Answer
$$ \frac{3}{16}$$
Work Step by Step
Given
$$\int_{0}^{1}\frac{x}{\left(x^{2}+1\right)^{3}} d x$$
Let
$$u= x^2+1 \ \ \ \ \Rightarrow \ \ \ du =2xdx $$
At $x=0 \to u= 1$ and at $x=1 \to u=2 $
Then
\begin{aligned} \int_{0}^{1} \frac{x}{\left(x^{2}+1\right)^{3}} d x &=\frac{1}{2} \int_{1}^{2} \frac{1}{u^{3}} d u \\ &=\left.\frac{1}{2}\left(\frac{1}{-2} u^{-2}\right)\right|_{1} ^{2} \\ &=-\left.\frac{1}{4}\left(\frac{1}{u^{2}}\right)\right|_{1} ^{2} \\ &=-\frac{1}{4}\left(\frac{1}{4}-1\right)= \frac{3}{16}
\end{aligned}
