$\underset{-1}{\overset{1}{\int }}\frac{1}{{(1+x^2)}^2}dx=$

$$\begin{gathered} \underset{-1}{\overset{1}{\int }}\frac{1}{{(1+x^2)}^2}dx \\ =2\underset{0}{\overset{1}{\int }}\frac{1}{{(1+x^2)}^2}dx \left[\therefore \frac{1}{{(1+x^2)}^2}is even function\right] \\ Put x=\tan \theta \\ dx=se{c}^2\theta d\theta \\ =2\underset{0}{\overset{\pi /4}{\int }}\frac{1}{{(1+{\tan }^2\theta )}^2}\left(se{c}^2\theta \right)\text{dθ} \\ =2\underset{0}{\overset{\pi /4}{\int }}\frac{1}{se{c}^4\theta }\left(se{c}^2\theta \right)\text{dθ} \\ =2\underset{0}{\overset{\pi /4}{\int }}{\text{\hspace{0.17em}cos}}^2\text{θdθ} \\ =\underset{0}{\overset{\pi /4}{\int }}\left(1+\cos 2\theta \right)d\theta \\ ={\left(\theta +\frac{\sin 2\theta }{2}\right)}_0^{\mathrm{\pi }/4} \\ =\frac{\pi }{4}+\frac{1}{2}\left(1\right)-0 \\ =\frac{\pi }{4}+\frac{1}{2} \end{gathered}$$