Question

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

Answer 1

We have,

$$I=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{(1+x)(1+x^2)}dx$$

$$\text{Putting }x=\tan \theta$$

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

$$\text{When }x\to 0;\theta \to 0$$

$$\text{and }x\to \mathrm{\infty };\theta \to \frac{\pi }{2}$$

$$\text{Now, integral becomes}$$

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

$$\text{Adding}\left(1\right)and\left(2\right),\text{we get}$$

$$2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin \theta +\cos \theta }{\sin \theta +\cos \theta }d\theta$$

$$\Rightarrow 2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}d\theta$$

$$\Rightarrow 2I=\frac{\pi }{2}$$

$$\Rightarrow I=\frac{\pi }{4}$$

$$\therefore \underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{(1+x)(1+x^2)}dx=\frac{\pi }{4}$$