CBSE Class 12 Mathematics Chapter 7 Integrals – NCERT Solutions Exercise 7.4 [2025-26]

Exercise 7.4

1. Find the integration of\[\dfrac{{3{x^2}}}{{{x^6} + 1}}\].

Ans: Let\[{x^3} = t\],

Now, differentiate both sides,

\[3{x^2}dx = dt\]

$\displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx}  = \displaystyle \int {\dfrac{{3{x^2}}}{{{{\left( {{x^3}} \right)}^2} + 1}}dx}  $

$ \displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx}  = \displaystyle \int {\dfrac{1}{{{t^2} + 1}}dt}  $

$ \displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx}  = {\tan ^{ - 1}}t + C $

$ \displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx}  = {\tan ^{ - 1}}\left( {{x^3}} \right) + C $ 

Where C is an arbitrary constant.

2. Find the integration of \[\dfrac{1}{{\sqrt {1 + 4{x^2}} }}\].

Ans: Let \[2x = t\],

Now, differentiate both sides,

$2dx = dt $

$dx = \dfrac{{dt}}{2} $ 

$ \displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {1 + {{\left( {2x} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {1 + {t^2}} }}\left( {\dfrac{{dt}}{2}} \right)}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt {1 + {t^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx}  = \dfrac{1}{2}\left( {\log |t + \sqrt {1 + {t^2}} |} \right) + C{\text{    }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx = \log |x + \sqrt {{x^2} + {a^2}} |} } \right] $

$ \displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx}  = \dfrac{1}{2}\log |2x + \sqrt {4{x^2} + 1} | + C $ 

Where C is an arbitrary constant.

3. Find the integration of \[\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}\].

Ans: Let \[2 - x = t\],

Now, differentiate both sides,

$ 0 - dx = dt $

$ dx =  - dt $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + 1} }}\left( { - dt} \right)}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx}  =  - \displaystyle \int {\dfrac{1}{{\sqrt {1 + {t^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx}  =  - \left( {\log |t + \sqrt {1 + {t^2}} |} \right) + C{\text{    }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx = \log |x + \sqrt {{x^2} + {a^2}} |} } \right] $

$ \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx}  =  - \log |\left( {2 - x} \right) + \sqrt {{{\left( {2 - x} \right)}^2} + 1} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx}  = \log \left( {\dfrac{1}{{\left( {2 - x} \right) + \sqrt {{x^2} - 4x + 5} }}} \right) + C $ 

Where C is an arbitrary constant.

4. Find the integration of \[\dfrac{1}{{\sqrt {9 - 25{x^2}} }}\].

Ans: Let \[5x = t\],

Now, differentiate both sides,

$5dx = dt $

$  dx = \dfrac{{dt}}{5} $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( {5x} \right)}^2}} }}} dx $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( t \right)}^2}} }}} \left( {\dfrac{{dt}}{5}} \right) $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \dfrac{1}{5}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( t \right)}^2}} }}}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \dfrac{1}{5}{\sin ^{ - 1}}\left( {\dfrac{t}{3}} \right) + C{\text{    }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}dx}  = {{\sin }^{ - 1}}\dfrac{x}{a}} \right] $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \dfrac{1}{5}{\sin ^{ - 1}}\left( {\dfrac{{5x}}{3}} \right) + C $ 

Where C is an arbitrary constant.

5. Find the integration of \[\dfrac{{3x}}{{1 + 2{x^4}}}\].

Ans: Let \[\sqrt 2 {x^2} = t\],

Now, differentiate both sides,

$2\sqrt 2 xdx = dt $

$xdx = \dfrac{{dt}}{{2\sqrt 2 }} $ 

Now,

$ \displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx}  = \displaystyle \int {\dfrac{{3x}}{{1 + {{\left( {\sqrt 2 {x^2}} \right)}^2}}}dx}  $

$ \displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx}  = \displaystyle \int {\dfrac{3}{{1 + {t^2}}}\left( {\dfrac{{dt}}{{2\sqrt 2 }}} \right)}  $

$ \displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx}  = \dfrac{3}{{2\sqrt 2 }}\displaystyle \int {\dfrac{1}{{1 + {t^2}}}dt}  $

$ \displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx}  = \dfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}t + C{\text{    }}\left[ {\displaystyle \int {\dfrac{1}{{1 + {x^2}}}dx}  = {{\tan }^{ - 1}}x + C} \right] $

$ \displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx}  = \dfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt 2 {x^2}} \right) + C $ 

Where C is an arbitrary constant.

6. Find the integration of \[\dfrac{{{x^2}}}{{1 - {x^6}}}\].

Ans: Let \[{x^3} = t\],

Now, differentiate both sides,

$3{x^2}dx = dt $

$ {x^2}dx = \dfrac{{dt}}{3} $ 

Now,

$ \displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx = } \displaystyle \int {\dfrac{{{x^2}}}{{1 - {{\left( {{x^3}} \right)}^2}}}dx}  $

$ \displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx}  = \displaystyle \int {\dfrac{1}{{1 - {t^2}}}\left( {\dfrac{{dt}}{3}} \right)}  $

$ \displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx}  = \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{1 - {t^2}}}dt}  $

$ \displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx}  = \dfrac{1}{3}\left( {\dfrac{1}{2}\log |\dfrac{{1 + t}}{{1 - t}}|} \right) + C{\text{    }}\left[ {\displaystyle \int {\dfrac{1}{{{a^2} - {x^2}}}dx}  = \dfrac{1}{{2a}}\log |\dfrac{{a + x}}{{a - x}}| + C} \right] $

$ \displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx}  = \dfrac{1}{3}\left( {\dfrac{1}{2}\log |\dfrac{{1 + {x^3}}}{{1 - {x^3}}}|} \right) + C $ 

Where C is an arbitrary constant.

7. Find the integration of \[\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}\].

Ans: \[  \displaystyle \int {\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx}  = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  - \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{    }}\left[ {eq.1} \right]\]

For \[\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} \], 

Let \[{x^2} - 1 = t\], 

Now differentiate both sides,

$2xdx = dt $

$xdx = \dfrac{{dt}}{2} $ 

Now,

$ \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt t }}\left( {\dfrac{{dt}}{2}} \right)}  $

$ \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\displaystyle \int {{t^{ - \dfrac{1}{2}}}dt}  $

$ \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\left( {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}} \right) + C $

$ \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\left( {2\sqrt t } \right) + C $

$ \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  = \sqrt t  + C $

$ \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  = \sqrt {{x^2} - 1}  + C $ 

Now, from equation \[1\],

$ \displaystyle \int {\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx}  = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  - \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{   }} $

$ \displaystyle \int {\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx}  = \sqrt {{x^2} - 1}  - \log |x + \sqrt {{x^2} - 1} | + C{\text{    }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx}  = \log |x + \sqrt {{x^2} - {a^2}} |} \right] $ 

Where C is an arbitrary constant.

8. Find the integration of \[\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}\].

Ans: Let \[{x^3} = t\],

Now, differentiate both sides,

$3{x^2}dx = dt $

$ {x^2}dx = \dfrac{{dt}}{3} $ 

Now,

$ \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx}  = \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{{\left( {{x^3}} \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}\left( {\dfrac{{dt}}{3}} \right)}  $

$ \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx}  = \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}dt}  $

$ \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx}  = \dfrac{1}{3}\log |t + \sqrt {{t^2} + {{\left( {{a^3}} \right)}^2}} | + C{\text{   }}\left[ {\displaystyle \int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = } \log |x + \sqrt {{x^2} + {a^2}} |} \right] $

$ \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx}  = \dfrac{1}{3}\log |{x^3} + \sqrt {{x^6} + {a^6}} | + C $ 

Where C is an arbitrary constant.

9. Find the integration of \[\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}\].

Ans: Let \[\tan x = t\],

Now, differentiate both sides,

\[{\sec ^2}xdx = dt\]

Now,

$ \displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\left( {\tan x} \right)}^2} + {{\left( 2 \right)}^2}} }}} dx $

$ \displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \displaystyle \int {\dfrac{{dt}}{{\sqrt {{t^2} + {{\left( 2 \right)}^2}} }}}  $

$ \displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \log |t + \sqrt {{t^2} + {{\left( 2 \right)}^2}} | + C{\text{   }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx}  = \log |x + \sqrt {{x^2} + {a^2}} |} \right] $

$ \displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \log |\tan x + \sqrt {{{\tan }^2}x + 4} | + C $ 

Where C is an arbitrary constant.

10. Find the integration of \[\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}\].

Ans: Let \[x + 1 = t\],

Now, differentiate both sides,

\[dx = dt\]

Now,

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

$ \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {{\left( 1 \right)}^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \log |t + \sqrt {{t^2} + 1} | + C{\text{   }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx}  = \log |x + \sqrt {{x^2} + {a^2}} |} \right] $

$ \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \log |\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 1} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \log |x + 1 + \sqrt {{x^2} + 2x + 2} | + C $ 

Where C is an arbitrary constant.

11. Find the integration of \[\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}\].

Ans: Let \[3x + 1 = t\],

Now, differentiate both sides,

$3dx = dt $

$ dx = \dfrac{{dt}}{3} $

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 1 + 4} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}\left( {\dfrac{{dt}}{3}} \right)}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\log |t + \sqrt {{t^2} + {2^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {9{x^2} + 6x + 5} | + C $ 

Where C is an arbitrary constant.

12. Find the integration of \[\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}\].

Ans: 

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x + 9 - 9} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {7 + 9 - {{\left( {x + 3} \right)}^3}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {16 - {{\left( {x + 3} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}  $ 

Let, 

$ x + 3 = t $

$ dx = dt $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {t^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = {\sin ^{ - 1}}\left( {\dfrac{t}{4}} \right) + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = {\sin ^{ - 1}}\left( {\dfrac{{x + 3}}{4}} \right) + C $ 

Where C is an arbitrary constant.

13. Find the integration of \[\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}\].

Ans: 

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + 2} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4} + 2} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \left( {\dfrac{1}{4}} \right)} }}} dx $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}  $ 

Now, let

 $x - \dfrac{3}{2} = t $

$ dx = dt $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \log |t + \sqrt {{t^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \log |\left( {x - \dfrac{3}{2}} \right) + \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \dfrac{1}{4}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \log |\left( {x - \dfrac{3}{2}} \right) + \sqrt {{x^2} - 3x + 2} | + C $ 

Where C is an arbitrary constant.

14. Find the integration of \[\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}\].

Ans: Let \[x - \dfrac{3}{2} = t\]

Now, differentiate both sides,

\[dx = dt\]

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4}} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {8 + \dfrac{9}{4} - \left( {{x^2} - 3x + \dfrac{9}{4}} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {\left( {\dfrac{{41}}{4}} \right) - {{\left( {{x^2} - \dfrac{3}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \dfrac{3}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {t^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = {\sin ^{ - 1}}\dfrac{t}{{\dfrac{{\sqrt {41} }}{2}}} + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = {\sin ^{ - 1}}\dfrac{{2\left( {x - \dfrac{3}{2}} \right)}}{{\sqrt {41} }} + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = {\sin ^{ - 1}}\dfrac{{2x - 3}}{{\sqrt {41} }} + C $ 

Where C is an arbitrary constant.

15. Find the integration of \[\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}\].

Ans: 

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

$ \left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + \dfrac{{{{\left( {a + b} \right)}^2}}}{4} - \dfrac{{{{\left( {a + b} \right)}^2}}}{4} + ab $

$ \left( {x - a} \right)\left( {x - b} \right) = {\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]^2} - \dfrac{{{{\left( {a - b} \right)}^2}}}{4} $ 

Now, let

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

$ dx = dt $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]}^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \log |t + \sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \log |\left\{ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right\} + \sqrt {\left( {x - a} \right)\left( {x - b} \right)} | + C $ 

Where C is an arbitrary constant.

16. Find the integration of \[\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}\].

Ans: Let \[2{x^2} + x - 3 = t\],

Now, differentiate both sides,

\[\left( {4x + 1} \right)dx = dt\]

Now,

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt t }}} dt $

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = \dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + C $

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = 2\sqrt t  + C $

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = 2\sqrt {2{x^2} + x - 3}  + C $ 

Where C is an arbitrary constant.

17. Find the integration of \[\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}\].

Ans: 

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  + \displaystyle \int {\dfrac{2}{{\sqrt {{x^2} - 1} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{   }}\left[ {eq.1} \right] $ 

Now, for \[\dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx} \], let \[{x^2} - 1 = t\]

Now, differentiate both sides,

\[2xdx = dt\]

Now, in equation \[1\],

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{dt}}{{\sqrt t }}}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\left( {2\sqrt t } \right) + 2\log |x + \sqrt {{x^2} - 1} | + C $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \sqrt {{x^2} - 1}  + 2\log |x + \sqrt {{x^2} - 1} | + C $ 

Where C is an arbitrary constant.

18. Find the integration of \[\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}\].

Ans: 

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \displaystyle \int {\dfrac{{5x}}{{3{x^2} + 2x + 1}}dx}  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x}}{{3{x^2} + 2x + 1}}dx}  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2 - 2}}{{3{x^2} + 2x + 1}}dx}  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx - \dfrac{5}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} }  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} {\text{   }}\left[ {eq.1} \right] $ 

Now, for\[\dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx} \]

Let \[3{x^2} + 2x + 1 = t\],

Now, differentiate both sides,

\[\left( {6x + 2} \right)dx = dt\]

Now, in eq. \[1\],

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x} \right) + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x + \dfrac{1}{9} - \dfrac{1}{9}} \right) + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} - \dfrac{1}{3} + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} + \dfrac{2}{3}}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{9}\displaystyle \int {\dfrac{1}{{{{\left( {x + \dfrac{1}{3}} \right)}^2} + {{\left( {\dfrac{{\sqrt 2 }}{3}} \right)}^2}}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\log |t| - \dfrac{{11}}{9}\left( {\dfrac{3}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\log |3{x^2} + 2x + 1| - \dfrac{{11}}{{3\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C $ 

Where C is an arbitrary constant.

19. Find the integration of \[\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}\].

Ans: 

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

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = \displaystyle \int {\dfrac{{6x}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}} dx $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9 + 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + \displaystyle \int {\dfrac{{27}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx} {\text{  }}\left[ {eq.1} \right] $ 

Now for \[3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} \],

Let \[{x^2} - 9x + 20 = t\],

Differentiate both sides,

\[\left( {2x - 9} \right)dx = dt\]

Now, from eq. \[1\],

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + \dfrac{{81}}{4} - \dfrac{{81}}{4} + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\left( {2\sqrt t } \right) + 34\log |\left( {x - \dfrac{9}{2}} \right) + \sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} | + C $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\left( {2\sqrt {{x^2} - 9x + 20} } \right) + 34\log |\left( {x - \dfrac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} | + C $ 

Where C is an arbitrary constant.

20. Find the integration of \[\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}\].

Ans:

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{x}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{ - 2x}}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x - 4}}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 2} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx} {\text{   }}\left[ {eq.1} \right] $ 

Now, for \[ - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx} \],

Let \[4x - {x^2} = t\],

Now, differentiate both sides,

\[\left( {4 - 2x} \right)dx = dt\]

Now, from eq. \[1\],

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt { - \left( {{x^2} - 4x + 4 - 4} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\left( {2\sqrt t } \right) + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \sqrt {4x - {x^2}}  + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C $ 

Where C is an arbitrary constant.

21. Find the integration of \[\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}\].

Ans: 

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} + 2x + 3} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} + 2x + 3} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2 - 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx - } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} {\text{  [eq}}{\text{.1]}} $ 

Now, for \[\dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} \]

Let \[{x^2} + 2x + 3 = t\],

Now, differentiate both sides,

\[\left( {2x + 2} \right)dx = dt\]

Now, from eq. \[1\]

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 1 + 2} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + } \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \dfrac{1}{2}\left( {2\sqrt t } \right) + \log |\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 2} | + C $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}  = \sqrt {{x^2} + 2x + 3}  + \log |\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} | + C $ 

Where C is an arbitrary constant.

22. Find the integration of \[\dfrac{{x + 3}}{{{x^2} - 2x - 5}}\].

Ans: 

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \displaystyle \int {\dfrac{x}{{{x^2} - 2x - 5}}dx}  + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}  $

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{{x^2} - 2x - 5}}dx}  + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}  $

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2 + 2}}{{{x^2} - 2x - 5}}dx}  + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}  $

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx}  + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx} {\text{  }}\left[ {eq.1} \right] $ 

Now, for \[\dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx} \]

Let \[{x^2} - 2x - 5 = t\],

Now, differentiate both sides,

\[\left( {2x - 2} \right)dx = dt\]

Now, from equation \[1\],

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx}  + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}  $

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{t}dt}  + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x + 1 - 1 - 5}}dx}  $

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{t}dt}  + 4\displaystyle \int {\dfrac{1}{{{{\left( {x - 1} \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}dx}  $

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\log |t| + 4\left( {\dfrac{1}{{2\sqrt 6 }}} \right)\log \left( {\dfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right) + C $

$ \displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}  = \dfrac{1}{2}\log |{x^2} - 2x - 5| + 4\left( {\dfrac{1}{{2\sqrt 6 }}} \right)\log \left( {\dfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right) + C $ 

Where C is an arbitrary constant.

23. Find the integration of \[\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}\].

Ans: 

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \displaystyle \int {\dfrac{{5x}}{{\sqrt {{x^2} + 4x + 10} }}dx}  + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}  $

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} + 4x + 10} }}dx}  + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}  $

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4 - 4}}{{\sqrt {{x^2} + 4x + 10} }}dx}  + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}  $

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx}  - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx} {\text{  }}\left[ {eq.1} \right] $ 

Now, for \[\dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} \],

Let \[{x^2} + 4x + 10 = t\],

Now, differentiate both sides,

\[\left( {2x + 4} \right)dx = dt\]

Now, from eq. \[1\],

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx}  - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}  $

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \dfrac{5}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt}  - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 4 + 6} }}dx}  $

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \dfrac{5}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt}  - 7\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = \dfrac{5}{2}\left( {2\sqrt t } \right) - 7\log |\left( {x + 2} \right) + \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} | + C $

$ \displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}  = 5\sqrt {{x^2} + 4x + 10}  - 7\log |\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 10} | + C $ 

Where C is an arbitrary constant.

24. The integration \[\displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}} \]is equals to:

A. \[x{\tan ^{ - 1}}\left( {x + 1} \right) + C\]

B. \[{\tan ^{ - 1}}\left( {x + 1} \right) + C\]

C. \[\left( {x + 1} \right){\tan ^{ - 1}}\left( x \right) + C\]

D. \[{\tan ^{ - 1}}\left( x \right) + C\]

Ans: 

$ \displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}}  = \displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 1 + 1}}}  $

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

$ \displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}}  = \dfrac{1}{1}{\tan ^{ - 1}}\left( {\dfrac{{x + 1}}{1}} \right) + C $

$ \displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}}  = {\tan ^{ - 1}}\left( {x + 1} \right) + C $ 

Thus, the correct answer is B.

25. The integration \[\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} \]is equals to:

A. \[\dfrac{1}{9}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{8}} \right) + C\]

B. \[\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{8x - 9}}{9}} \right) + C\]

C. \[\dfrac{1}{3}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{8}} \right) + C\]

D. \[\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{9}} \right) + C\]

Ans: 

$ \displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}  = \displaystyle \int {\dfrac{{dx}}{{\sqrt { - 4\left( {{x^2} - \dfrac{9}{4}x} \right)} }}}  $

$ \displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}  = \displaystyle \int {\dfrac{1}{{\sqrt { - 4\left( {{x^2} - \dfrac{9}{4}x + \dfrac{{81}}{{64}} - \dfrac{{81}}{{64}}} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}  = \displaystyle \int {\dfrac{1}{{\sqrt { - 4\left[ {{{\left( {x - \dfrac{9}{8}} \right)}^2} - {{\left( {\dfrac{9}{8}} \right)}^2}} \right]} }}dx}  $

$ \displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}  = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{9}{8}} \right)}^2} - {{\left( {x - \dfrac{9}{8}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}  = \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{x - \dfrac{9}{8}}}{{\dfrac{9}{8}}}} \right) + C $

$ \displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}  = \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{8x - 9}}{9}} \right) + C $ 

Thus, the correct answer is B.

Conclusion

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4 by Vedantu covers the integration of specific functions using various methods such as substitution, trigonometric identities, integration by parts, and partial fractions. This exercise is crucial for mastering the calculation of indefinite integrals, which is a fundamental concept in calculus. Students should focus on understanding these integration techniques as they form the basis for solving complex problems and are frequently tested in exams.

Class 12 Maths Chapter 7: Exercises Breakdown

CBSE Class 12 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

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