Question

$\underset{0}{\overset{1}{\int }}\frac{\log \left(1+x\right)}{1+x^2}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\underset{0}{\overset{1}{\int }}\frac{\log \left(1+x\right)}{1+x^2}dx \\ \mathrm{Putting}x=\tan \theta \\ \Rightarrow dx={\sec }^2\theta d\theta \\ \mathrm{When}x\to 0;\theta \to 0 \\ \mathrm{and}x\to 1;\theta \to \frac{\mathrm{\pi }}{4} \\ \mathrm{Now},\mathrm{integral}\mathrm{becomes} \end{gathered}$$

$$\begin{gathered} I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{\log \left(1+\tan \theta \right)}{{\sec }^2\theta }{\sec }^2\theta d\theta \\ \Rightarrow I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left[\log \left(1+\tan \theta \right)\right]d\theta .....\left(1\right) \\ \Rightarrow I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left[\log \left\{1+\tan \left(\frac{\mathrm{\pi }}{4}-\theta \right)\right\}\right]d\theta \left[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left[\log \left\{1+\frac{\tan {\displaystyle \frac{\pi }{4}}-\tan \theta }{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan \theta }\right\}\right]d\theta \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left[\log \left\{1+\frac{1-\tan \theta }{1+\tan \theta }\right\}\right]d\theta \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left[\log \left\{\frac{2}{1+\tan \theta }\right\}\right]d\theta \\ I={\int }_0^{\frac{\mathrm{\pi }}{4}}\left[\log 2-\log \left(1+\tan \theta \right)\right]d\theta .....\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{4}}\left(\log 2\right)d\theta \\ \Rightarrow 2I=\left(\log 2\right){\left[\theta \right]}_0^{\frac{\mathrm{\pi }}{4}} \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{4}\log 2 \\ \Rightarrow I=\frac{\mathrm{\pi }}{8}\log 2 \\ \therefore \underset{0}{\overset{1}{\int }}\frac{\log \left(1+x\right)}{1+x^2}dx=\frac{\mathrm{\pi }}{8}\log 2 \end{gathered}$$