Find the $\int \frac{(1-y^2)}{(1+y^2)}dy$ $\int \frac{(1-y^2)}{(1+y^2)}dy$ first I tried to divide then I got 1-$\frac{2y^2}{1+y^2}$ and i still can't integrate it. 

Let $y = \tan \theta \Rightarrow I = \displaystyle \int \dfrac{\cos^2\theta - \sin^2 \theta}{\cos^2 \theta}d\theta= \displaystyle \int (1-\tan^2 \theta)d\theta = \displaystyle \int (2-\sec^2\theta)d\theta= 2\theta - \tan \theta + C=2\tan^{-1}y-y+C$
