# NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4

## Chapter 7 Ex.7.4 Question 1

Integrate $$\frac{{3{x^2}}}{{{x^6} + 1}}$$

### Solution

Put,$${x^3} = t$$

\begin{align}&\therefore 3{x^2}dx = dt\\& \Rightarrow \; \; \frac{{3{x^2}}}{{{x^6} + 1}}dx = \int {\frac{{dt}}{{{t^2} + 1}}} \\&= {\tan ^{ - 1}}t + C\\&= {\tan ^{ - 1}}\left( {{x^3}} \right) + C\end{align}

## Chapter 7 Ex.7.4 Question 2

Integrate $$\frac{1}{{\sqrt {1 + 4{x^2}} }}$$

### Solution

Put, $$2x = t$$

\begin{align}&\therefore 2dx = dt\\& \Rightarrow \; \; \int {\frac{1}{{\sqrt {1 + 4{x^2}} }}dx = \frac{1}{2}\int {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} } \\&= \frac{1}{2}\left[ {\log \left| {t + \sqrt {{t^2} + 1} } \right|} \right] + C\,\,\,\,\,\,\,\,\,\left[ {\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dt = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right|} } \right]\\& = \frac{1}{2}\log \left| {2x + \sqrt {4{x^2} + 1} } \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 3

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

### Solution

Put, $$2 - x = t$$

\begin{align}& \Rightarrow \; \; - dx = dt\\& \Rightarrow \; \; \int {\frac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx = - \int {\frac{1}{{\sqrt {{t^2} + 1} }}dt} } \\& = - \log \left| {t + \sqrt {{t^2} + 1} } \right| + C\,\,\,\,\,\,\,\,\,\,\,\left[ {\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dt = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right|} } \right]\\ &= - \log \left| {2 - x + \sqrt {{{\left( {2 - x} \right)}^2} + 1} } \right| + C\\& = \log \left| {\frac{1}{{\left( {2 - x} \right) + \sqrt {{x^2} - 4x + 5} }}} \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 4

Integrate $$\frac{1}{{\sqrt {9 - 25{x^2}} }}$$

### Solution

Put, $$5x = t$$

\begin{align}&\therefore 5dx = dt\\ &\Rightarrow \; \; \int {\frac{1}{{\sqrt {9 - 25{x^2}} }}dx = \frac{1}{5}\int {\frac{1}{{\sqrt {9 - {t^2}} }}dt} } \\ &= \frac{1}{5}\int {\frac{1}{{\sqrt {{3^2} - {t^2}} }}dt} \\ &= \frac{1}{5}{\sin ^{ - 1}}\left( {\frac{t}{3}} \right) + C\\ &= \frac{1}{5}{\sin ^{ - 1}}\left( {\frac{{5x}}{3}} \right) + C\end{align}

## Chapter 7 Ex.7.4 Question 5

Integrate $$\frac{{3x}}{{1 + 2{x^4}}}$$

### Solution

Let $$\sqrt 2 {x^2} = t$$

\begin{align}&\therefore 2\sqrt 2 xdx = dt\\& \Rightarrow \; \; \int {\frac{{3x}}{{1 + 2{x^4}}}dx = \frac{3}{{2\sqrt 2 }}\int {\frac{{dt}}{{1 + {t^2}}}} } \\& = \frac{3}{{2\sqrt 2 }}\left[ {{{\tan }^{ - 1}}t} \right] + C\\ &= \frac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt 2 {x^2}} \right) + C\end{align}

## Chapter 7 Ex.7.4 Question 6

Integrate $$\frac{{{x^2}}}{{1 - {x^6}}}$$

### Solution

Put, $${x^3} = t$$

\begin{align}&\therefore 3{x^2}dx = dt\\ &\Rightarrow \; \; \int {\frac{{{x^2}}}{{1 - {x^6}}}dx = \frac{1}{3}\int {\frac{{dt}}{{1 - {t^2}}}} } \\& = \frac{1}{3}\left[ {\frac{1}{2}\log \left| {\frac{{1 + t}}{{1 - t}}} \right|} \right] + C\\& = \frac{1}{6}\log \left| {\frac{{1 + {x^3}}}{{1 - {x^3}}}} \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 7

Integrate $$\frac{{x - 1}}{{\sqrt {{x^2} - 1} }}$$

### Solution

$$\int {\frac{{x - 1}}{{\sqrt {{x^2} - 1} }}} dx = \int {\frac{x}{{\sqrt {{x^2} - 1} }}dx - \int {\frac{1}{{\sqrt {{x^2} - 1} }}dx} } \,\,\,\, \ldots \left( 1 \right)$$

For $$\int {\frac{x}{{\sqrt {{x^2} - 1} }}dx,}$$ let $${x^2} - 1 = t \Rightarrow \; \; 2xdx = dt$$

\begin{align}&\therefore \int {\frac{x}{{\sqrt {{x^2} - 1} }}dx = \frac{1}{2}\int {\frac{{dt}}{{\sqrt t }}} } \\ &= \frac{1}{2}\int {{t^{ - \frac{1}{2}}}dt} \\& = \frac{1}{2}\left[ {2{t^{\frac{1}{2}}}} \right]\\ &= \sqrt t \\ &= \sqrt {{x^2} - 1} \end{align}

From (1), we get

\begin{align}&\int {\frac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx = \int {\frac{x}{{\sqrt {{x^2} - 1} }}dx} - \int {\frac{1}{{\sqrt {{x^2} - 1} }}dx} } \,\,\,\,\,\,\,\,\,\,\,\left[ {\int {\frac{x}{{\sqrt {{x^2} - {a^2}} }}dt = \log \left| {x + \sqrt {{x^2} - {a^2}} } \right|} } \right]\\ &= \sqrt {{x^2} - 1} - \log \left| {x + \sqrt {{x^2} - 1} } \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 8

Integrate $$\frac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}$$

### Solution

Put, $${x^3} = t \Rightarrow \; \; 3{x^2}dx = dt$$

\begin{align}\therefore \int {\frac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}} dx &= \frac{1}{3}\int {\frac{{dt}}{{\sqrt {{t^2} + {{\left( {{a^3}} \right)}^2}} }}} \\ &= \frac{1}{3}\log \left| {t + \sqrt {{t^2} + {a^6}} } \right| + C\\ &= \frac{1}{3}\log \left| {{x^3} + \sqrt {{x^6} + {a^6}} } \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 9

Integrate $$\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}$$

### Solution

Put, $$\tan x = t$$

\begin{align}&\therefore {\sec ^2}xdx = dt\\ &\Rightarrow \; \; \int {\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx = \int {\frac{{dt}}{{\sqrt {{t^2} + {2^2}} }}} } \\& = \log \left| {t + \sqrt {{t^2} + 4} } \right| + C\\ &= \log \left| {\tan x + \sqrt {{{\tan }^2}x + 4} } \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 10

Integrate $$\frac{1}{{\sqrt {{x^2} + 2x + 2} }}$$

### Solution

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

Let $$x + 1 = t$$

\begin{align}&\therefore dx = dt\\ &\Rightarrow \; \; \int {\frac{1}{{\sqrt {{x^2} + 2x + 2} }}dx = \int {\frac{1}{{\sqrt {{t^2} + 1} }}dt} } \\& = \log \left| {t + \sqrt {{t^2} + 1} } \right|C\\& = \log \left| {\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 1} } \right| + C\\& = \log \left| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 2} } \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 11

Integrate $$\frac{1}{{\sqrt {9{x^2} + 6x + 5} }}$$

### Solution

\begin{align}&\int {\frac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx = \int {\frac{1}{{{{\left( {3x + 1} \right)}^2} + {{\left( 2 \right)}^2}}}} dx} \\&{\rm{Let }}\left( {3x + 1} \right) = t\\ &\Rightarrow \; \; 3dx = dt\\ &\Rightarrow \; \; \int {\frac{1}{{{{\left( {3x + 1} \right)}^2} + {{\left( 2 \right)}^2}}}dx = \frac{1}{3}\int {\frac{1}{{{t^2} + {2^2}}}dt} } \\ &= \frac{1}{3}\left[ {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{t}{2}} \right)} \right] + C\\ &= \frac{1}{6}\left[ {{{\tan }^{ - 1}}\left( {\frac{{3x + 1}}{2}} \right)} \right] + C\end{align}

## Chapter 7 Ex.7.4 Question 12

Integrate $$\frac{1}{{\sqrt {7 - 6x - {x^2}} }}$$

### Solution

$$7 - 6x - {x^2}$$ can be written as $$7 - \left( {{x^2} + 6x + 9 - 9} \right)$$

Thus,

\begin{align}&7 - \left( {{x^2} + 6x + 9 - 9} \right)\\ &= 16 - \left( {{x^2} + 6x + 9} \right)\\& = 16 - {\left( {x + 3} \right)^2}\\& = {\left( 4 \right)^2} - {\left( {x + 3} \right)^2}\\&\therefore \int {\frac{1}{{\sqrt {7 - 6x - {x^2}} }}dx = \int {\frac{1}{{\sqrt {{{\left( 4 \right)}^2} - {{\left( {x + 3} \right)}^2}} }}dx} } \end{align}

\begin{align}&{\rm{Let }}x + 3 = t\\&\Rightarrow \; \; dx = dt\\&\Rightarrow \; \; \int {\frac{1}{{\sqrt {{{\left( 4 \right)}^2} - {{\left( {x + 3} \right)}^2}} }}dx = \int {\frac{1}{{\sqrt {{{\left( 4 \right)}^2} - {{\left( t \right)}^2}} }}dt} } \\ &= {\sin ^{ - 1}}\left( {\frac{t}{4}} \right) + C\\&= {\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + C\end{align}

## Chapter 7 Ex.7.4 Question 13

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

### Solution

$$\left( {x - 1} \right)\left( {x - 2} \right)$$can be written as $${x^2} - 3x + 2$$

Thus,

\begin{align}&{x^2} - 3x + 2\\ &= {x^2} - 3x + \frac{9}{4} - \frac{9}{4} + 2\\& = {\left( {x - \frac{3}{2}} \right)^2} - \frac{1}{4}\\ &= {\left( {x - \frac{3}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2}\\&\therefore \int {\frac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx = \int {\frac{1}{{\sqrt {{{\left( {x - \frac{3}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} }}dx} } \end{align}

Let $$\left( {x - \frac{3}{2}} \right) = t$$

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

## Chapter 7 Ex.7.4 Question 14

Integrate $$\frac{1}{{\sqrt {8 + 3x - {x^2}} }}$$

### Solution

$$8 + 3x - {x^2} = 8 - \left( {{x^2} - 3x + \frac{9}{4} - \frac{9}{4}} \right)$$

Thus,

\begin{align}&8 - \left( {{x^2} - 3x + \frac{9}{4} - \frac{9}{4}} \right)\\& = \frac{{41}}{4} - {\left( {x - \frac{3}{2}} \right)^2}\\& = \int {\frac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = \int {\frac{1}{{\sqrt {\frac{{41}}{4} - {{\left( {x - \frac{3}{2}} \right)}^2}} }}dx} \\&{\rm{Let }}\left( {x - \frac{3}{2}} \right) = t\\&\therefore dx = dt\\& \Rightarrow \; \; \int {\frac{1}{{\sqrt {\frac{{41}}{4} - {{\left( {x - \frac{3}{2}} \right)}^2}} }}dx} = \int {\frac{1}{{\sqrt {\left( {\frac{{41}}{4}} \right) - {t^2}} }}dt} \\ &= {\sin ^{ - 1}}\left( {\frac{t}{{\frac{{\sqrt {41} }}{2}}}} \right) + C\\ &= {\sin ^{ - 1}}\left( {\frac{{x - \frac{3}{2}}}{{\frac{{\sqrt {41} }}{2}}}} \right) + C\\ &= {\sin ^{ - 1}}\left( {\frac{{2x - 3}}{{\sqrt {41} }}} \right) + C\end{align}

## Chapter 7 Ex.7.4 Question 15

Integrate $$\frac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}$$

### Solution

$$\left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + ab$$

Thus,

\begin{align}&{x^2} - \left( {a + b} \right)x + ab\\ &= {x^2} - \left( {a + b} \right)x + \frac{{{{\left( {a + b} \right)}^2}}}{4} - \frac{{{{\left( {a + b} \right)}^2}}}{4} + ab\\ &= {\left[ {x - \left( {\frac{{a + b}}{2}} \right)} \right]^2} - \frac{{{{\left( {a - b} \right)}^2}}}{4}\\& \Rightarrow \; \; \int {\frac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx = \int {\frac{1}{{\sqrt {{{\left\{ {x - \left( {\frac{{a + b}}{2}} \right)} \right\}}^2} - {{\left( {\frac{{a + b}}{2}} \right)}^2}} }}dx} } \end{align}

Let $$x - \left( {\frac{{a + b}}{2}} \right) = t$$

\begin{align}&\therefore dx = dt\\& \Rightarrow \; \; \int {\frac{1}{{\sqrt {{{\left\{ {x - \left( {\frac{{a + b}}{2}} \right)} \right\}}^2} - {{\left( {\frac{{a + b}}{2}} \right)}^2}} }}dx = \int {\frac{1}{{\sqrt {{t^2} - {{\left( {\frac{{a + b}}{2}} \right)}^2}} }}dt} } \\& = \log \left| {t + \sqrt {{t^2} - {{\left( {\frac{{a + b}}{2}} \right)}^2}} } \right| + C\\ &= \log \left| {\left\{ {x - \left( {\frac{{a + b}}{2}} \right)} \right\} + \sqrt {\left( {x - a} \right)\left( {x - b} \right)} } \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 16

Integrate $$\frac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}$$

### Solution

\begin{align}{\rm{Let,}}\,4x + 1 &= {\rm{A}}\frac{d}{{dx}}\left( {2{x^2} + x - 3} \right) + {\rm{B}}\\& \Rightarrow \; \; {\rm{4}}x + 1 = {\rm{A}}\left( {4x + 1} \right) + {\rm{B}}\\ &\Rightarrow \; \; {\rm{4}}x + 1 = 4{\rm{A}}x + {\rm{A + B}}\end{align}

Equating the coefficients of x and constant term on both sides, we get

\begin{align}&4{\rm{A}} = {\rm{4}} \Rightarrow \; \; {\rm{A}} = {\rm{1}}\\&{\rm{A + B}} = {\rm{1}} \Rightarrow \; \; {\rm{B}} = {\rm{0}}\\&{\rm{Let}}\,{\rm{ 2}}{x^2} + x - 3 = t\\&\therefore \left( {4x + 1} \right)dx = dt\\& \Rightarrow \; \; \int {\frac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} = \int {\frac{1}{{\sqrt t }}dt} \\ &= 2\sqrt t + C\\ &= 2\sqrt {2{x^2} + x - 3} + C\end{align}

## Chapter 7 Ex.7.4 Question 17

Integrate $$\frac{{x + 2}}{{\sqrt {{x^2} - 1} }}$$

### Solution

$${\rm{Put,}}\;x + 2 = {\rm{A}}\frac{d}{{dx}}\left( {{x^2} - 1} \right) + {\rm{B}}\,\,\, \ldots \left( 1 \right)$$

$$\Rightarrow \; \; x + 2 = {\rm{A}}\left( {2x} \right) + {\rm{B}}$$

Equating the coefficients of x and constant term on both sides, we get

\begin{align}2{\rm{A}} &= 1 \Rightarrow \; \; {\rm{A}} = \frac{1}{2}\\{\rm{B}} &= 2\end{align}

From ($$1$$) we get

\begin{align}& \Rightarrow \; \; \int {\frac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \int {\frac{{\frac{1}{2}\left( {2x} \right) + 2}}{{\sqrt {{x^2} - 1} }}} dx\\ &= \frac{1}{2}\int {\frac{{2x}}{{\sqrt {{x^2} - 1} }}dx} + \int {\frac{2}{{\sqrt {{x^2} - 1} }}dx} \,\,\,\,\,\,\, \ldots \left( 2 \right)\\&{\text{ In }}\frac{1}{2}\int {\frac{{2x}}{{\sqrt {{x^2} - 1} }}dx} {\text{, Let }}{x^2} - 1 = t \Rightarrow \; \; 2xdx = dt\\&\frac{1}{2}\int {\frac{{2x}}{{\sqrt {{x^2} - 1} }}dx = \frac{1}{2}\int {\frac{{dt}}{{\sqrt t }}} } \\ &= \frac{1}{2}\left[ {2\sqrt t } \right]\\& = \frac{1}{2}\left[ {2\sqrt {{x^2} - 1} } \right]\\ &= \sqrt {{x^2} - 1} \end{align}

Then, $$\int {\frac{2}{{\sqrt {{x^2} - 1} }}dx = 2\int {\frac{1}{{\sqrt {{x^2} - 1} }}dx = 2\log \left| {x + \sqrt {{x^2} - 1} } \right|} }$$

From equation ($$2$$) we get

$$\int {\frac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \sqrt {{x^2} - 1} + 2\log \left| {x + \sqrt {{x^2} - 1} } \right| + C$$

## Chapter 7 Ex.7.4 Question 18

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

### Solution

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

$$\Rightarrow \; \; 5x - 2 = A\left( {2 + 6x} \right) + B$$

Equating the coefficients of x and constant term on both sides, we get

\begin{align}&5 = 6A \Rightarrow \; \; A = \frac{5}{6}\\&2A + B = - 2 \Rightarrow \; \; B = - \frac{{11}}{3}\\&\therefore 5x - 2 = \frac{5}{6}\left( {2 + 6x} \right) + \left( { - \frac{{11}}{3}} \right)\\& \Rightarrow \; \; \int {\frac{{5x - 2}}{{1 + 2x + 3{x^2}}}dx} = \int {\frac{{\frac{5}{6}\left( {2 + 6x} \right) - \frac{{11}}{3}}}{{1 + 2x + 3{x^2}}}dx} \\ &\Rightarrow \; \; \frac{5}{6}\int {\frac{{2 + 6x}}{{1 + 2x + 3{x^2}}}dx - \frac{{11}}{3}\int {\frac{1}{{1 + 2x + 3{x^2}}}dx} } \end{align}

Let

\begin{align}{I_1} &= \int {\frac{{2 + 6x}}{{1 + 2x + 3{x^2}}}dx} \,\,{\text{and }}{I_2} = \int {\frac{1}{{1 + 2x + 3{x^2}}}dx} \\&\therefore \int {\frac{{5x - 2}}{{1 + 2x + 3{x^2}}}dx = \frac{5}{6}{I_1} - \frac{{11}}{3}{I_2}\,\,\,\,\,\,\,\, \ldots \left( 1 \right)} \\&{I_1} = \int {\frac{{2 + 6x}}{{1 + 2x + 3{x^2}}}dx} \\&{\text{Put }}1 + 2x + 3{x^2} = t\\& \Rightarrow \; \; \left( {2 + 6x} \right)dx = dt\\&\therefore {I_1} = \int {\frac{{dt}}{t}} \\&{I_1} = \log \left| t \right|\\&{I_1} = \log \left| {1 + 2x + 3{x^2}} \right|\,\,\,\,\,\, \ldots \left( 2 \right)\\&{I_2} = \int {\frac{1}{{1 + 2x + 3{x^2}}}dx} \end{align}

$$1 + 2x + 3{x^2}$$ can be written as $$1 + 3\left( {{x^2} + \frac{2}{3}x} \right)$$

Thus,

\begin{align}&1 + 3\left( {{x^2} + \frac{2}{3}x} \right)\\ &= 1 + 3\left( {{x^2} + \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}} \right)\\& = 1 + 3{\left( {x + \frac{1}{3}} \right)^2} - \frac{1}{3}\\& = \frac{2}{3} + 3{\left( {x + \frac{1}{3}} \right)^2}\\& = 3\left[ {{{\left( {x + \frac{1}{3}} \right)}^2} + \frac{2}{9}} \right]\\& = 3\left[ {{{\left( {x + \frac{1}{3}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{3}} \right)}^2}} \right]\\&{I_2} = \frac{1}{3}\int {\frac{1}{{\left[ {{{\left( {x + \frac{1}{3}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{3}} \right)}^2}} \right]}}} dx\\ &= \frac{1}{3}\left[ {\frac{1}{{\frac{{\sqrt 2 }}{3}}}{{\tan }^{ - 1}}\left( {\frac{{x + \frac{1}{3}}}{{\frac{{\sqrt 2 }}{3}}}} \right)} \right]\\ &= \frac{1}{3}\left[ {\frac{3}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{{3x + 1}}{{\sqrt 2 }}} \right)} \right]\\ &= \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{3x + 1}}{{\sqrt 2 }}} \right)\,\,\,\,\,\,\, \ldots \left( 3 \right)\end{align}

Substituting equations (2) and (3) in equation (1), we get

\begin{align}\int {\frac{{5x - 2}}{{1 + 2x + 3{x^2}}}dx} &= {\frac{5}{6}\left[ {\log \left| {1 + 2x + 3{x^2}} \right|} \right]} - \frac{{11}}{3}\left[ {\frac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{{3x + 1}}{{\sqrt 2 }}} \right)} \right] + C\\ &= \frac{5}{6}\log \left| {1 + 2x + 3{x^2}} \right| - \frac{{11}}{{3\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{3x + 1}}{{\sqrt 2 }}} \right) + C\end{align}

## Chapter 7 Ex.7.4 Question 19

Integrate $$\frac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}$$

### Solution

$$\frac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }} = \frac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }}$$

Put, $$6x + 7 = A\frac{d}{{dx}}\left( {{x^2} - 9x + 20} \right) + B$$

$$\Rightarrow \; \; 6x + 7 = A\left( {2x - 9} \right) + B$$

Equating the coefficients of x and constant term, we get

\begin{align}&2A = 6 \Rightarrow \; \; A = 3\\ &- 9A + B = 7 \Rightarrow \; \; B = 34\\&\therefore 6x + 7 = 3\left( {2x - 9} \right) + 34\\&\int {\frac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }} = \int {\frac{{3\left( {2x - 9} \right) + 34}}{{\sqrt {{x^2} - 9x + 20} }}dx} } \\& = 3\int {\frac{{\left( {2x - 9} \right)}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 34\int {\frac{1}{{\sqrt {{x^2} - 9x + 20} }}dx} \end{align}

Let $${I_1} = \int {\frac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx\,\,{\rm{and }}{I_2} = \int {\frac{1}{{\sqrt {{x^2} - 9x + 20} }}dx} }$$

$$\therefore \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}}=3 I_{1}+34 I_{2} \qquad \dots (1)$$

Then,

\begin{align}&{I_1} = \int {\frac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}} dx\\&{\rm{Let }}{x^2} - 9x + 20 = t\\ &\Rightarrow \; \; \left( {2x - 9} \right)dx = dt\\& \Rightarrow \; \; {I_1} = \frac{{dt}}{{\sqrt t }}\\&{I_1} = 2\sqrt t \\&{I_1} = 2\sqrt {{x^2} - 9x + 20} \,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}

and

\begin{align}{I_2} &= \int {\frac{1}{{\sqrt {{x^2} - 9x + 20} }}dx} \\{x^2} - 9x + 20 &= {x^2} - 9x + 20 + \frac{{81}}{4} - \frac{{81}}{4}\end{align}

Thus,

\begin{align}&{x^2} - 9x + 20 + \frac{{81}}{4} - \frac{{81}}{4}\\ &= {\left( {x - \frac{9}{2}} \right)^2} - \frac{1}{4}\\& = {\left( {x - \frac{9}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2}\\& \Rightarrow \; \; {I_2} = \int {\frac{1}{{{{\left( {x - \frac{9}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}}}dx} \\{I_2}& = \log \left| {\left( {x - \frac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} } \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)\end{align}

Substituting equations (2) and (3) in (1), we get

\begin{align}&\int {\frac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }}dx = 3\left[ {2\sqrt {{x^2} - 9x + 20} } \right]} + 34\log \left[ {\left( {x - \frac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} } \right] + C\\ &= 6\sqrt {{x^2} - 9x + 20} + 34\log \left[ {\left( {x - \frac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} } \right] + C\end{align}

## Chapter 7 Ex.7.4 Question 20

Integrate $$\frac{{x + 2}}{{\sqrt {4x - {x^2}} }}$$

### Solution

Consider, $$x + 2 = A\frac{d}{{dx}}\left( {4x - {x^2}} \right) + B$$

$$\Rightarrow \; \; x + 2 = A\left( {4 - 2x} \right) + B$$

Equating the coefficients of x and constant term on both sides, we get

\begin{align} &- 2A = 1 \Rightarrow \; \; A = - \frac{1}{2}\\&4A + B = 2 \Rightarrow \; \; B = 4\\ &\Rightarrow \; \; \left( {x + 2} \right) = - \frac{1}{2}\left( {4 - 2x} \right) + 4\\&\therefore \int {\frac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = \int {\frac{{ - \frac{1}{2}\left( {4 - 2x} \right) + 4}}{{\sqrt {\left( {4x - {x^2}} \right)} }}} dx\\& = - \frac{1}{2}\int {\frac{{\left( {4 - 2x} \right)}}{{\sqrt {\left( {4x - {x^2}} \right)} }}} dx + 4\int {\frac{1}{{\sqrt {\left( {4x - {x^2}} \right)} }}} dx\end{align}

Let $${I_1} = \int {\frac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}} dx\,\,{\rm{and }}{I_2} = \int {\frac{1}{{\sqrt {4x - {x^2}} }}dx}$$

$$\therefore \int {\frac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \frac{1}{2}{I_1} + {\rm{4}}{I_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$$

Then,

\begin{align}&{I_1} = \int {\frac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx} \\&{\text{Let }}4x - {x^2} = t\\& \Rightarrow \; \; \left( {4 - 2x} \right)dx = dt\\ &\Rightarrow \; \; {I_1} = \int {\frac{{dt}}{{\sqrt t }} = 2\sqrt t = 2\sqrt {4x - {x^2}} } \,\,\,\,\,\,\, \ldots \left( 2 \right)\\&{I_2} = \int {\frac{1}{{\sqrt {4x - {x^2}} }}dx} \\& \Rightarrow \; \; 4x - {x^2} = - \left( { - 4x + {x^2}} \right)\\ &= \left( { - 4x + {x^2} + 4 - 4} \right)\\ &= 4 - {\left( {x - 2} \right)^2}\\& = {\left( 2 \right)^2} - {\left( {x - 2} \right)^2}\\&\therefore {I_2} = \int {\frac{1}{{\sqrt {{{\left( 2 \right)}^2} - {{\left( {x - 2} \right)}^2}} }}dx} = {\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right)\,\,\,\,\, \ldots \left( 3 \right)\end{align}

Using equations (2) and (3) in (1), we get

\begin{align}\int {\frac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} &= - \frac{1}{2}\left( {2\sqrt {4x - {x^2}} } \right) + 4{\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) + C\\&= - \sqrt {4x - {x^2}} + 4{\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) + C\end{align}

## Chapter 7 Ex.7.4 Question 21

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

### Solution

\begin{align}\int {\frac{{x + 2}}{{\sqrt {{x^2} + 2x} + 3}}dx} &= \frac{1}{2}\int {\frac{{2\left( {x + 2} \right)}}{{\sqrt {{x^2} + 2x + 3} }}} dx\\& = \frac{1}{2}\int {\frac{{2x + 4}}{{\sqrt {{x^2} + 2x + 3} }}dx} \\ &= \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} + \frac{1}{2}\int {\frac{2}{{\sqrt {{x^2} + 2x + 3} }}dx} \\ &= \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} + \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} \end{align}

Let $${I_1} = \int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} \,\,\,{\rm{and }}{I_2} = \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}} dx$$

\begin{align}&\therefore \int {\frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \frac{1}{2}{I_1} + {I_2}\,\,\,\,\,\,\, \ldots \left( 1 \right)\\&{\text{Then, }}{I_1} = \int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} \end{align}

Put, $${x^2} + 2x + 3 = t$$

\begin{align}& \Rightarrow \; \; \left( {2x + 2} \right)dx = dt\\{I_1} &= \int {\frac{{dt}}{{\sqrt t }} = 2\sqrt t } = 2\sqrt {{x^2} + 2x + 3} \,\,\,\,\,\, \ldots \left( 2 \right)\\{I_2} &= \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} \\ &\Rightarrow \; \; {x^2} + 2x + 3 = {x^2} + 2x + 1 + 2 = {\left( {x + 1} \right)^2} + {\left( {\sqrt 2 } \right)^2}\\&\therefore {I_2} = \int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} }}dx} = \log \left| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} } \right|\,\,\,\,\,\, \ldots \left( 3 \right)\end{align}

Using equations (2) and (3) in (1), we get

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

## Chapter 7 Ex.7.4 Question 22

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

### Solution

Let $$\left( {x + 3} \right) = A\frac{d}{{dx}}\left( {{x^2} - 2x - 5} \right) + B$$

$$\left( {x + 3} \right) = A\left( {2x - 2} \right) + B$$

Equating the coefficients of x and constant term on both sides, we get

\begin{align}&2A = 1 \Rightarrow \; \; A = \frac{1}{2}\\& - 2A + B = 3 \Rightarrow \; \; B = 4\\&\therefore \left( {x + 3} \right) = \frac{1}{2}\left( {2x - 2} \right) + 4\\&\Rightarrow \; \; \int {\frac{{x + 3}}{{{x^2} - 2x - 5}}} dx = \int {\frac{{\frac{1}{2}\left( {2x - 2} \right) + 4}}{{{x^2} - 2x - 5}}} dx\\ &= \frac{1}{2}\int {\frac{{2x - 2}}{{{x^2} - 2x - 5}}dx} + 4\int {\frac{1}{{{x^2} - 2x - 5}}} dx\end{align}

Let $${I_1} = \int {\frac{{2x - 2}}{{{x^2} - 2x - 5}}dx\,\,{\rm{and }}{I_2} = \int {\frac{1}{{{x^2} - 2x - 5}}} dx}$$

$$\therefore \int {\frac{{x + 3}}{{{x^2} - 2x - 5}}} dx = \frac{1}{2}{I_1} + 4{I_2}\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$$

Then, $${I_1} = \int {\frac{{2x - 2}}{{{x^2} - 2x - 5}}} dx$$

Put, $${x^2} - 2x - 5 = t$$

\begin{align}& \Rightarrow \; \; \left( {2x - 2} \right)dx = dt\\& \Rightarrow \; \; {I_1} = \int {\frac{{dt}}{t}} = \log \left| t \right| = \log \left| {{x^2} - 2x - 5} \right|\,\,\,\,\,\,\, \ldots \left( 2 \right)\\{I_2} &= \int {\frac{1}{{{x^2} - 2x - 5}}} dx\\ &= \int {\frac{1}{{\left( {{x^2} - 2x + 1} \right) - 6}}} dx\\ &= \int {\frac{1}{{{{\left( {x - 1} \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}dx} \\& = \frac{1}{{2\sqrt 6 }}\log \left( {\frac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)\end{align}

Substituting (2) and (3) in (1), we get

\begin{align}\int {\frac{{x + 3}}{{{x^2} - 2x - 5}}dx} &= \frac{1}{2}\log \left| {{x^2} - 2x - 5} \right| + \frac{4}{{2\sqrt 6 }}\log \left| {\frac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right| + C\\ &= \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| + C\end{align}

## Chapter 7 Ex.7.4 Question 23

Integrate $$\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}$$

### Solution

Let $$5x + 3 = A\frac{d}{{dx}}\left( {{x^2} + 4x + 10} \right) + B$$

$$\Rightarrow \; \; 5x + 3 = A\left( {2x + 4} \right) + B$$

Equating the coefficients of x and constant term, we get

\begin{align}&2A = 5 \Rightarrow \; \; A = \frac{5}{2}\\&4A + B = 3 \Rightarrow \; \; B = - 7\\&\therefore 5x + 3 = \frac{5}{2}\left( {2x + 4} \right) - 7\\ &\Rightarrow \; \; \int {\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}} dx = \int {\frac{{\frac{5}{2}\left( {2x + 4} \right) - 7}}{{\sqrt {{x^2} + 4x + 10} }}dx} \\& = \frac{5}{2}\int {\frac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} - 7\int {\frac{1}{{\sqrt {{x^2} + 4x + 10} }}dx} \end{align}

Let $${I_1} = \int {\frac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} \,\,{\rm{and }}{I_2} = \int {\frac{1}{{\sqrt {{x^2} + 4x + 10} }}} dx$$

$$\therefore \int {\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \frac{5}{2}{I_1} - 7{I_2}\,\,\,\,\,\,\, \ldots \left( 1 \right)$$

Then,

$${I_1} = \int {\frac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx}$$

Put, $${x^2} + 4x + 10 = t$$

\begin{align}&\therefore \left( {2x + 4} \right)dx = dt\\ &\Rightarrow \; \; {I_1} = \int {\frac{{dt}}{t} = 2\sqrt t } = 2\sqrt {{x^2} + 4x + 10} \,\,\,\,\,\, \ldots \left( 2 \right)\\{I_2}& = \int {\frac{1}{{\sqrt {{x^2} + 4x + 10} }}dx} \\ &= \int {\frac{1}{{\sqrt {\left( {{x^2} + 4x + 4} \right) + 6} }}} dx\\ &= \int {\frac{1}{{\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} }}dx} \\ &= \log \left| {\left( {x + 2} \right)\sqrt {{x^2} + 4x + 10} } \right|\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)\end{align}

Using equations (2) and (3) in (1), we get

\begin{align}\int {\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} &= \frac{5}{2}\left[ {2\sqrt {{x^2} + 4x + 10} } \right] - 7\log \left| {\left( {x + 2} \right)\sqrt {{x^2} + 4x + 10} } \right| + C\\& = 5\sqrt {{x^2} + 4x + 10} - 7\log \left| {\left( {x + 2} \right)\sqrt {{x^2} + 4x + 10} } \right| + C\end{align}

## Chapter 7 Ex.7.4 Question 24

$$\int {\frac{{dx}}{{{x^2} + 2x + 2}}}$$ equals

\begin{align}&\left( A \right)\,\,\,x{\tan ^{ - 1}}\left( {x + 1} \right) + C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( B \right)\,\,\,{\tan ^{ - 1}}\left( {x + 1} \right) + C\\&\left( C \right)\,\,\left( {x + 1} \right)\,{\tan ^{ - 1}}x + C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( D \right)\,\,\,{\tan ^{ - 1}}x + C\end{align}

### Solution

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

Hence, the correct option is B.

## Chapter 7 Ex.7.4 Question 25

$$\int {\frac{{dx}}{{\sqrt {9x - 4{x^2}} }}}$$ equals

\begin{align}&\left( A \right)\,\,\frac{1}{9}{\sin ^{ - 1}}\left( {\frac{{9x - 8}}{8}} \right)\, + C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( B \right)\,\,\,\frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{8x - 9}}{8}} \right)\, + C\\&\left( C \right)\,\,\frac{1}{3}{\sin ^{ - 1}}\left( {\frac{{9x - 8}}{8}} \right)\, + C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( D \right)\,\,\,\frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{9x - 8}}{9}} \right)\, + C\end{align}

### Solution

\begin{align}\int {\frac{{dx}}{{\sqrt {9x - 4{x^2}} }}} \\& = \int {\frac{1}{{\sqrt { - 4\left( {{x^2} - \frac{9}{4}x} \right)} }}} dx\\ &= \int {\frac{1}{{\sqrt { - 4\left( {{x^2} - \frac{9}{4}x + \frac{{81}}{{64}} - \frac{{81}}{{64}}} \right)} }}} dx\\ &= \int {\frac{1}{{\sqrt { - 4\left[ {{{\left( {x - \frac{9}{8}} \right)}^2} - {{\left( {\frac{9}{8}} \right)}^2}} \right]} }}} dx\\&= \frac{1}{2}\int {\frac{1}{{{{\left( {\frac{9}{8}} \right)}^2} - {{\left( {x - \frac{9}{8}} \right)}^2}}}} dx\\&= \frac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {\frac{{x - \frac{9}{8}}}{{\frac{9}{8}}}} \right)} \right] + C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {\frac{{dy}}{{\sqrt {{a^2} - {y^2}} }} = {{\sin }^{ - 1}}\frac{y}{a} + C} } \right)\\&= \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{8x - 9}}{9}} \right) + C\end{align}

Hence, the correct option is B.

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