Answer
The given improper integral
$$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} d x =0 \\
\end{aligned}
$$
Work Step by Step
$$
\int_{-\infty}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} d x
$$
The given improper integral
$$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} d x &=\int_{-\infty}^{0} \frac{x}{\left(1+x^{2}\right)^{2}} d x+\int_{0}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} d x \\
&=\lim _{a \rightarrow-\infty} \int_{a}^{\infty} \frac{x d x}{\left(1+x^{2}\right)^{2}}+\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{x d x}{\left(1+x^{2}\right)^{2}} \\
&=\left.\lim _{a \rightarrow-\infty}\left(\frac{-1}{2\left(1+x^{2}\right)}\right)\right|_{a} ^{0}+\left.\lim _{b \rightarrow \infty}\left(\frac{-1}{2\left(1+x^{2}\right)}\right)\right|_{0} ^{b} \\
&= \lim _{a \rightarrow-\infty}\left|\frac{-1}{2\left(1+0^{2}\right)}+\frac{1}{2\left(1+a^{2}\right)}\right| \\ &+\lim _{b \rightarrow \infty}\left[\frac{-1}{2\left(1+b^{2}\right)}+\frac{1}{2\left(1+0^{2}\right)}\right] \\
&=-\frac{1}{2}+0+0+\frac{1}{2}\\
&=0 \\
\end{aligned}
$$
