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

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