Answer
$\frac{\pi }{4}$
Work Step by Step
Take the partial fraction:
$\frac{1}{\left(x+1\right)\left(x^2+1\right)}$ = $\frac{\frac{1}{2}}{x+1}+\frac{\frac{-1}{2}x+\frac{1}{2}}{x^2+1}$
Now, integrate the partial fraction:
$\int _0^{\infty }\:\frac{1}{\left(x+1\right)\left(x^2+1\right)}dx$ = $\int _{0}^{\infty }\:\:\frac{\frac{1}{2}}{x+1}dx+\int _{0}^{\infty }\:\frac{\frac{-1}{2}x+\frac{1}{2}}{x^2+1}dx$
=$\frac{1}{2}\lim _{a\to \infty}\left(\int _0^a\frac{1}{x+1}dx\:\right)$ - $\frac{1}{4}\lim _{a\to \infty}\left(\int _0^a\:\frac{2x}{x^2+1}dx\right)$+$\frac{1}{2}\lim _{a\to \infty}\left(\int _0^a\:\frac{1}{x^2+1}dx\right)$
= $\frac{1}{2}\lim _{a\to \infty }\left(ln\left(\frac{a+1}{0+1}\right)\right)-\frac{1}{4}\lim _{a\to \infty }\left(ln\left(\frac{a^2+1}{0^2+1}\right)\right)$+$\frac{1}{2}\lim _{a\to \infty }\left(\arctan \left(a\right)-arctan\left(0\right)\right)$
=$\frac{1}{2}\lim _{a\to \infty }\left(\arctan \left(a\right)-arctan\left(0\right)\right)$+$\frac{1}{4}\lim _{a\to \infty }\left(2\ln \left(a+1\right)-\ln \left(a^2+1\right)\right)$
=$\frac{1}{2}\times\frac{\pi }{2}$+$\frac{1}{4}\lim _{a\to \infty }\left(\ln \left(\frac{\left(a+1\right)^2}{a^2+1}\right)\right)$
=$\frac{1}{2}\times\frac{\pi }{2}$+0
=$\frac{\pi }{4}$
