The value of \(\int_{0}^{\infty} \frac{x}{(1+x)\left(x^{2}+1\right)} d x\) is

Let \(I=\int_{0}^{\infty} \frac{x d x}{(1+x)\left(x^{2}+1\right)}\)
By partial fraction, \(\frac{x}{(1+x)\left(x^{2}+1\right)}=\frac{A}{(1+x)}+\frac{B x+C}{\left(x^{2}+1\right)}\)
\(\Rightarrow x=A\left(x^{2}+1\right)+(1+x)(B x+C)\)
\(\Rightarrow x=A\left(x^{2}+1\right)+\left(B x+B x^{2}+C+C x\right)\)
\(\Rightarrow x=(A+B) x^{2}+(B+C) x+(A+C)\)
On comparing both sides, we get \(A+B=0, \quad\)\(B+C=1, \quad A+C=0 \quad \ldots(\mathrm{i})\)
On adding all these equations, we get
\(A+B+C =\frac{1}{2}...(ii) \)
\( \therefore A=\frac{1}{2}-1 =-\frac{1}{2}, C=\frac{1}{2} \)
\( \text { and } B=\frac{1}{2} \)
\( \therefore I=\int_{0}^{\infty}\left\{\frac{-1}{2(1+x)}+\frac{1}{2} \frac{(x+1)}{\left(x^{2}+1\right)}\right\} d x \)
\(=-\frac{1}{2} \int_{0}^{\infty} \frac{d x}{1+x}+\frac{1}{2} \int_{0}^{\infty} \frac{x}{x^{2}+1} d x \)
\(+\frac{1}{2} \int_{0}^{\infty} \frac{d x}{1+x^{2}} \)
\(=-\frac{1}{2}[\log (1+x)]_{0}^{\infty}+\frac{1}{4}\left[\log \left(x^{2}+1\right)\right]_{0}^{\infty}\)
\(+\frac{1}{2} \times \frac{\pi}{2}\)
\(=-\frac{1}{2} \lim _{x \rightarrow \infty} \log (1+x)+\frac{1}{4}\)
\(\lim _{x \rightarrow \infty} \log \left(1+x^{2}\right)+\frac{\pi}{4}\)
\(=\lim _{x \rightarrow \infty} \log \left[\frac{\left(1+x^{2}\right)^{1 / 4}}{(1+x)^{1 / 2}}\right]+\frac{\pi}{4}\)
\(=\lim _{x \rightarrow \infty} \log \left[\frac{\sqrt{x}\left(\frac{1}{x^{2}}+1\right)^{1 / 4}}{\sqrt{x}\left(\frac{1}{x}+1\right)^{1 / 2}}\right]+\frac{\pi}{4}\)
\(=\log \frac{(0+1)^{1 / 4}}{(0+1)^{1 / 2}}+\frac{\pi}{4}\)
\(=\log (1)+\frac{\pi}{4}=0+\frac{\pi}{4}=\frac{\pi}{4}\)