12th NCERT INTEGRALS Exercise 7.5 Questions 23
Do or do not
There is no try
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
\[\frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}dx} = \frac{1}{2}\log \left| {{x^2} + 1} \right| + c\] \[\therefore \quad I = \log \left| x \right| - \frac{1}{2}\log \left| {{x^2} + 1} \right| + c\]
Exercise 7.4 ⇐
⇒ Exercise 7.6