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NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.4 - 2025-26

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Maths Class 12 Chapter 7 Questions and Answers - Free PDF Download

In Ncert Solutions Class 12 Maths Chapter 7 Exercise 7 4, you will get step-by-step help to solve all the integration problems from this part of the Integrals chapter. This exercise is all about learning smart ways to integrate, like substitution, using trigonometric identities, and handling expressions with square roots or tricky denominators.

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If you ever get stuck on a question, Vedantu’s simple NCERT Solutions break every step down so you’re never lost. Downloadable PDFs make revision easy—just pick a question or run through the solutions for practice. Stressed about the full syllabus? You can also check the Class 12 Maths syllabus here.


These NCERT Solutions help clear your biggest doubts—like when to use substitution or split a complex fraction. Use them for last-minute prep or daily homework and grab a free PDF of all Class 12 Maths NCERT Solutions as a reliable revision buddy. This chapter carries 9 marks in your CBSE exam.


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Access PDF for Maths NCERT Chapter 7 Integrals Exercise 7.4 Class 12

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

S.No.

Chapter 7 - Integrals Exercises in PDF Format

1

Class 12 Maths Chapter 7 Exercise 7.1 - 22 Questions & Solutions (21 Short Answers, 1 MCQs)

2

Class 12 Maths Chapter 7 Exercise 7.2 - 39 Questions & Solutions (37 Short Answers, 2 MCQs)

3

Class 12 Maths Chapter 7 Exercise 7.3 - 24 Questions & Solutions (22 Short Answers, 2 MCQs)

4

Class 12 Maths Chapter 7 Exercise 7.5 - 23 Questions & Solutions (21 Short Answers, 2 MCQs)

5

Class 12 Maths Chapter 7 Exercise 7.6 - 24 Questions & Solutions (22 Short Answers, 2 MCQs)

6

Class 12 Maths Chapter 7 Exercise 7.7 - 11 Questions & Solutions (9 Short Answers, 2 MCQs)

7

Class 12 Maths Chapter 7 Exercise 7.8 - 6 Questions & Solutions (6 Short Answers)

8

Class 12 Maths Chapter 7 Exercise 7.9 - 22 Questions & Solutions (20 Short Answers, 2 MCQs)

9

Class 12 Maths Chapter 7 Exercise 7.10 - 10 Questions & Solutions (8 Short Answers, 2 MCQs)

10

Class 12 Maths Chapter 7 Miscellaneous Exercise - 40 Questions & Solutions



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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FAQs on NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.4 - 2025-26

1. What are the main integration methods covered in the NCERT Solutions for Class 12 Maths Chapter 7, Integrals?

The NCERT Solutions for Class 12 Maths Chapter 7 primarily focus on mastering three fundamental integration techniques as per the CBSE 2025-26 syllabus:

  • Integration by Substitution: Used to simplify complex integrands by substituting a part of the function with a new variable.
  • Integration using Partial Fractions: A method to break down complex rational functions into simpler fractions that are easier to integrate.
  • Integration by Parts: Applied when the integrand is a product of two different types of functions, often using the ILATE rule to select the functions.

2. Why is it essential to add the constant of integration '+ C' when solving indefinite integrals in Chapter 7?

Adding the constant of integration '+ C' is crucial because the derivative of a constant is zero. This means that for any given function, there exists an entire family of antiderivatives, or integrals, that differ from each other only by a constant. For example, the derivative of x² is 2x, but the derivative of x² + 5 is also 2x. The '+ C' represents this entire family of possible solutions, which is a core concept in indefinite integration.

3. What is the correct step-by-step method to solve integrals using substitution, as shown in NCERT solutions?

The NCERT solutions follow a systematic approach for integration by substitution:

  • Step 1: Identify a part of the integrand, usually a function within another function (like g(x) in f(g(x))), to set as a new variable, 't'.
  • Step 2: Differentiate the substitution, t = g(x), to find dt in terms of dx (i.e., dt = g'(x)dx).
  • Step 3: Rewrite the entire integral in terms of 't' and 'dt'. The original variable 'x' must be completely eliminated.
  • Step 4: Solve the new, simpler integral with respect to 't'.
  • Step 5: Substitute the original function g(x) back in place of 't' and add the constant of integration, C.

4. How do I apply the 'Integration by Parts' formula correctly using the ILATE rule for NCERT questions?

To apply integration by parts, you use the formula: ∫u v dx = u ∫v dx - ∫(u' ∫v dx) dx. The key is choosing the first function (u) and second function (v) correctly using the ILATE rule, which stands for:

  • I - Inverse Trigonometric Functions
  • L - Logarithmic Functions
  • A - Algebraic Functions
  • T - Trigonometric Functions
  • E - Exponential Functions
  • The function that appears first in the ILATE order should be chosen as 'u'. This method simplifies the process and is consistently used in the NCERT solutions to ensure a standard approach.

5. In Chapter 7 exercises, how do I decide whether to use substitution or integration by parts?

Choosing the correct method is a key problem-solving skill. Here’s a general guide:

  • Choose Substitution when the integrand contains a function and its derivative (or a multiple of its derivative). For example, in ∫2x sin(x²) dx, the derivative of x² is 2x.
  • Choose Integration by Parts when the integrand is a product of two functions from different categories in the ILATE rule, such as an algebraic function multiplied by a trigonometric one (e.g., ∫x cos(x) dx). It is not suitable when one function is a clear derivative of another's part.

6. How do NCERT solutions handle integrals of rational functions using partial fractions?

NCERT solutions teach a structured method to decompose a rational function into partial fractions based on the nature of the denominator's factors. The first step is to ensure the fraction is proper (degree of numerator < degree of denominator). Then, the decomposition follows specific rules for different types of factors in the denominator, such as non-repeated linear factors, repeated linear factors, and non-factorizable quadratic factors.

7. What is the standard approach to solving integrals of the form ∫(px + q) / √(ax² + bx + c) dx in NCERT exercises?

For integrals of this specific form, the NCERT method involves expressing the linear numerator (px + q) in terms of the derivative of the quadratic expression in the denominator. The standard form is:
px + q = A * d/dx(ax² + bx + c) + B
By solving for constants A and B, the integral is split into two parts: one that can be solved by a simple substitution and another that reduces to a standard integral form.

8. What is the fundamental difference between solving a definite integral and an indefinite integral in Chapter 7?

The key difference lies in the result. An indefinite integral gives a general function (e.g., F(x) + C) representing a family of curves. In contrast, a definite integral, which has upper and lower limits, yields a specific numerical value. This value represents the area under the curve between those limits. Consequently, you apply the limits of integration after finding the antiderivative and do not add the constant of integration '+ C' in the final answer for definite integrals.

9. How do the problems in the Miscellaneous Exercise for Chapter 7 differ from those in other exercises?

The Miscellaneous Exercise of Chapter 7 is designed to be more challenging and comprehensive. Unlike regular exercises that focus on a single integration technique at a time, the miscellaneous problems often require a combination of multiple methods (e.g., using substitution first and then applying integration by parts). They test a student's overall understanding and ability to identify the correct sequence of steps for complex integrals.

10. What is the main benefit of using Vedantu's step-by-step NCERT solutions for integrals instead of just checking the final answer?

Using detailed, step-by-step NCERT solutions helps you learn the correct problem-solving methodology, which is critical for scoring well in CBSE board exams. It allows you to understand the reasoning behind each step, identify common errors in your own work, and master the standard forms and techniques prescribed by the NCERT curriculum. This builds a stronger conceptual foundation than simply verifying a final answer.