NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.4

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.

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Learn More about Integrals and Integration

Integral is a tricky chapter, and students will get a good understanding of the types of sums covered in integration by going through the Ex 7.4 Class 12 NCERT Solutions PDF available on Vedantu. However, before going through Class 12 Maths NCERT Solutions Chapter 7 Exercise 7.4, there are some concepts that students should be familiar with. In this section, we will talk about those important concepts.

Class 12 Maths Ex 7.4 Solutions, cover solutions to all the sums given in this exercise. In this chapter you will learn that integration can be explained as the method for finding out the areas of the two-dimensional region and computing volumes of three-dimensional objects.

This is why if an individual wants to find out the integral of a function with respect to x, then he or she needs to find the area to the x-axis from the curve. There are many experts who also refer to integrals as antiderivatives. This is because integration is the reverse process of differentiation.

There are also different types of integrals. Students will get to know about the two types of integrals in Maths NCERT Solutions Class 12 Chapter 7 Exercise 7.4. The two types of integrals are as follows.

Definite Integral

Definite integrals can be defined as the types of integrals that have definite limits, the upper limit and lower limit. While working on the NCERT Solution of Maths Class 12 Chapter 7 Exercise 7.4, students should remember that definite integrals are also known as Riemann integrals. These integrals can be expressed as:

∫(b upper limit and a lower limit) f (x) d (x).

Indefinite Integral

Indefinite integrals, on the other hand, are defined as integrals in which the upper and lower limits are not defined. This means that while solving the sums of Exercise 7.4 Class 12th Maths answers, you should represent indefinite integrals as:

∫ f (x) d (x) = F (x) + c

In this equation, c is the constant value.

It should be noted that the topic of indefinite integrals is more important for solving maths Exercise 7.4 Class 12 questions. Also, if integration is the inverse of differentiation, then

a / ax F (x) = f (x), then ∫ f (x) dx = F (x) + c

Indefinite integrals are also known as general integrals. These integrals differ in the constant term. From a geometrical perspective, an indefinite integral can be seen as a compilation of curves that are obtained through the translation of one of the curves parallel to itself download and upward along with the y-axis.

There are also several applications of integrals. In mathematics, the applications of integrals are:

• To find out the average value of a curve.

• To calculate the area under a curve.

• To find out the area between any two curves.

• To calculate the center of mass of the centroid of an area that has curved sides.

The applications of integrals in the subject of Physics are:

• To calculate the amount of thrust.

• To calculate the center of gravity.

• To find out the trajectory of a satellite at the time of placing it in an orbit.

• To find out the momentum and mass of inertia of vehicles.

• To calculate the momentum and mass of satellites.

• To find the center of mass.

• To calculator the velocity of a satellite at the time of placing it inside the orbit.

It should also be noted that integration denotes the summation of discrete data. The integral can also be calculated to find the functions that will describe the displacement, area, and volume that occurs because of a collection of small data. This value cannot be measured singularly.

In a broad sense, in the field of calculus, the idea of limit is used for geometry and algebra. The limits help in studying the result of points on a graph like how the points get closer to one another until the distance almost measures to be almost zero.

There are also two major types of calculus. These types of calculus are:

• Integral Calculus.

• Differential Calculus.

The concept of integration can also help in finding out the problem function when the derivatives are mentioned. Also, the concept can be used to find the area bounded by the graph of a function under various constraints.

These points are also important for the development of the concept of ‘Integral Calculus.’ Integral calculus consists of indefinite and definite integrals. Also, in calculus, the concept of differentiating a function and integrating a function is linked through a theorem, which is known as the Fundamental Theorem of Calculus.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

Concepts covered in the exercise: 

• Methods of integration

• Integration by substitution

• Integration using trigonometric identities

• Integrals for some particular functions

• Integration by partial fractions

Importance of Integrals Chapter:

Integral calculus helps in the identification of a function's antiderivatives. The function's integrals are another name for these antiderivatives. Integration is the process of determining the antiderivative of a function. The method of determining integrals is the inverse of that of finding derivatives. A family of curves will be represented by the integral of a function. The fundamental calculus means finding both derivatives and integrals. The concepts are important to understand since they help us solve problems quickly and efficiently. They also result in better performance in examinations. 

Download Ex 7.4 Class 12 Maths NCERT Solutions PDF from Vedantu

Integral is an important chapter in mathematics. You can download and refer to the solved NCERT Solutions PDF from Vedantu.

NCERT Solution Class 12 Maths of Chapter 7 All Exercises

Students should solve all the questions given in the other exercises. We at Vedantu provide solutions of all the exercises of chapter 7 class 12th. The PDF files of these solutions are available for free.

Benefits of Downloading NCERT Maths Ex 7.4 Class 12 Solutions From Vedantu

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• All solutions are present in a PDF format.

• The PDF file can be downloaded at any time and anywhere on-the-go.

• All sums are solved by our subject-matter experts explaining every step clearly.

• The solutions are prepared according to the guidelines set by CBSE.

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