Chapter 7: Integrals

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.

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.

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.

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.

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.