = \int {\left( {\frac{{\frac{1}{2}}}{{x - 1}} + \frac{{ - \frac{1}{2}x + \frac{1}{2}}}{{{x^2} + 1}}} \right)dx}$ $I = \frac{1}{2}\int {\frac{1}{{x - 1}}dx} - \frac{1}{2}\int {\frac{{x - 1}}{{{x^2} + 1}}dx}$ $I = \frac{1}{2}\log \left| {x - 1} \right| - \frac{1}{2}\int {\frac{x}{{{x^2} + 1}}dx + \frac{1}{2}\int {\frac{1}{{{x^2} + 1}}dx} }$ $I = \frac{1}{2}\log \left| {x - 1} \right| - \frac{1}{{2 \times 2}}\int {\frac{{2x}}{{{x^2} + 1}}dx + \frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{x}{1}} \right) + c}$ $I = \frac{1}{2}\log \left| {x - 1} \right| - \frac{1}{4}\log \left| {{x^2} + 1} \right| + \frac{1}{2}{\tan ^{ - 1}}x + c$

Question (8)

$\frac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}$

Solution

$I = \int {\frac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}dx}$ $\text{let} \quad \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{x + 2}} = \frac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}$ $A\left( {x - 1} \right)\left( {x + 2} \right) + B\left( {x + 2} \right) + C{\left( {x - 1} \right)^2} = x$ If x =1 ⇒ 3B = 1 ⇒ B = 1/3
If x = -2 ⇒ 9C = -2 ⇒ C= -2/9
If x = 0 ⇒ -2A + 2B + C = 0
$- 2A + \frac{2}{3} - \frac{2}{9} = 0$ $- 2A = \frac{{ - 4}}{9}$ A = 2/9
Replacing values of A, B, and C. We get
$\frac{{\frac{2}{9}}}{{x - 1}} + \frac{{\frac{1}{3}}}{{{{\left( {x - 1} \right)}^2}}} + \frac{{\frac{{ - 2}}{9}}}{{x + 2}} = \frac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}$ $I = \int {\left( {\frac{{\frac{2}{9}}}{{x - 1}} + \frac{{\frac{1}{3}}}{{{{\left( {x - 1} \right)}^2}}} + \frac{{\frac{{ - 2}}{9}}}{{x + 2}}} \right)} dx$ $= \frac{2}{9}\int {\frac{1}{{x - 1}}dx + \frac{1}{3}\int {{{\left( {x - 1} \right)}^{ - 2}}dx - \frac{2}{9}\int {\frac{1}{{x + 2}}dx} } }$
$= \frac{2}{9}\log \left| {x - 1} \right| + \frac{1}{3} \times \frac{{ - 1}}{{\left( {x - 1} \right)}} - \frac{2}{9}\log \left| {x + 2} \right| + c$ $= \frac{2}{9}\log \left| {\frac{{x - 1}}{{x + 2}}} \right| - \frac{1}{{3\left( {x - 1} \right)}} + c$

Question (9)

$\frac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}$

Solution

$I = \int {\frac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}dx}$
$I = \int {\frac{{3x + 5}}{{{x^2}\left( {x - 1} \right) - \left( {x - 1} \right)}}dx}$ $I = \int {\frac{{3x + 5}}{{\left( {x - 1} \right)\left( {{x^2} - 1} \right)}}dx}$ $I = \int {\frac{{3x + 5}}{{\left( {x - 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}dx}$ $I = \int {\frac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}dx}$ $\text{Let}\quad \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{x + 1}} = \frac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}$ ∴ A(x-1)(x+1) + B(x+1) + C(x-1)2=(3x+5)
If x =1 ⇒ 2B = 8 ⇒ B =4
If x = -1 ⇒ 4C =2 ⇒ C = ½
If x = 0 ⇒ -A + B + C =5
∴ -A + 4 + ½ = 5
A = -½ $I = \left( {\int {\frac{{\frac{{ - 1}}{2}}}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}}} + \frac{{\frac{1}{2}}}{{x + 1}}} \right)dx$
$I = \frac{{ - 1}}{2}\int {\frac{1}{{x - 1}}dx + 4\int {{{\left( {x - 1} \right)}^{ - 2}}} } dx + \frac{1}{2}\int {\frac{1}{{x + 1}}dx}$
$I = \frac{1}{2}\log \left| {\frac{{x + 1}}{{x - 1}}} \right| - \frac{4}{{x - 1}} + c$

Question (10)

$\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}$

Solution

$I = \int {\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}dx}$ $I = \int {\frac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + 3} \right)}}dx}$ $\text{Let}\quad \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{C}{{2x + 3}} = \frac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + 3} \right)}}$ ∴ A(x+1)(2x+3) + B(x-1)(2x+3) + C(x-1)(x+1) = 2x-3
If x = 1 ⇒ A(2)(5) = 2-3 ⇒ A = -1/10
If x = -1 ⇒ B(-2)(1) = -2-3 ⇒ B = 5/2
If x=-3/2 ⇒ C(-5/2)(-1/2) = -6 ⇒ C= -24/5
$I = \int {\frac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + 3} \right)}}dx}$ $I = \int {\frac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + 3} \right)}}dx}$ $I = \frac{{ - 1}}{{10}}\int {\frac{1}{{x - 1}}} dx + \frac{5}{2}\int {\frac{1}{{x + 1}}dx - \frac{{24}}{5}\int {\frac{1}{{2x + 3}}dx} }$ $= \frac{{ - 1}}{{10}}\log \left| {x - 1} \right| + \frac{5}{2}\log \left| {x + 1} \right| - \frac{{\cancel{24}}^{12}}{5}\frac{{\log \left| {2x + 3} \right|}}{\cancel2} + c$ $= \frac{{ - 1}}{{10}}\log \left| {x - 1} \right| + \frac{5}{2}\log \left| {x + 1} \right| - \frac{{12}}{5}\log \left| {2x + 3} \right| + c$

Question (11)

$\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$

Solution

$I = \int {\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}dx}$ $I = 5\int {\frac{{x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}dx}$ $\text{Let}\quad \frac{A}{{x + 1}} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} = \frac{{x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}$ ∴ A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2) = x
If x = -1 ⇒ A(1)(-3) = -1 ⇒ A = 1/3
If x = -2 ⇒ B(-1)(-4) = -2 ⇒ B = -1/2
If x = 2 ⇒ C(3)(4) = 2 ⇒ C = 1/6
$I = 5\int {\frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}dx}$ $I = 5\int {\left( {\frac{{\frac{1}{3}}}{{x + 1}} - \frac{{\frac{{ - 1}}{2}}}{{x + 2}} + \frac{{\frac{1}{6}}}{{x - 2}}} \right)} dx$ $= \frac{5}{3}\int {\frac{1}{{x + 1}}dx - \frac{5}{2}\int {\frac{1}{{x + 2}}dx} + \frac{5}{6}\int {\frac{1}{{x - 2}}dx} }$ $= \frac{5}{3}\log \left| {x + 1} \right| - \frac{5}{2}\log \left| {x + 2} \right| + \frac{5}{6}\log \left| {x - 2} \right| + c$

Question (12)

$\frac{{{x^3} + x + 1}}{{{x^2} - 1}}$

Solution

$I = \int {\frac{{{x^3} + x + 1}}{{{x^2} - 1}}dx}$
Power of numerator is more than denominator to first we will divide to reduce power of numerator
$$\require{enclose} \begin{array}{r1} x \quad\; \\ {x^2} - 1 \enclose{longdiv}{\; {x^3}+{x} +\;1}\ \\ \underline {-{x^3} \mp \ x\mp \ 0}\ \\ \;2x +1\phantom{00} \\ \end{array}$$
$\therefore \frac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \frac{{2x + 1}}{{{x^2} - 1}}$
$I = \int {\frac{{{x^3} + x + 1}}{{{x^2} - 1}}} dx$ $I = \int {\left( {x + \frac{{2x + 1}}{{{x^2} - 1}}} \right)} dx$ $I = \int {\left( {x + \frac{{2x + 1}}{{{x^2} - 1}}} \right)} dx$ $I = \int {xdx + \int {\frac{{2x + 1}}{{{x^2} - 1}}dx} }$ $I = \frac{{{x^2}}}{2} + {I_1} \qquad equation (1)$ $\text{where} \quad {I_1} = \int {\frac{{2x + 1}}{{{x^2} - 1}}dx}$ ${I_1} = \int {\frac{{2x + 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}dx}$
$\text{Let} \quad \frac{A}{{x + 1}} + \frac{B}{{x - 1}} = \frac{{2x + 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}$ ∴ A(x-1) + B(x+1) = 2x+1
If x = -1 ⇒ -2A = -1 ⇒ A = 1/2
If x = 1 ⇒ 2B = 3 ⇒ B = 3/2
${I_1} = \int {\frac{{2x+1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}dx}$
${I_1} = \int {\left( {\frac{{\frac{1}{2}}}{{x + 1}} + \frac{{\frac{3}{2}}}{{x - 1}}} \right)dx}$
${I_1} = \frac{1}{2}\int {\frac{1}{{x + 1}}dx + \frac{3}{2}\int {\frac{1}{{x - 1}}dx} }$
${I_1} = \frac{1}{2}\int {\frac{1}{{x + 1}}dx + \frac{3}{2}\int {\frac{1}{{x - 1}}dx} }$
Now from equation 1
$I = \frac{{{x^2}}}{2} + \frac{1}{2}\log \left| {x + 1} \right| + \frac{3}{2}\log \left| {x - 1} \right| + c$

Question (13)

$\frac{2}{{\left( {1 - x} \right)\left( {1 + {x^2}} \right)}}$

Solution

$I = \int {\frac{2}{{\left( {1 - x} \right)\left( {1 + {x^2}} \right)}}dx}$ $\text{Let}\quad \frac{A}{{1 - x}} + \frac{{Bx + c}}{{1 + {x^2}}} = \frac{2}{{\left( {1 - x} \right)\left( {1 + {x^2}} \right)}}$ ∴ A(1+x2) + (Bx+C)(1-x) = 2
If x = 1 ⇒ 2A=2 ⇒ A = 1
If x= 0 ⇒ A+C = 2 ⇒ 1+C = 2 ⇒ C=1
If x = -1 ⇒ 2A + (-B+C)(2) = 2
2A-2B+2C = 2
2-2B+2 = 2
B = 2 ⇒ B = 1
$I = \int {\left( {\frac{1}{{1 - x}} + \frac{{x + 1}}{{1 + {x^2}}}} \right)} dx$ $I = \int {\frac{1}{{1 - x}}dx + \int {\frac{{x + 1}}{{1 + {x^2}}}dx} }$ $I = \int {\frac{1}{{1 - x}}dx + \int {\frac{x}{{1 + {x^2}}}dx + \int {\frac{1}{{1 + {x^2}}}dx} } }$ $I = \frac{{\log \left| {1 - x} \right|}}{{ - 1}} + \frac{1}{2}\int {\frac{{2x}}{{1 + {x^2}}}dx} + {\tan ^{ - 1}}x + c$ $I = - \log \left| {1 - x} \right| + \frac{1}{2}\log \left| {{x^2} + 1} \right| + {\tan ^{ - 1}}x + c$

Question (14)

$\frac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}}$

Solution

$I = \int {\frac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}}dx}$ $\text{Let}\quad \frac{A}{{x + 2}} + \frac{B}{{{{\left( {x + 2} \right)}^2}}} = \frac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}}$ ∴ A(x+2) + B = 3x -1
If x=-2 ⇒ B = -7
If x = 0 ⇒ 2A+B = -1
⇒ 2A-7 = -1
∴ A=3
Replacing Values of A and B we get
$I = \int {\left( {\frac{3}{{x + 2}} + \frac{{ - 7}}{{{{\left( {x + 2} \right)}^2}}}} \right)} dx$ $I = 3\int {\frac{1}{{x + 2}}dx} - 7\int {\frac{1}{{{{\left( {x + 2} \right)}^2}}}} dx$ $I = 3\int {\frac{1}{{x + 2}}dx - } 7\int {{{\left( {x + 2} \right)}^{ - 2}}} dx$ $I = 3\log \left| {x + 2} \right| - 7\frac{{{{\left( {x + 2} \right)}^{ - 1}}}}{{ - 1}} + c$ $I = 3\log \left| {x + 2} \right| + 7\frac{1}{{x + 2}} + c$

Question (15)

$\frac{1}{{{x^4} - 1}}$

Solution

$I = \int {\frac{1}{{{x^4} - 1}}dx}$ $I = \int {\frac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}dx}$
As both factors of denominator is quadric we can write directly as
$Let\frac{A}{{{x^2} - 1}} + \frac{B}{{{x^2} + 1}} = \frac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}$ ∴ A(x2 +1) + B(x2 - 1) = 1
If x2 = 1 ⇒2A = 1 ⇒ 1/2
If x2 = -1 ⇒ -2B = 1 ⇒ B = -1/2
By replacing values of A and B we get $I = \int {\frac{{\frac{1}{2}}}{{{x^2} - 1}}} dx + \int {\frac{{\frac{{ - 1}}{2}}}{{{x^2} + 1}}} dx$ $= \frac{1}{2} \times \frac{1}{{2\left( 1 \right)}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2} \times \frac{1}{1}{\tan ^{ - 1}}\left( {\frac{x}{1}} \right) + c$ $= \frac{1}{4}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2}{\tan ^{ - 1}}x + c$

Question (16)

$\frac{1}{{x\left( {{x^n} + 1} \right)}}$

Solution

$I = \int {\frac{1}{{x\left( {{x^n} + 1} \right)}}dx}$
Multiply numerator and denominator by xn-1 as it is derivative of xn
$I = \int {\frac{{{x^{n - 1}}}}{{x \cdot {x^{n - 1}}\left( {{x^n} + 1} \right)}}dx}$ Let xn = t
On differentiating
nxn-1dx = dt
∴ xn-1 dx = dt/n
Replacing values we get
$I = \int {\frac{{\frac{{dt}}{n}}}{{t\left( {t + 1} \right)}}}$ $I = \frac{1}{n}\int {\frac{{dt}}{{t\left( {t + 1} \right)}}}$
Fractor are denominator. equation can be solved by partial fraction as well, but as coefficient of its same for both factorss we can adjust it directly
$I = \frac{1}{n}\int {\left[ {\frac{{t + 1}}{{t\left( {t + 1} \right)}} - \frac{t}{{t\left( {t + 1} \right)}}} \right]} dt$ $I = \frac{1}{n}\left[ {\int {\frac{1}{t}d} t - \frac{1}{{t + 1}}dt} \right]$ $I = \frac{1}{n}\left[ {\log \left| t \right| - \log \left| {t + 1} \right|} \right] + c$ $I = \frac{1}{n}\log \left| {\frac{t}{{t + 1}}} \right| + c$ Substitutng value of t
$I = \frac{1}{n}\log \left| {\frac{{{x^n}}}{{{x^n} + 1}}} \right| + c$

Question (17)

$\frac{{\cos x}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}$

Solution

$I = \int {\frac{{\cos x}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}dx}$ Let sinx = t
∴ cosxdx = dt
Replacing there value we get
$I = \int {\frac{{dt}}{{\left( {1 - t} \right)\left( {2 - t} \right)}}}$
As function in denominator is we can solve using partial fraction as
$\frac{A}{{1 - t}} + \frac{B}{{2 - t}} = \frac{1}{{\left( {1 - t} \right)\left( {2 - t} \right)}}$ But as coefficient of t is same for both the brackets we can adjust numerator 1= (2-t) -(1-t)
Now replace numarator
$I = \int {\frac{{\left( {2 - t} \right) - \left( {1 - t} \right)}}{{\left( {1 - t} \right)\left( {2 - t} \right)}}} dt$ $I = \int {\frac{{\left(\cancel{ {2 - t}} \right)}}{{\left( {1 - t} \right)\left( \cancel{{2 - t}} \right)}}} dt - \int {\frac{{\left( \cancel{{1 - t}} \right)}}{{\left( \cancel{{1 - t}} \right)\left( {2 - t} \right)}}} dt$ $I = \int {\frac{1}{{1 - t}}dt - \int {\frac{1}{{2 - t}}dt} }$ $I = \frac{{\log \left| {1 - t} \right|}}{{ - 1}} - \frac{{\log \left| {2 - t} \right|}}{{ - 1}} + c$ $I = - \log \left| {1 - t} \right| + \log \left| {2 - t} \right| + c$ $I = \log \left| {\frac{{2 - t}}{{1 - t}}} \right| + c$ Substitute value of t
$I = \log \left| {\frac{{2 - \sin x}}{{1 - \sin x}}} \right| + c$

Question (18)

$\frac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Solution

$I = \int {\frac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}dx}$
Power of x in numerator and denominator is same. For integration power in numerator must be less than denominator. To reduce the power first. We can achieve by doing actual division
$I = \int {\frac{{{x^4} + 3{x^2} + 2}}{{{x^4} + 7{x^2} + 12}}dx} \qquad equation (1)$ $$\require{enclose} \begin{array}{r1} 1 \quad\; \\ {x^4}+ {7x^2} + 12 \enclose{longdiv}{\; {x^4}+{3x^2} +\;2}\ \\ \underline {-{x^4} \pm \ {7x^2}\pm \;12}\ \\ \;-4x^2 - 10\phantom{00} \\ \end{array}$$
Now we have
$\frac{{{x^4} + 3{x^2} + 2}}{{{x^4} + 7{x^2} + 12}} = 1 + \frac{{ - 4{x^2} - 10}}{{{x^4} + 7{x^2} + 12}}$ Replacing in equation 1 $I = \int {\left( {1 + \frac{{ - 4{x^2} - 10}}{{{x^4} + 7{x^2} + 12}}} \right)} dx$ $I = \int {dx - 2\int {\frac{{2{x^2} + 5}}{{{x^4} + 7x + 12}}dx} }$ $I = x - 2{I_1}$ $\text{Here} \quad{I_1} = \int {\frac{{2{x^2} + 5}}{{{x^4} + 7x + 12}}dx}$
As denominator has factor. We solve by partial fraction
$\frac{A}{{{x^2} + 3}} + \frac{B}{{{x^2} + 4}} = \frac{{2{x^2} + 5}}{{{x^4} + 7x + 12}}$ ⇒ A(x2+4) + B(x2+3) = 2x2 +5
If x2 = -3 ⇒ A = -1
If x2 = -4 ⇒ -B = -8+5 ⇒B =3
Replacing value of A and B we get
$\frac{{ - 1}}{{{x^2} + 3}} + \frac{3}{{{x^2} + 4}} = \frac{{2{x^2} + 5}}{{{x^4} + 7x + 12}}$ $I_1 = \int {\frac{{2{x^2} + 5}}{{{x^4} + 7x + 12}}dx}$ $\therefore I = \int {\left( {\frac{{ - 1}}{{{x^2} + 3}} + \frac{3}{{{x^2} + 4}}} \right)} dx$ $I_1 = - \int {\frac{1}{{{x^2} + {{\left( {\sqrt 3 } \right)}^2}}}dx + 3\int {\frac{1}{{{x^2} + {2^2}}}} } dx$
Use formula
$\int {\frac{1}{{{x^2} + {a^2}}}dx = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} + c$
$I_1 = \frac{{ - 1}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + 3 \cdot \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + c$ $I_1 = \frac{{ - 1}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + c_1$ Now I=x-2I1
Replacing the value of I1
$I = x - 2\left( {\frac{{ - 1}}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + \frac{3}{2}{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right) + {c_1}} \right)$ $I = x + \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) - 3{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + c$

Question (19)

$\frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Solution

$I = \int {\frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}dx}$
Denominator both fractions are quadratic. and numerator is linear.
We can assume x2 = t, to change denominator as linear
∴ 2x dx = dt
$I = \int {\frac{1}{{\left( {t + 1} \right)\left( {t + 3} \right)}}dt}$
Here denominator has factors we can use partial fractions to solve as $\frac{A}{{t + 1}} + \frac{B}{{t + 3}} = \frac{1}{{\left( {t + 1} \right)\left( {t + 3} \right)}}$ As coefficient of t is same. So we can adjust number 1 as
$1 = \frac{1}{2}\left[ {\left( {t + 3} \right) - \left( {t + 1} \right)} \right]$ and replace 1 in given equation
$I = \int {\frac{{dt}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}}$ $I = \int {\frac{{\frac{1}{2}\left[ {\left( {t + 3} \right) - \left( {t + 1} \right)} \right]}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}}$ $I = \frac{1}{2}\left[ {\int {\frac{\cancel{t + 3}}{{\left( {t + 1} \right)\left( \cancel{t + 3} \right)}}} dt - \int {\frac{\cancel{t + 1}}{{\left( \cancel{t + 1} \right)\left( {t + 3} \right)}}} dt} \right]$ $I = \frac{1}{2}\left[ {\log \left| {t + 1} \right| - \log \left| {t + 3} \right|} \right] + c$ $I = \frac{1}{2}\log \left| {\frac{{t + 1}}{{t + 3}}} \right| + c$ Substituting value of t
$I = \frac{1}{2}\log \left| {\frac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + c$ $I = \frac{1}{2}\log \left| {\frac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + c$

Question (20)

$\frac{1}{{x\left( {{x^4} - 1} \right)}}$

Solution

$I = \int {\frac{1}{{x\left( {{x^4} - 1} \right)}}dx}$
factors of denominator are x(x-1)(x+1)(x2+1) we can use factors as partial factor as
$\frac{1}{{x\left( {{x^4} - 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{x + 1}} + \frac{{Dx + E}}{{{x^2} + 1}}$ Solution by this method will be very lengthy
Insted we will take x4 = t
Derivative of x4 is 4x3.
By multiplying and dividing given equation by x3, and replacing x4 by t , equation will become in t and dt terms
$I = \int {\frac{1}{{x\left( {{x^4} - 1} \right)}}dx}$ $I = \int {\frac{{{x^3}}}{{x \cdot {x^3}\left( {{x^4} - 1} \right)}}dx}$ Let x4 = t
taking derivative
4x3dx = dt
x3dx = dt/4
Replacing values
$I = \int {\frac{{\frac{{dt}}{4}}}{{t\left( {t - 1} \right)}}}$ $I = \frac{1}{4}\int {\frac{{dt}}{{t\left( {t - 1} \right)}}}$
As denominator has factors. We can solve by partial factor as $\frac{1}{{t\left( {t - 1} \right)}} = \frac{A}{t} + \frac{B}{{t - 1}}$ Now coefficient of t is same in both the factor, then we can write numerator 1 = t-(t-1) which will be simple than partial factor method
$I = \frac{1}{4}\int {\frac{1}{{t\left( {t - 1} \right)}}dt}$ $I = \frac{1}{4}\int {\frac{{t - \left( {t - 1} \right)}}{{t\left( {t - 1} \right)}}dt}$ $I =\frac{1}{4} \int {\frac{\cancel{t}}{{\cancel{t}\left( {t - 1} \right)}}dt - \frac{1}{4}\int {\frac{{\left( \cancel{{t - 1}} \right)}}{{t\left(\cancel{ {t - 1}} \right)}}} } dt$ $I = \frac{1}{4}\int {\frac{1}{{\left( {t - 1} \right)}}dt - \frac{1}{4}\int {\frac{1}{t}} } dt$ $I =\frac{1}{4} \log \left| {t - 1} \right| - \frac{1}{4}\log \left| t \right| + c$ $I =\frac{1}{4} \log \left| {\frac{{t - 1}}{t}} \right| + c$ substituting value of t
$I =\frac{1}{4} \log \left| {\frac{{{x^4} - 1}}{{{x^4}}}} \right| + c$

Question (21)

$\frac{1}{{{e^x} - 1}}$

Solution

$I = \int {\frac{1}{{{e^x} - 1}}dx}$ Let ex = t
∴ exdx = dt
$dx = \frac{{dt}}{{{e^x}}}$ Replacing ex= t
$dx = \frac{{dt}}{t}$ Now replace values in given equation
$I = \int {\frac{1}{{{e^x} - 1}}dx}$ $I = \int {\frac{1}{{t - 1}} \cdot \frac{{dt}}{t}}$ $I = \int {\frac{1}{{t\left( {t - 1} \right)}}dt}$ Adjust numerator 1=t-t(t-1)
$I = \int {\frac{{t - \left( {t - 1} \right)}}{{t\left( {t - 1} \right)}}} dt$ $I = \int {\frac{\cancel{t}}{{\cancel{t}\left( {t - 1} \right)}}dt - \int {\frac{\cancel{{t - 1}}}{{t\left( \cancel{{t - 1}} \right)}}dt} }$ $I = \int {\frac{1}{{\left( {t - 1} \right)}}dt - \int {\frac{1}{t}dt} }$ $I = \log \left| {\frac{{t - 1}}{t}} \right| + c$ Substitute value of t
$I = \log \left| {\frac{{{e^x} - 1}}{{{e^x}}}} \right| + c$ Choose the correct answer in each of the Exercises 22 and 23

Question (22)

$\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}}$ Options
$\text{(A)}\quad \log \left| {\frac{{{{\left( {x - 1} \right)}^2}}}{{x - 2}}} \right| + c$ $\text{(B)}\quad \log \left| {\frac{{{{\left( {x - 2} \right)}^2}}}{{x - 1}}} \right| + c$ $\text{(C)}\quad\log \left| {{{\left( {\frac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + c$ $\text{(D)}\quad\log \left| {\left( {x - 1} \right)(x - 2)} \right| + c$

Solution

$I = \int {\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}}dx}$ $\text{Let}\quad\frac{A}{{x - 1}} + \frac{B}{{x - 2}} = \frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}}$ A(x - 2) + B(x - 1) = x
If x = 1 ⇒ -A = 1 ⇒A = -1
If x=2 ⇒ B = 2
Replace value of A and B
$I = \int {\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}}dx}$ $I = \int {\left( {\frac{{ - 1}}{{x - 1}} + \frac{2}{{x - 2}}} \right)dx}$ $I = - \int {\frac{1}{{x - 1}}dx + 2\int {\frac{1}{{x - 2}}dx} }$ $I = - \log \left| {x - 1} \right| + 2\log \left| {x - 2} \right| + c$ $I = - \log \left| {x - 1} \right| + \log {\left( {x - 2} \right)^2} + c$ $I = \log \left| {\frac{{{{\left( {x - 2} \right)}^2}}}{{x - 1}}} \right| + c$ The option "B" is correct

Question (23)

$\int {\frac{1}{{x\left( {{x^2} + 1} \right)}}} dx$ $\text{(A)}\quad\log \left| x \right| - \frac{1}{2}\log \left( {{x^2} + 1} \right) + c$ $\text{(B)}\quad\log \left| x \right| + \frac{1}{2}\log \left( {{x^2} + 1} \right) + c$ $\text{(C)}\quad - \log \left| x \right| + \frac{1}{2}\log \left( {{x^2} + 1} \right) + c$ $\text{(D)}\quad\frac{1}{2}\log \left| x \right| + \log \left( {{x^2} + 1} \right) + c$

Solution

We will multiply numerator and denominator by x
$I = \int {\frac{x}{{{x^2}\left( {{x^2} + 1} \right)}}dx}$ Let x2 =t
2xdx = dt
xdx= dt/2
Replace values
$I = \frac{1}{2}\int {\frac{{dt}}{{t\left( {t + 1} \right)}}}$ Adjust numerator 1= (t+1) - (t)
$I = \frac{1}{2}\int {\frac{{\left( {t + 1} \right) - t}}{{t\left( {t + 1} \right)}}} dt$ $I = \frac{1}{2}\left[ {\int {\frac{{\left( \cancel{{t + 1}} \right)}}{{t\left( \cancel{{t + 1}} \right)}}dt - \int {\frac{\cancel{t}}{{\cancel{t}\left( {t + 1} \right)}}dt} } } \right]$ $I = \frac{1}{2}\left[ {\int {\frac{1}{t}dt - \int {\frac{1}{{\left( {t + 1} \right)}}dt} } } \right]$ $I = \frac{1}{2}\left[ {\log \left| t \right| - \log \left| {t + 1} \right|} \right] + c$ $I = \frac{1}{2}\log \left| {{x^2}} \right| - \frac{1}{2}\log \left| {{x^2} + 1} \right| + c$ $I = \frac{1}{2} \cdot 2\log \left| x \right| - \frac{1}{2}\log \left| {{x^2} + 1} \right| + c$ $I = \log \left| x \right| - \frac{1}{2}\log \left| {{x^2} + 1} \right| + c$ Option "A" correct ANOTHER METHOD $\int {\frac{1}{{x\left( {{x^2} + 1} \right)}}dx}$ $\text{Let}\quad\frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} = \frac{1}{{x\left( {{x^2} + 1} \right)}}$ A(x2+1) + (Bx+C)x = 1
If x = 0 ⇒ A = 1
If x=1 ⇒ 2A+B+C=1
If x=-1 ⇒ 2A+B-C = 1
⇒ 1-c-c= 1
⇒ -2C = 0 ⇒ C = 0;
2A+B+C = 1 ⇒ 2+B+O =1 ⇒B=-1
Replacing value we get
$I = \int {\left( {\frac{1}{x} + \frac{{ - x + 0}}{{{x^2} + 1}}} \right)dx}$ $I = \int {\frac{1}{x}dx - \int {\frac{x}{{{x^2} + 1}}dx} }$ $I = \log \left| x \right| - \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}dx}$ Let x2 +1 = t
2xdx = dt Now on substituting values $\frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}dx} = \int {\frac{{dt}}{t}} = \log \left| t \right|$ Substituting value of t
$\frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}dx} = \frac{1}{2}\log \left| {{x^2} + 1} \right| + c$ $\therefore \quad I = \log \left| x \right| - \frac{1}{2}\log \left| {{x^2} + 1} \right| + c$