Answer
$$\frac{1}{4}+\frac{3 \pi}{32}$$
Work Step by Step
Given $$\int_{0}^{1} \frac{d x}{\left(x^{2}+1\right)^{3}}$$ Let $$x= \tan u\ \ \ \ \ \to \ \ \ dx= \sec^2 udu $$ At $x=0\ \ \to \ \ u=0$, at $x=1\ \ \to \ \ u=\pi/4$, then \begin{align*} \int_{0}^{1} \frac{d x}{\left(x^{2}+1\right)^{3}}&=\int_{0}^{\pi/4} \frac{\sec^2 udu}{\left(\tan ^{2}u+1\right)^{3}}\\ &=\int_{0}^{\pi/4} \frac{du}{\sec^4 u}\\ &=\int_{0}^{\pi/4} \cos^4 udu\\ &= \frac{1}{4}\int_{0}^{\pi/4} (1+\cos2 u)^2du\\ &= \frac{1}{4}\int_{0}^{\pi/4}\left(1+2 \cos 2 u+\cos ^{2} 2 u\right) du\\ &=\int_{0}^{\pi/4}\left(\frac{1}{2} \cos 2 u+\frac{3}{8}+\frac{1}{8} \cos 4 u\right) du\\ &=\frac{1}{4} \sin 2 u+\frac{3}{8}u+\frac{1}{32} \sin 4 u\bigg|_{0}^{\pi/4}\\ &= \frac{1}{4}+\frac{3 \pi}{32} \end{align*}
