The value of the integral _0^1( x,dx)/((1+x)(1+3x)(3+x)) is

Question

The value of the integral $\displaystyle\int_0^1\frac{\sqrt x\,dx}{(1+x)(1+3x)(3+x)}$ is

Accepted Answer

$$I = \int\limits_0^1 {{{\sqrt x dx} \over {(1 + x)(1 + 3x)(3 + x)}}} dx$$

Let x = t2 $$\Rightarrow$$ dx = 2t . dt

$$I = \int\limits_0^1 {{{t(2t)} \over {({t^2} + 1)(1 + 3{t^2})(3 + {t^2})}}} dt$$

$$I = \int\limits_0^1 {{{(3{t^2} + 1) - ({t^2} + 1)} \over {(3{t^2} + 1)({t^2} + 1)(3 + {t^2})}}} dt$$

$$I = \int\limits_0^1 {{{dt} \over {({t^2} + 1)(3 + {t^2})}}}  - \int\limits_0^1 {{{dt} \over {(1 + 3{t^2})(3 + {t^2})}}} $$

$$ = {1 \over 2}\int\limits_0^1 {{{(3 + {t^2}) - ({t^2} + 1)} \over {({t^2} + 1)(3 + {t^2})}}} dt + {1 \over 8}\int\limits_0^1 {{{(1 + 3{t^2}) - 3(3 + {t^2})} \over {(1 + 3{t^2})(3 + {t^2})}}} dt$$

$$ = {1 \over 2}\int\limits_0^1 {{{dt} \over {1 + {t^2}}} - {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} + {1 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {(1 + 3{t^2})}}} } } } $$

$$ = {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 1}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {1 + 3{t^2}}}} } } $$

$$ = {1 \over 2}({\tan ^{ - 1}}(t))_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {{t \over {\sqrt 3 }}} \right)} \right)_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {\sqrt 3 t} \right)} \right)_0^1$$

$$ = {1 \over 2}\left( {{\pi  \over 4}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi  \over 6}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi  \over 3}} \right)$$

$$ = {\pi  \over 8} - {{\sqrt 3 } \over {16}}\pi $$

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

Solutions · 1