Explanation:
(B): Given,
\(I=\int_{0}^{\infty} \frac{x}{(1+x)\left(x^{2}+1\right)} d x\)
\(=\int_{0}^{\infty} \frac{(1+x)-1}{(1+x)\left(1+x^{2}\right)} d x\)
\(=\int_{0}^{\infty} \frac{(1+x)}{(1+x)\left(1+x^{2}\right)} d x-\int_{0}^{\infty} \frac{d x}{(1+x)\left(1+x^{2}\right)}\)
\(=\int_{0}^{\infty} \frac{d x}{\left(1+x^{2}\right)}-\int_{0}^{\infty} \frac{d x}{(1+x)\left(1+x^{2}\right)} \ldots .(i) \tag{i}\)
Let, \(\quad I=\int_{0}^{\infty} \frac{d x}{(1+x)\left(1+x^{2}\right)}\)
Put, \(\mathrm{x}=\tan \theta\)
\(\mathrm{dx}=\sec ^{2} \theta \mathrm{d} \theta\)
When, \(x=0, \theta=0\) and when \(x=\infty, \quad \theta=\frac{\pi}{2}\)
\(\therefore \quad I=\int_{0}^{\pi / 2} \frac{\sec ^{2} \theta d \theta}{(1+\tan \theta)\left(1+\tan ^{2} \theta\right)}\)
\(=\int_{0}^{\pi / 2} \frac{\mathrm{d} \theta}{1+\tan \theta}\)
\(=\int_{0}^{\pi / 2} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta \tag{ii}\)
\(=\int_{0}^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\sin \left(\frac{\pi}{2}-\theta\right)+\cos \left(\frac{\pi}{2}-\theta\right)} \mathrm{d} \theta\)
\(=\int_{0}^{\pi / 2} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta \tag{iii}\)
Adding equation (ii) and (iii), we get -
\(2 \mathrm{I}=\int_{0}^{\pi / 2} \frac{\sin \theta+\cos \theta}{\sin \theta+\cos \theta} d \theta\)
\(2 \mathrm{I}=[\theta]_{0}^{\pi / 2}\)
\(2 \mathrm{I}=\frac{\pi}{2}-0 \quad 2 \mathrm{I}=\frac{\pi}{2}\)
\(I=\frac{\pi}{4}\)
On putting \(\mathrm{I}=\frac{\pi}{4}\) in equation (i), we get -
\(I=\left[\tan ^{-1} x\right]_{0}^{\infty}-\frac{\pi}{4}\)
\(I=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\)
