The value of _0^1 8(1+x)1+x^2,dx is:

Question

The value of $\displaystyle\int_0^1 \dfrac{8\log(1+x)}{1+x^2}\,dx$ is:

Accepted Answer

$$I = \int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx$$

put $$x = \tan \,\theta ,$$

$$\therefore$$ $${{dx} \over {d\theta }} = {\sec ^2}\theta  \Rightarrow dx = {\sec ^2}\theta d\theta $$

$$\therefore$$ $$I = 8\int\limits_0^{\pi /4} {{{\log \left( {1 + \tan \theta } \right)} \over {1 + {{\tan }^2}\theta }}} .{\sec ^2}\theta d\theta $$

$$I = 8\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta \,...\left( i \right)$$

$$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {{\pi  \over 4} - \theta } \right)} \right]} d\theta $$

$$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + {{1 - \tan \theta } \over {1 + \tan \theta }}} \right]} d\theta $$

$$ = 8\int\limits_0^{\pi /4} {\log \left[ {{2 \over {1 + \tan \theta }}} \right]} d\theta $$

$$ = 8\int\limits_0^{\pi /4} {\left[ {\log 2 - \log \left( {1 + \tan \theta } \right)} \right]d\theta } $$

$$I = 8.\left( {\log 2} \right)\left[ x \right]_0^{\pi /4} - 8$$

$$\,\,\,\,\,\,\,\,\,\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta $$

$$I = 8.{\pi  \over 4}.\log 2 - I$$ $$\left[ {} \right.$$ From equation $(i)$ $$\left. {} \right]$$

$$ \Rightarrow 2I = 2\pi \log 2,$$

$$\therefore$$ $$I = \pi \log 2$$

Solutions · 1