Question

The value of the integral ${\int }_0^{\infty }\frac{x\mathrm{dx}}{\left(1+x\right)\left(1+x^2\right)}$ is equal to

• A. $\pi$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $2\pi$

Answer 1

The correct option is B $\frac{\pi }{4}$

We have,
$\mathrm{Let}I={\int }_0^{\infty }\frac{x\mathrm{dx}}{\left(1+x\right)\left(1+x^2\right)}\mathrm{Let}x=\tan \theta \Rightarrow \mathrm{dx}={\sec }^2\theta \mathrm{d\theta }\mathrm{When}x=\infty ,\theta =\frac{\pi }{2}\mathrm{and}x=0,\theta =0I={\int }_0^{\frac{\pi }{2}}\frac{\tan \theta {\sec }^2\theta }{\left(1+\tan \theta \right)\left(1+{\tan }^2\theta \right)}\mathrm{d\theta }\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\frac{\sin \theta }{\cos \theta }{\sec }^2\theta }{\left(1+\frac{\sin \theta }{\cos \theta }\right){\sec }^2\theta }\mathrm{d\theta }\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sin \theta }{\sin \theta +\cos \theta }\mathrm{d\theta }....\left(1\right]\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(\frac{\pi }{2}-\theta \right)}{\sin \left(\frac{\pi }{2}-\theta \right)+\cos \left(\frac{\pi }{2}-\theta \right)}\mathrm{d\theta }\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\cos \theta }{\sin \theta +\cos \theta }\mathrm{d\theta }....\left(2\right]\mathrm{Adding}\left(1\right]\mathrm{and}\left(2\right],\mathrm{we}\mathrm{get}\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\frac{\sin \theta +\cos \theta }{\sin \theta +\cos \theta }\mathrm{d\theta }\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}1\cdot \mathrm{d\theta }\Rightarrow 2I={\left(x\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}-0\Rightarrow I=\frac{\pi }{4}$