Answer
$$\displaystyle{\int_0^1\frac{1}{\left(x^2+1\right)^{2}}dx=\frac{1}{4}+\frac{\pi }{8}}\\ $$
Work Step by Step
$\displaystyle{I=\int_0^1\frac{1}{\left(x^2+1\right)^{2}}dx}\\$
$\displaystyle \left[\begin{array}{ll} x=\tan\theta & x^2=\tan^2\theta \\ & \\ \frac{dx}{d\theta}=\sec^2\theta & dx=\sec^2\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int_0^{\frac{\pi}{4}}\frac{1}{\left(\tan^2\theta+1\right)^2}\sec^2\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{4}}\frac{\sec^2\theta}{\sec^4\theta}\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{4}}\frac{1}{\sec^2\theta}\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{4}}\cos^2\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{4}}\frac{1}{2}\left(\cos2\theta+1\right)\ d\theta}\\ \displaystyle{I=\frac{1}{2}\int_0^{\frac{\pi}{4}}\cos2\theta+1\ d\theta}\\ \displaystyle{I=\frac{1}{2}\left[\frac{1}{2}\sin2\theta+\theta\right]_0^{\frac{\pi}{4}}}\\ \displaystyle{I=\frac{1}{4}\sin\left(\frac{\pi }{2}\right)+\frac{\pi }{4}\times\frac{1}{2}-0-0}\\ \displaystyle{I=\frac{1}{4}+\frac{\pi }{8}}\\ $
