Chapter 7: Integrals

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.