Aloha :)
$$I=\int\frac{1}{(1+x^2)^2}\,dx=\int\frac{(1\pink{+x^2})\pink{-x^2}}{(1+x^2)^2}\,dx=\int\left(\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}\right)dx$$$$\phantom I=\int\frac{1}{1+x^2}\,dx-\int\underbrace{\frac x2}_{=u}\cdot\underbrace{\frac{2x}{(1+x^2)^2}}_{=v'}dx$$$$\phantom I=\int\frac{1}{1+x^2}\,dx-\left(\underbrace{\frac x2}_{=u}\cdot\underbrace{\frac{-1}{1+x^2}}_{=v}-\int\underbrace{\frac 12}_{=u'}\cdot\underbrace{\frac{-1}{1+x^2}}_{=v}dx\right)$$$$\phantom I=\frac12\underbrace{\int\frac{1}{1+x^2}\,dx}_{=\arctan(x)}+\frac12\cdot\frac{x}{1+x^2}=\frac12\left(\arctan(x)+\frac{x}{1+x^2}\right)$$
