NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.5

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Chapter 7 Ex.7.5 Question 1

Integrate the rational function \(\frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}\)

Solution

Let \(\frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{A}{{\left( {x + 1} \right)}} + \frac{B}{{\left( {x + 2} \right)}}\)

\( \Rightarrow x = A\left( {x + 2} \right) + B\left( {x + 1} \right)\)

Equating the coefficients of x and constant term, we get

\[\begin{align}A + B = 1\\2A + B = 0\end{align}\]

On solving, we get

\[\begin{align}A& = - 1\,\,{\rm{and}}\,\;B = 2\\\therefore \frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}& = \frac{{ - 1}}{{\left( {x + 1} \right)}} + \frac{2}{{\left( {x + 2} \right)}}\end{align}\]

\[\begin{align}& \Rightarrow \int {\frac{x}{{\left( {x + 1} \right)\left( {x + 2} \right)}}dx = } \int {\frac{{ - 1}}{{\left( {x + 1} \right)}} + \frac{2}{{\left( {x + 2} \right)}}} dx\\ &= - \log \left| {x + 1} \right| + 2\log \left| {x + 2} \right| + C\\ &= \log {\left( {x + 2} \right)^2} - \log \left( {x + 1} \right) + C\\& = \log \frac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x + 1} \right)}} + C\end{align}\]

Chapter 7 Ex.7.5 Question 2

Integrate the rational function \(\frac{1}{{{x^2} - 9}}\)

Solution

Let \(\frac{1}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{A}{{\left( {x + 3} \right)}} + \frac{B}{{\left( {x - 3} \right)}}\)

\(1 = A\left( {x - 3} \right) + B\left( {x + 3} \right)\)

Equating the coefficients of \(x\) and constants term, we get

\[\begin{align}A + B &= 0\\ - 3A + 3B &= 1\end{align}\]

On solving, we get

\[\begin{align}&A = - \frac{1}{6}\,\,{\rm{and }}B = \frac{1}{6}\\&\therefore \frac{1}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{ - 1}}{{6\left( {x + 3} \right)}} + \frac{1}{{6\left( {x - 3} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{1}{{\left( {{x^2} - 9} \right)}}dx} = \int {\left( {\frac{{ - 1}}{{6\left( {x + 3} \right)}} + \frac{1}{{6\left( {x - 3} \right)}}} \right)dx} \\& = - \frac{1}{6}\log \left| {x + 3} \right| + \frac{1}{6}\log \left| {x - 3} \right| + C\\ &= \frac{1}{6}\log \frac{{\left| {\left( {x - 3} \right)} \right|}}{{\left| {\left( {x + 3} \right)} \right|}} + C\end{align}\]

Chapter 7 Ex.7.5 Question 3

Integrate the rational function \(\frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}\)

Solution

Let \(\frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{A}{{\left( {x - 1} \right)}} + \frac{B}{{\left( {x - 2} \right)}} + \frac{C}{{\left( {x - 3} \right)}}\)

\(3x - 1 = A\left( {x - 2} \right)\left( {x - 3} \right) + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right)\,\,\,\,\, \ldots \left( 1 \right)\)

Equating the coefficients of \({x^2},x\)and constant terms, we get

\[\begin{align} A+B+C&=0 \\ -5A-4B-3C&=3 \\ 6A+3B+2C&=-1\end{align}\]

Solving these equations, we get

\[\begin{align}&A = 1,B = - 5\,\,{\rm{and }}C = 4\\&\therefore \frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{1}{{\left( {x - 1} \right)}} - \frac{5}{{\left( {x - 2} \right)}} + \frac{4}{{\left( {x - 3} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx} = \int {\left\{ {\frac{1}{{\left( {x - 1} \right)}} - \frac{5}{{\left( {x - 2} \right)}} + \frac{4}{{\left( {x - 3} \right)}}} \right\}} dx\\ &= \log \left| {x - 1} \right| - 5\log \left| {x - 2} \right| + 4\log \left| {x - 3} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 4

Integrate the rational function \(\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}\)

Solution

Let \(\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{A}{{\left( {x - 1} \right)}} + \frac{B}{{\left( {x - 2} \right)}} + \frac{C}{{\left( {x - 3} \right)}}\)

\(x = A\left( {x - 2} \right)\left( {x - 3} \right) + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right)\,\,\,\,\, \ldots \left( 1 \right)\)

Equating the coefficients of \({x^2},x\) and constant terms, we get

\[\begin{align}A + B + C &= 0\\ - 5A - 4B - 3C &= 1\\6A + 4B + 2C &= 0\end{align}\]

Solving these equations, we get

\[\begin{align}&A = \frac{1}{2},B = - 2\,\,{\rm{and }}C = \frac{3}{2}\\&\therefore \frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{1}{{2\left( {x - 1} \right)}} - \frac{2}{{\left( {x - 2} \right)}} + \frac{3}{{2\left( {x - 3} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx} = \int {\left\{ {\frac{1}{{2\left( {x - 1} \right)}} - \frac{2}{{\left( {x - 2} \right)}} + \frac{3}{{2\left( {x - 3} \right)}}} \right\}} dx\\ &= \frac{1}{2}\log \left| {x - 1} \right| - 2\log \left| {x - 2} \right| + \frac{3}{2}\log \left| {x - 3} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 5

Integrate the rational function \(\frac{{2x}}{{{x^2} + 3x + 2}}\)

Solution

\[\begin{align} {\text{Let }}\frac{{2x}}{{{x^2} + 3x + 2}} &= \frac{A}{{\left( {x + 1} \right)}} + \frac{B}{{\left( {x + 2} \right)}} \hfill \\ 2x &= A\left( {x + 2} \right) + B\left( {x + 1} \right)\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \hfill \\ \end{align} \]

Equating the coefficients of x and constant terms, we get

\[\begin{align}A + B &= 2\\2A + B &= 0\end{align}\]

Solving these equations, we get

\[\begin{align}&A = - 2\,\,{\text{and }}B = 4\\&\therefore \frac{{2x}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{{ - 2}}{{\left( {x + 1} \right)}} + \frac{4}{{\left( {x + 2} \right)}}\\& \Rightarrow \; \; \; \; \int {\frac{{2x}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}dx} = \int {\left\{ {\frac{4}{{\left( {x + 2} \right)}} - \frac{2}{{\left( {x + 1} \right)}}} \right\}} dx\\ &= 4\log \left| {x + 2} \right| - 2\log \left| {x + 1} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 6

Integrate the rational function \(\frac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}\)

Solution

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing \(\left( 1-{{x}^{2}} \right)\)by \(x\left( 1-2x \right)\), we get

\[\frac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}= \frac{1}{2} + \frac{1}{2}\left( {\frac{{2 - x}}{{x\left( {1 - 2x} \right)}}} \right) \ldots \left( 1 \right)\]

Let \(\frac{{2 - x}}{{x\left( {1 - 2x} \right)}} = \frac{A}{x} + \frac{B}{{\left( {1 - 2x} \right)}}\)

\( \Rightarrow \; \; \; \; \left( {2 - x} \right) = A\left( {1 - 2x} \right) + Bx\)

Equating the coefficients of x and constant term, we get

Solving these equations, we get

\(A = 2\) and \(B = 3\)

\(\therefore \frac{{2 - x}}{{x\left( {1 - 2x} \right)}} = \frac{2}{x} + \frac{3}{{\left( {1 - 2x} \right)}}\)

Substituting in equation (1), we get

\[\begin{align}&\frac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}} = \frac{1}{2} + \frac{1}{2}\left\{ {\frac{2}{x} + \frac{3}{{\left( {1 - 2x} \right)}}} \right\}\\& \Rightarrow \;\int {\frac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}} dx = \int {\left\{ {\frac{1}{2} + \frac{1}{2}\left( {\frac{2}{x} + \frac{3}{{\left( {1 - 2x} \right)}}} \right)} \right\}} dx\\& = \frac{x}{2} + \log \left| x \right| + \frac{3}{{2\left( { - 2} \right)}}\log \left| {1 - 2x} \right| + C\\& = \frac{x}{2} + \log \left| x \right| - \frac{3}{4}\log \left| {1 - 2x} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 7

Integrate the rational function \(\frac{x}{\left( {{x}^{2}}+1 \right)\left( x-1 \right)}\)

Solution

Let \(\frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} = \frac{{Ax + B}}{{\left( {{x^2} + 1} \right)}} + \frac{C}{{\left( {x - 1} \right)}}\;\;\;\;\;\;\;\;\;\;......\left( 1 \right)\)

Equating the coefficients of \({x^2},x\)and constant term, we get

\[\begin{align}A + C &= 0\\ - A + B &= 1\\ - B + C &= 0\end{align}\]

On solving these equations, we get

\(A = - \frac{1}{2},B = \frac{1}{2}\,\,{\text{and }}C = \frac{1}{2}\)

From equation (1), we get

\[\begin{align}&\therefore \frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} = \frac{{\left( { - \frac{1}{2}x + \frac{1}{2}} \right)}}{{{x^2} + 1}} + \frac{{\frac{1}{2}}}{{\left( {x - 1} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} = - \frac{1}{2}\int {\frac{x}{{{x^2} + 1}}dx} } + \frac{1}{2}\int {\frac{1}{{{x^2} + 1}}} dx + \frac{1}{2}\int {\frac{1}{{x - 1}}} dx\\ &= - \frac{1}{4}\int {\frac{{2x}}{{{x^2} + 1}}} dx + \frac{1}{2}{\tan ^{ - 1}}x + \frac{1}{2}\log \left| {x - 1} \right| + C\end{align}\]

Consider \(\int{\frac{2x}{{{x}^{2}}+1}dx}\), let \(\left( {{x^2} + 1} \right) = t \Rightarrow 2xdx = dt\)

\[\begin{align}& \Rightarrow \int {\frac{{2x}}{{{x^2} + 1}}dx = \int {\frac{{dt}}{t}} = \log \left| t \right| = \log \left| {{x^2} + 1} \right|} \\&\therefore \int {\frac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} = - \frac{1}{4}} \log \left| {{x^2} + 1} \right| + \frac{1}{2}{\tan ^{ - 1}}x + \frac{1}{2}\log \left| {x - 1} \right| + C\\ &= \frac{1}{2}\log \left| {x - 1} \right| - \frac{1}{4}\log \left| {{x^2} + 1} \right| + \frac{1}{2}{\tan ^{ - 1}}x + C\end{align}\]

Chapter 7 Ex.7.5 Question 8

Integrate the rational function \(\frac{x}{{{\left( x-1 \right)}^{2}}\left( x+2 \right)}\)

Solution

Let \(\frac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}} = \frac{A}{{\left( {x - 1} \right)}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{\left( {x + 2} \right)}}\)

\(x = A\left( {x - 1} \right)\left( {x + 2} \right) + B\left( {x + 2} \right) + C{\left( {x - 1} \right)^2}\)

Equating the coefficients of \({x^2},x\)and constant term, we get

\[\begin{align}A + C &= 0\\A + B - 2C &= 1\\ - 2A + 2B + C &= 0\end{align}\]

On solving these equations, we get

\(A=\frac{2}{9},B=\frac{1}{3}\,\,\text{and }C=-\frac{2}{9}\)

\[\begin{align}&\therefore \frac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}} = \frac{2}{{9\left( {x + 1} \right)}} + \frac{1}{{3{{\left( {x - 1} \right)}^2}}} - \frac{2}{{9\left( {x + 2} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}dx} = \frac{2}{9}\int {\frac{1}{{\left( {x - 1} \right)}}} dx + \frac{1}{3}\int {\frac{1}{{{{\left( {x - 1} \right)}^2}}}} dx - \frac{2}{9}\int {\frac{1}{{\left( {x - 2} \right)}}dx} \\ &= \frac{2}{9}\log \left| {x - 1} \right| + \frac{1}{3}\left( {\frac{{ - 1}}{{x - 1}}} \right) - \frac{2}{9}\log \left| {x + 2} \right| + C\\ &= \frac{2}{9}\log \left| {\frac{{x - 1}}{{x + 2}}} \right| - \frac{1}{{3\left( {x - 1} \right)}} + C\end{align}\]

Chapter 7 Ex.7.5 Question 9

Integrate the rational function \(\frac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}\)

Solution

\(\frac{{3x + 5}}{{{x^3} - {x^2} - x + 1}} = \frac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\)

Let \(\frac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \frac{A}{{\left( {x - 1} \right)}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{\left( {x + 1} \right)}}\)

\[\begin{align}3x + 5 &= A\left( {x - 1} \right)\left( {x + 1} \right) + B\left( {x + 1} \right) + C{\left( {x - 1} \right)^2}\\3x + 5 &= A\left( {x - 1} \right)\left( {x + 1} \right) + B\left( {x + 1} \right) + C\left( {{x^2} - 2x + 1} \right)\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{align}\]

Equating the coefficients of \({{x}^{2}},x\)and constant term, we get

\[\begin{align}A + C &= 0\\B - 2C &= 3\\ - A + B + C &= 5\end{align}\]

On solving these equations, we get

\[\begin{align}&A = - \frac{1}{2},B = 4\,\,{\text{and }}C = \frac{1}{2}\\&\therefore \frac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \frac{{ - 1}}{{2\left( {x - 1} \right)}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}} + \frac{1}{{2\left( {x + 1} \right)}}\\& \Rightarrow \; \; \; \; \int {\frac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}dx} = - \frac{1}{2}\int {\frac{1}{{\left( {x - 1} \right)}}} dx + 4\int {\frac{1}{{{{\left( {x - 1} \right)}^2}}}dx} + \frac{1}{2}\int {\frac{1}{{\left( {x + 1} \right)}}} dx\\ &= - \frac{1}{2}\log \left| {x - 1} \right| + 4\left( {\frac{{ - 1}}{{x - 1}}} \right) + \frac{1}{2}\log \left| {x + 1} \right| + C\\ &= \frac{1}{2}\log \left| {\frac{{x + 1}}{{x - 1}}} \right| - \frac{4}{{\left( {x - 1} \right)}} + C\end{align}\]

Chapter 7 Ex.7.5 Question 10

Integrate the rational function \(\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}\)

Solution

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

Let \(\frac{2x-3}{\left( x-1 \right)\left( x+1 \right)\left( 2x+3 \right)}=\frac{A}{\left( x+1 \right)}+\frac{B}{\left( x-1 \right)}+\frac{C}{\left( 2x+3 \right)}\)

\[\begin{align} &\Rightarrow \; \; \; \; \left( {2x - 3} \right) = A\left( {x - 1} \right)\left( {2x - 3} \right) + B\left( {x + 1} \right)\left( {2x + 3} \right) + C\left( {x + 1} \right)\left( {x - 1} \right)\\ &\Rightarrow \; \; \; \; \left( {2x - 3} \right) = A\left( {2{x^2} + x - 3} \right) + B\left( {2{x^2} + 5x + 3} \right) + C\left( {{x^2} - 1} \right)\\ &\Rightarrow \; \; \; \; \left( {2x - 3} \right) = \left( {2A + 2B + C} \right){x^2} + \left( {A + 5B} \right)x + \left( { - 3A + 3B - C} \right)\end{align}\]

Equating the coefficients of \({x^2},x\)and constant term, we get

\[\begin{align}&2A + 2B + C = 0\\&A + 5B = 2\\& - 3A + 3B - C = - 3\end{align}\]

On solving, we get

\[\begin{align}&A = \frac{5}{2},B = - \frac{1}{{10}}\,\,{\rm{and }}C = - \frac{{24}}{5}\\&\therefore \frac{{2x - 3}}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {2x + 3} \right)}} = \frac{5}{{2\left( {x + 1} \right)}} - \frac{1}{{10\left( {x - 1} \right)}} - \frac{{24}}{{5\left( {2x + 3} \right)}}\\& \Rightarrow \; \; \; \; \int {\frac{{2x - 3}}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}dx} = \frac{5}{2}\int {\frac{1}{{\left( {x + 1} \right)}}} dx - \frac{1}{{10}}\int {\frac{1}{{\left( {x - 1} \right)}}dx} - \frac{{24}}{5}\int {\frac{1}{{\left( {2x + 3} \right)}}} dx\\& = \frac{5}{2}\log \left| {x + 1} \right| - \frac{1}{{10}}\log \left| {x - 1} \right| - \frac{{24}}{{5 \times 2}}\log \left| {2x + 3} \right| + C\\& = \frac{5}{2}\log \left| {x + 1} \right| - \frac{1}{{10}}\log \left| {x - 1} \right| - \frac{{12}}{5}\log \left| {2x + 3} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 11

Integrate the rational function \(\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}\)

Solution

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

Let \(\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}} = \frac{A}{{\left( {x + 1} \right)}} + \frac{B}{{\left( {x + 2} \right)}} + \frac{C}{{\left( {x - 2} \right)}}\)

\(5x=A\left( x+2 \right)\left( x-2 \right)+B\left( x+1 \right)\left( x-2 \right)+C\left( x+1 \right)\left( x+2 \right)\,\,\,\,\,\,\ldots \left( 1 \right)\)

Equating the coefficients of \({x^2},x\) and constant term, we get

\[\begin{align} A + B + C &= 0 \hfill \\ - B + 3C &= 5 \hfill \\ - 4A - 2B + 2C &= 0 \hfill \end{align}\]

On solving, we get

\[\begin{align}&A = \frac{5}{3},B = - \frac{5}{2}\,\,{\rm{and }}C = \frac{5}{6}\\&\therefore \frac{{5x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}} = \frac{5}{{3\left( {x + 1} \right)}} - \frac{5}{{2\left( {x + 2} \right)}} + \frac{5}{{6\left( {x - 2} \right)}}\\& \Rightarrow \; \; \; \; \int {\frac{{5x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}dx} = \frac{5}{3}\int {\frac{1}{{\left( {x + 1} \right)}}} dx - \frac{5}{2}\int {\frac{1}{{\left( {x + 2} \right)}}dx} + \frac{5}{6}\int {\frac{1}{{\left( {x - 2} \right)}}} dx\\ &= \frac{5}{3}\log \left| {x + 1} \right| - \frac{5}{2}\log \left| {x + 2} \right| + \frac{5}{6}\log \left| {x - 2} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 12

Integrate the rational function \(\frac{{{x^3} + x + 1}}{{{x^2} - 1}}\)

Solution

On dividing \(\left( {{x^3} + x + 1} \right)\)by \({x^2} - 1\), we get

\(\frac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \frac{{2x + 1}}{{{x^2} - 1}}\)

Let \(\frac{{2x + 1}}{{{x^2} - 1}} = \frac{A}{{\left( {x + 1} \right)}} + \frac{B}{{\left( {x + 1} \right)}}\)

\(2x + 1 = A\left( {x - 1} \right) + B\left( {x + 1} \right)\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

Equating the coefficients of \(x\) and constant term, we get

\[\begin{align}A + B &=2 \\ - A + B &= 1\end{align}\]

On solving, we get

\[\begin{align}&A = \frac{1}{2}\;{\text{and}}\;B = \frac{3}{2}\\&\therefore \frac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \frac{1}{{2\left( {x + 1} \right)}} + \frac{3}{{2\left( {x - 1} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{{{x^3} + x + 1}}{{{x^2} - 1}}} dx = \int {xdx} + \frac{1}{2}\int {\frac{1}{{\left( {x + 1} \right)}}dx} + \frac{3}{2}\int {\frac{1}{{\left( {x - 1} \right)}}dx} \\& = \frac{{{x^2}}}{2} + \frac{1}{2}\log \left| {x + 1} \right| + \frac{3}{2}\log \left| {x - 1} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 13

Integrate the rational function \(\frac{2}{{\left( {1 - x} \right)\left( {1 + {x^2}} \right)}}\)

Solution

Let \(\frac{2}{{\left( {1 - x} \right)\left( {1 + {x^2}} \right)}} = \frac{A}{{\left( {1 - x} \right)}} + \frac{{Bx + C}}{{\left( {1 + {x^2}} \right)}}\)

\[\begin{align} 2 &= A\left( {1 + {x^2}} \right) + \left( {Bx + C} \right)\left( {1 - x} \right) \hfill \\ 2 &= A + A{x^2} + Bx - B{x^2} + C - Cx \hfill \\ \end{align}\]

Equating the coefficients of \({{x}^{2}},x\)and constant term, we get

\[\begin{align}A - B = 0\\B - C = 0\\A + C = 2\end{align}\]

On solving these equations, we get

\[\begin{align} & A=1,B=1\,\,\text{and }C=1 \\ & \therefore \frac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}=\frac{1}{1-x}+\frac{x+1}{1+{{x}^{2}}} \\ & \Rightarrow \; \; \; \; \int{\frac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}dx}=\int{\frac{1}{1-x}dx}+\int{\frac{x}{1+{{x}^{2}}}dx}+\int{\frac{1}{1+{{x}^{2}}}dx} \\ & =-\int{\frac{1}{1-x}dx}+\frac{1}{2}\int{\frac{2x}{1+{{x}^{2}}}dx}+\int{\frac{1}{1+{{x}^{2}}}dx} \\ & =-\log \left| x-1 \right|+\frac{1}{2}\log \left| 1+{{x}^{2}} \right|+{{\tan }^{-1}}x+C \\ \end{align}\]

Chapter 7 Ex.7.5 Question 14

Integrate the rational function \(\frac{3x-1}{{{\left( x+2 \right)}^{2}}}\)

Solution

Let \(\frac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}} = \frac{A}{{\left( {x + 2} \right)}} + \frac{B}{{{{\left( {x + 2} \right)}^2}}}\)

\( \Rightarrow \; \; \; \; 3x - 1 = A\left( {x + 2} \right) + B\)

Equating the coefficient of x and constant term, we get

\[\begin{align}&A = 3\\&2A + B = - 1 \Rightarrow \; \; \; \; B = - 7\\&\therefore \frac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}} = \frac{3}{{\left( {x + 2} \right)}} - \frac{7}{{{{\left( {x + 2} \right)}^2}}}\\ &\Rightarrow \; \; \; \; \int {\frac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}}} dx = 3\int {\frac{1}{{\left( {x + 2} \right)}}} dx - 7\int {\frac{x}{{{{\left( {x + 2} \right)}^2}}}dx} \\ &= 3\log \left| {x + 2} \right| - 7\left( {\frac{{ - 1}}{{\left( {x + 2} \right)}}} \right) + C\\&= 3\log \left| {x + 2} \right| + \frac{7}{{\left( {x + 2} \right)}} + C\end{align}\]

Chapter 7 Ex.7.5 Question 15

Integrate the rational function \(\frac{1}{{{x^4} - 1}}\)

Solution

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

Let \(\frac{1}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{\left( {x + 1} \right)}} + \frac{B}{{\left( {x - 1} \right)}} + \frac{{Cx + D}}{{\left( {{x^2} + 1} \right)}}\)

\[\begin{align}1 &= A\left( {x - 1} \right)\left( {1 + {x^2}} \right) + B\left( {x + 1} \right)\left( {1 + {x^2}} \right) + \left( {Cx + D} \right)\left( {{x^2} - 1} \right)\\1& = A\left( {{x^3} + x - {x^2} - 1} \right) + B\left( {{x^3} + x + {x^2} + 1} \right) + C{x^3} + D{x^2} - Cx - D\\1 &= \left( {A + B + C} \right){x^3} + \left( { - A + B + D} \right){x^2} + \left( {A + B - C} \right)x + \left( { - A + B - D} \right)\end{align}\]

Equating the coefficients of \({x^3},{x^2},x\)and constant term, we get

\[\begin{align}A + B + C &= 0\\ - A + B + D &= 0\\A + B - C& = 0\\ - A + B - D& = 1\end{align}\]

On solving, we get

\[\begin{align}&A = \frac{{ - 1}}{4},B = \frac{1}{4},C = 0\,\,{\rm{and D}} = - \frac{1}{2}\\&\therefore \frac{1}{{\left( {{x^4} - 1} \right)}} = \frac{{ - 1}}{{4\left( {x + 1} \right)}} + \frac{1}{{4\left( {x - 1} \right)}} + \frac{1}{{2\left( {1 + {x^2}} \right)}}\\& \Rightarrow \; \; \; \; \int {\frac{1}{{\left( {{x^4} - 1} \right)}}dx} = \int {\frac{{ - 1}}{{4\left( {x + 1} \right)}}dx} + \int {\frac{1}{{4\left( {x - 1} \right)}}dx} - \int {\frac{1}{{2\left( {1 + {x^2}} \right)}}dx} \\ &\Rightarrow \; \; \; \; \int {\frac{1}{{\left( {{x^4} - 1} \right)}}dx} = - \frac{1}{4}\log \left| {x + 1} \right| + \frac{1}{4}\log \left| {x - 1} \right| - \frac{1}{2}{\tan ^{ - 1}}x + C\\ &= \frac{1}{4}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2}{\tan ^{ - 1}}x + C\end{align}\]

Chapter 7 Ex.7.5 Question 16

Integrate the rational function \(\frac{1}{{x\left( {{x^n} + 1} \right)}}\)

[Hint: multiply numerator and denominator by \({x^{n - 1}}\)and put \({x^n} = t\)]

Solution

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

Multiplying numerator and denominator by\({x^{n - 1}}\), we get

\[\begin{align}&\frac{1}{{x\left( {{x^n} + 1} \right)}} = \frac{{{x^{n - 1}}}}{{{x^{n - 1}}x\left( {{x^n} + 1} \right)}} = \frac{{{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}\\&{\text{Let }}{x^n} = t \Rightarrow \; \; \; \; n{x^{n - 1}}dx = dt\\&\therefore \int {\frac{1}{{x\left( {{x^n} + 1} \right)}}dx} = \int {\frac{{{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}dx} = \frac{1}{n}\int {\frac{1}{{t\left( {t + 1} \right)}}} dt\\&{\text{Let }}\frac{1}{{t\left( {t + 1} \right)}} = \frac{A}{t} + \frac{B}{{\left( {t + 1} \right)}}\\&1 = A\left( {1 + t} \right) + Bt\,\,\,\, \ldots \left( 1 \right)\end{align}\]

Equating the coefficients of t and constant term, we get

\[\begin{align}&A = 1\,\,{\text{and }}B = - 1\\&\therefore \frac{1}{{t\left( {t + 1} \right)}} = \frac{1}{t} - \frac{1}{{\left( {1 + t} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{1}{{x\left( {{x^n} + 1} \right)}}dx} = \frac{1}{n}\int {\left\{ {\frac{1}{t} - \frac{1}{{\left( {1 + t} \right)}}} \right\}dx} \\& = \frac{1}{n}\left[ {\log \left| t \right| - \log \left| {t + 1} \right|} \right] + C\\ &= \frac{1}{n}\left[ {\log \left| {{x^n}} \right| - \log \left| {{x^n} + 1} \right|} \right] + C\\& = \frac{1}{n}\log \left| {\frac{{{x^n}}}{{{x^n} + 1}}} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 17

Integrate the rational function \(\frac{{\cos x}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}\) [Hint: Put\({\rm{sin }}x = t\)]

Solution

\(\frac{{\cos x}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}\) Put, \(\sin x = t \Rightarrow \; \; \; \; \cos xdx = dt\)

\(\therefore \int {\frac{{\cos x}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}dx = \int {\frac{{dt}}{{\left( {1 - t} \right)\left( {2 - t} \right)}}} } \)

Let \(\frac{1}{{\left( {1 - t} \right)\left( {2 - t} \right)}} = \frac{A}{{\left( {1 - t} \right)}} + \frac{B}{{\left( {2 - t} \right)}}\)

\(1 = A\left( {2 - t} \right) + B\left( {1 - t} \right)\,\,\,\,\,\, \ldots \left( 1 \right)\)

Equating the coefficients of t and constant, we get

\( - 2A - B = 0,\,\,{\rm{and }}2A + B = 1\)

On solving, we get

\(A = 1\,\,{\rm{and }}B = - 1\)

\[\begin{align}&\therefore \frac{1}{{\left( {1 - t} \right)\left( {2 - t} \right)}} = \frac{1}{{\left( {1 - t} \right)}} - \frac{1}{{\left( {2 - t} \right)}}\\& \Rightarrow \; \; \; \; \int {\frac{{\cos x}}{{\left( {1 - \sin x} \right)\left( {2 - \sin x} \right)}}dx} = \int {\left\{ {\frac{1}{{1 - t}} - \frac{1}{{\left( {2 - t} \right)}}} \right\}dt} \\ &= - \log \left| {1 - t} \right| + \log \left| {2 - t} \right| + C\\& = \log \left| {\frac{{2 - t}}{{1 - t}}} \right| + C\\& = \log \left| {\frac{{2 - \sin x}}{{1 - \sin x}}} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 18

Integrate the rational function \(\frac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}\)

Solution

\(\frac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \frac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}\)

Let \(\frac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \frac{{Ax + B}}{{\left( {{x^2} + 3} \right)}} + \frac{{Cx + D}}{{\left( {{x^2} + 4} \right)}}\)

\[\begin{align}4{x^2} + 10 &= \left( {Ax + B} \right)\left( {{x^2} + 4} \right) + \left( {Cx + D} \right)\left( {{x^2} + 3} \right)\\4{x^2} + 10 &= A{x^3} + 4Ax + B{x^2} + 4B + C{x^3} + 3Cx + D{x^2} + 3D\\4{x^2} + 10 &= \left( {A + C} \right){x^3} + \left( {B + D} \right){x^2} + \left( {4A + 3C} \right)x + \left( {4B + 3D} \right)\end{align}\]

Equating the coefficients of \({x^3},{x^2},x\)and constant term, we get

\[\begin{align}A + C = 0\\B + D = 4\\4A + 3C = 0\\4B + 3D = 10\end{align}\]

On solving these equations, we get

\[\begin{align}&A = 0,B = - 2,C = 0\,\,{\rm{and}}\,D = 6\\&\therefore \frac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \frac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \frac{6}{{\left( {{x^2} + 4} \right)}}\\&\frac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \left( {\frac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \frac{6}{{\left( {{x^2} + 4} \right)}}} \right)\\ &\Rightarrow \; \; \; \; \int {\frac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}dx} = \int {1 - \left\{ {\frac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \frac{6}{{\left( {{x^2} + 4} \right)}}} \right\}} dx\\&= \int {\left\{ {1 + \frac{2}{{{x^2} + {{\left( {\sqrt 3 } \right)}^2}}} - \frac{6}{{{x^2} + {2^2}}}} \right\}} dx\\ &= x + 2\left( {\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\frac{x}{{\sqrt 3 }}} \right) - 6\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{x}{2}} \right) + C\\ &= x + \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 3 }} - 3{\tan ^{ - 1}}\frac{x}{2} + C\end{align}\]

Chapter 7 Ex.7.5 Question 19

Integrate the rational function \(\frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}\)

Solution

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

Put,\({x^2} = t \Rightarrow \; \; \; \; 2xdx = dt\)

\[\begin{align}&\therefore \int {\frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}dx} = \int {\frac{{dt}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}\,\,\,\,\, \ldots \left( 1 \right)} \\&{\rm{Let }}\frac{1}{{\left( {t + 1} \right)\left( {t + 3} \right)}} = \frac{A}{{\left( {t + 1} \right)}} + \frac{B}{{\left( {t + 3} \right)}}\\&1 = A\left( {t + 3} \right) + B\left( {t + 1} \right)\,\,\,\, \ldots \left( 2 \right)\end{align}\]

Equating the coefficients of t and constant, we get

\(A + B = 0\,\,{\rm{and}}\,\,3A + B = 1\)

On solving, we get

\[\begin{align}&A = \frac{1}{2}\,\,{\rm{and }}B = - \frac{1}{2}\\&\therefore \frac{1}{{\left( {t + 1} \right)\left( {t + 3} \right)}} = \frac{1}{{2\left( {t + 1} \right)}} + \frac{1}{{2\left( {t + 3} \right)}}\\& \Rightarrow \; \; \; \; \int {\frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\left\{ {\frac{1}{{2\left( {t + 1} \right)}} - \frac{1}{{2\left( {t + 3} \right)}}} \right\}} dt\\ &= \frac{1}{2}\log \left| {\left( {t + 1} \right)} \right| - \frac{1}{2}\log \left| {t + 3} \right| + C\\ &= \frac{1}{2}\log \left| {\frac{{t + 1}}{{t + 3}}} \right| + C = \frac{1}{2}\log \left| {\frac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 20

Integrate the rational function \(\frac{1}{{x\left( {{x^4} - 1} \right)}}\)

Solution

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

Multiplying Nr and Dr by \({x^3}\), we get

\[\begin{align}&\frac{1}{{x\left( {{x^4} - 1} \right)}} = \frac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}\\&\therefore \int {\frac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \int {\frac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}} dx\\&{\rm{Put, }}{x^4} = t \Rightarrow \; \; \; \; 4{x^3} = dt\\&\therefore \int {\frac{1}{{x\left( {{x^4} - 1} \right)}}dx} = \frac{1}{4}\int {\frac{{dt}}{{t\left( {t - 1} \right)}}} \\&{\rm{Let }}\frac{1}{{t\left( {t - 1} \right)}} = \frac{A}{t} + \frac{B}{{\left( {t - 1} \right)}}\\&1 = A\left( {t - 1} \right) + Bt\,\,\,\, \ldots \left( 1 \right)\end{align}\]

Equating the coefficients of t and constant, we get

\[\begin{align}&A = - 1\,\,{\rm{and }}B = 1\\ &\Rightarrow \; \; \; \; \frac{1}{{t\left( {t - 1} \right)}} = \frac{{ - 1}}{t} + \frac{1}{{t - 1}}\\ &\Rightarrow \; \; \; \; \int {\frac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \frac{1}{4}\int {\left\{ {\frac{{ - 1}}{t} + \frac{1}{{t - 1}}} \right\}} dt\\ &= \frac{1}{4}\left[ { - \log \left| t \right| + \log \left| {t - 1} \right|} \right] + C\\ &= \frac{1}{4}\log \left| {\frac{{t - 1}}{t}} \right| + C = \frac{1}{4}\log \left| {\frac{{{x^4} - 1}}{{{x^4}}}} \right| + C\end{align}\]

Chapter 7 Ex.7.5 Question 21

Integrate the rational function \(\frac{1}{{\left( {{e^x} - 1} \right)}}\) [Hint: Put \({e^x} = t\)]

Solution

Put \({e^x} = t \Rightarrow \; \; \; \; {e^x}dx = dt\)

\[\begin{align}& \Rightarrow \; \; \; \; \int {\frac{1}{{\left( {{e^x} - 1} \right)}}dx} = \int {\frac{1}{{t - 1}} \times \frac{{dt}}{t}} = \int {\frac{1}{{t\left( {t - 1} \right)}}} dt\\&{\text{Let }}\frac{1}{{t\left( {t - 1} \right)}} = \frac{A}{t} + \frac{B}{{t - 1}}\\&1 = A\left( {t - 1} \right) + Bt\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{align}\]

Equating the coefficients of t and constant, we get

\(\begin{align} & A=-1\,\,\text{and }B=1 \\ & \therefore \frac{1}{t\left( t-1 \right)}=\frac{-1}{t}+\frac{1}{t-1} \\ & \Rightarrow \; \; \; \; \int{\frac{1}{t\left( t-1 \right)}}dt=\log \left| \frac{t-1}{t} \right|+C \\ & =\log \left| \frac{{{e}^{x}}-1}{{{e}^{x}}} \right|+C \\ \end{align}\)

Chapter 7 Ex.7.5 Question 22

\(\int {\frac{{xdx}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}} \) equals

\(\begin{align}&A.\,\,\log \left| {\frac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 2} \right)}}} \right| + C\\&B.\,\,\log \left| {\frac{{{{\left( {x - 2} \right)}^2}}}{{\left( {x - 1} \right)}}} \right| + C\\&C.\,\,\log \left| {{{\left( {\frac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + C\\&D.\,\,\log \left| {\left( {x - 1} \right)\left( {x - 2} \right)} \right| + C\end{align}\)

Solution

Let \(\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \frac{A}{{\left( {x - 1} \right)}} + \frac{B}{{\left( {x - 2} \right)}}\)

\(x = A\left( {x - 2} \right) + B\left( {x - 1} \right)\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

Equating the coefficients of x and constant, we get

\[\begin{align}&A = - 1\,\,{\text{and }}B = 2\\&\therefore \frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \frac{{ - 1}}{{\left( {x - 1} \right)}} + \frac{2}{{\left( {x - 2} \right)}}\\ &\Rightarrow \; \; \; \; \int {\frac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}}dx} = \int {\left\{ {\frac{{ - 1}}{{\left( {x - 1} \right)}} + \frac{2}{{\left( {x - 2} \right)}}} \right\}} dx\\ &= - \log \left| {x - 1} \right| + 2\log \left| {x - 2} \right| + C\\& = \log \left| {\frac{{{{\left( {x - 2} \right)}^2}}}{{x - 1}}} \right| + C\end{align}\]

Thus, the correct option is B.

Chapter 7 Ex.7.5 Question 23

\(\int {\frac{{dx}}{{x\left( {{x^2} + 1} \right)}}} \)equals

\(\begin{align} &A.\,\,\log \left| x \right| - \frac{1}{2}\log \left( {{x^2} + 1} \right) + C \hfill \\ &B.\,\,\log \left| x \right| + \frac{1}{2}\log \left( {{x^2} + 1} \right) + C \hfill \\& C.\,\, - \log \left| x \right| + \frac{1}{2}\log \left( {{x^2} + 1} \right) + C \hfill \\ &D.\,\,\frac{1}{2}\log \left| x \right| + \log \left( {{x^2} + 1} \right) + C \hfill \\ \end{align}\)

Solution

Let \(\frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}}\)

\(1 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x\)

Equating the coefficients of \({x^2},x\) and constant terms, we get

\[\begin{align}&A + B = 0\\&C = 0\\&A = 1\\&{\text{On solving these equations, we get}}\\&A = 1\,\,B = - 1{\text{ and}}\,\,C = 0\\&\therefore \frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{1}{x} + \frac{{ - x}}{{{x^2} + 1}}\\ &\Rightarrow \; \; \; \; \int {\frac{1}{{x\left( {{x^2} + 1} \right)}}dx = \int {\left\{ {\frac{1}{x} - \frac{x}{{{x^2} - 1}}} \right\}} dx} \\ &= \log \left| x \right| - \frac{1}{2}\log \left| {{x^2} + 1} \right| + C\end{align}\]

Thus, the correct option is A.

  
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