Question

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

Answer 1

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

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

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