I have specific problem for solving this type of integral :
$$\int \frac{x-y}{x^2-2y^2}\, dx $$
I tried applying partial fractions, making the denominator complete square but the sum keeps on getting lengthy. Any hints Appreciated to make it simpler.
On
By setting $x= y z $ we have: $$ I = \int \frac{x-y}{x^2-2y^2}\,dx = y \int\frac{y z-y}{y^2 z^2 - 2 y^2}\,dz= \int \frac{z-1}{z^2-2}\,dz$$ and the last integral may be computed through partial fraction decomposition: $$ \frac{z-1}{z^2-2} = \frac{1}{4}\cdot\frac{2-\sqrt{2}}{z-\sqrt{2}}+\frac{1}{4}\cdot\frac{2+\sqrt{2}}{z+\sqrt{2}} $$ leads to: $$ I = \frac{1}{2}\log(2-z^2)+\frac{1}{2\sqrt{2}}\log\left(\frac{\sqrt{2}+z}{\sqrt{2}-z}\right).$$
Try this $$ \frac{x-y}{x^2-2y^2}=\frac{x-\sqrt{2}y+\sqrt{2}y-y}{(x-\sqrt{2}y)(x+\sqrt{2}y)}= \frac{1}{x+\sqrt{2}y}+\frac{(\sqrt{2}-1)y}{(x-\sqrt{2}y)(x+\sqrt{2}y)}=\\ =\frac{1}{x+\sqrt{2}y}+\frac{\sqrt{2}-1}{2\sqrt{2}}\left(\frac{1}{x-\sqrt{2}y}-\frac{1}{x+\sqrt{2}y}\right). $$