A calculus problem by Aareyan Manzoor

putting y=1 makes this problem easy, but i will try to generalise the given integral. $\int \dfrac{xy+1}{(x^2+2xy+1)^2} dx=\int \dfrac{y(x+y)}{(x^2+2xy+1)^2} dx-(y^2-1)\int \dfrac{1}{(x^2+2xy+1)^2} dx$ In the former integral make a sub of $u=x^2+2xy+1 \to du= (2x+2y)dx$ $\frac{y}{2}\int \dfrac{1}{u^2} du=-\dfrac{y}{2u}+C=-\dfrac{y}{2(x^2+2xy+1)}+C$ The latter part is a bit tedious but still easy: $\int \dfrac{1}{(x^2+2xy+1)^2} dx=\int \left( \dfrac{1}{(x+y+\sqrt{y^2-1})(x+y-\sqrt{y^2-1})}\right)^2dx\\ =\dfrac{1}{4(y^2-1)}\int \left( \dfrac{1}{x+y-\sqrt{y^2-1}}- \dfrac{1}{x+y+\sqrt{y^2-1}}\right)^2dx\\ =\dfrac{1}{4(y^2-1)}\int \left( \dfrac{1}{(x+y-\sqrt{y^2-1})^2}+ \dfrac{1}{(x+y+\sqrt{y^2-1})^2}-2\dfrac{1}{(x-y+\sqrt{y^2-1})(x-y-\sqrt{y^2-1})}\right)dx\\ =\dfrac{1}{4(y^2-1)}\int \left( \dfrac{1}{(x+y-\sqrt{y^2-1})^2}+ \dfrac{1}{(x+y+\sqrt{y^2-1})^2}-2\left(\dfrac{1}{x+y-\sqrt{y^2-1}}- \dfrac{1}{x+y+\sqrt{y^2-1}}\right)\right)dx$ its easy to integrate all the parts differently now, we will endup with $\dfrac{1}{4(y^2-1)}\left(-\dfrac{1}{x+y-\sqrt{y^2-1}}- \dfrac{1}{x+y+\sqrt{y^2-1}}-2\ln\left|\dfrac{x+y-\sqrt{y^2-1}}{x+y+\sqrt{y^2-1}}\right|\right)$ we can simplify all of this (includding the former integral) to get $\dfrac{x}{2(x^2+2xy+1)}+\dfrac{1}{2}\ln\left|\dfrac{x+y-\sqrt{y^2-1}}{x+y+\sqrt{y^2-1}}\right|$ and we are done(note that the ln can be turned into arctan by some identities)