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The value of the integral $$\int\limits_0^1 {{{\sqrt x dx} \over {(1 + x)(1 + 3x)(3 + x)}}} $$ is :

JEE · Math · previous-year question

  1. A.$${\pi \over 8}\left( {1 - {{\sqrt 3 } \over 2}} \right)$$correct
  2. B.$${\pi \over 4}\left( {1 - {{\sqrt 3 } \over 6}} \right)$$
  3. C.$${\pi \over 8}\left( {1 - {{\sqrt 3 } \over 6}} \right)$$
  4. D.$${\pi \over 4}\left( {1 - {{\sqrt 3 } \over 2}} \right)$$

Answer

A. $${\pi \over 8}\left( {1 - {{\sqrt 3 } \over 2}} \right)$$

Explanation

$$I = \int\limits_0^1 {{{\sqrt x dx} \over {(1 + x)(1 + 3x)(3 + x)}}} dx$$ Let x = t2 $$\Rightarrow$$ dx = 2t . dt $$I = \int\limits_0^1 {{{t(2t)} \over {({t^2} + 1)(1 + 3{t^2})(3 + {t^2})}}} dt$$ $$I = \int\limits_0^1 {{{(3{t^2} + 1) - ({t^2} + 1)} \over {(3{t^2} + 1)({t^2} + 1)(3 + {t^2})}}} dt$$ $$I = \int\limits_0^1 {{{dt} \over {({t^2} + 1)(3 + {t^2})}}} - \int\limits_0^1 {{{dt} \over {(1 + 3{t^2})(3 + {t^2})}}} $$ $$ = {1 \over 2}\int\limits_0^1 {{{(3 + {t^2}) - ({t^2} + 1)} \over {({t^2} + 1)(3 + {t^2})}}} dt + {1 \over 8}\int\limits_0^1 {{{(1 + 3{t^2}) - 3(3 + {t^2})} \over {(1 + 3{t^2})(3 + {t^2})}}} dt$$ $$ = {1 \over 2}\int\limits_0^1 {{{dt} \over {1 + {t^2}}} - {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} + {1 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {(1 + 3{t^2})}}} } } } $$ $$ = {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 1}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {1 + 3{t^2}}}} } } $$ $$ = {1 \over 2}({\tan ^{ - 1}}(t))_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {{t \over {\sqrt 3 }}} \right)} \right)_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {\sqrt 3 t} \right)} \right)_0^1$$ $$ = {1 \over 2}\left( {{\pi \over 4}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi \over 6}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi \over 3}} \right)$$ $$ = {\pi \over 8} - {{\sqrt 3 } \over {16}}\pi $$ $$ = {\pi \over 8}\left( {1 - {{\sqrt 3 } \over 2}} \right)$$

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