Answer
$$\frac{1}{2} $$
Work Step by Step
Given $$\int_{0}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} d x $$
Let $ u=1+x^2 \ \ \to \ du=2xdx$ and at $x=0\to u=1$, $ x=\infty\to u=\infty$, then
\begin{aligned} \int_{0}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} d x &=\int_{1}^{\infty} u^{-2} \cdot \frac{d u}{2} \\ &=\frac{1}{2} \lim _{R \rightarrow \infty}\left[\frac{u^{-1}}{-1}\right]_{1}^{R} \\ &=\frac{1}{2} \lim _{R \rightarrow \infty}\left(1-R^{-1}\right) \\ &=\frac{1}{2} \end{aligned}
