Differential equation with integral that is hard to compute Task and my attempt to solve it
$$xy'=\sqrt{2x^2+y^2}-y$$
I have tried substitution like $y=z·x$ but finally got very difficult integral 
$$ \int \frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{dx}{x}$$
at the left side.
Maybe i have to do something different to avoid difficult computation?
Please explain me how to solve it.

*

*Multiply by $\frac{\sqrt{2+z^2}+2z}{\sqrt{2+z^2}+2z}$:
$$
\int\frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{\sqrt{2+z^2}+2z}{(2+z^2)-4z^2}\,dz=\int\frac{\sqrt{2+z^2}}{2-3z^2}\,dz+\int\frac{2z}{2-3z^2}\,dz.
$$
The second integral is simple.

*In the first integral do the change of variables $t=z+\sqrt{z^2+2}$:
$$
\int\frac{\sqrt{2+z^2}}{2-3z^2}\,dz=\int\frac{(t^2+2)^2}{t(-3t^4+20t^2-12)}\,dt.
$$

*Another change of variables $u=t^2$ gives
$$
\int\frac{(t^2+2)^2}{t(-3t^4+20t^2-12)}\,dt=\int\frac{\frac12(u+2)^2}{u(-3u^2+20u-12)}\,du=\frac12\int\frac{(u+2)^2}{u(u-6)(2-3u)}\,du
$$
which can be calculated by partial fractions.


However, I wouldn't wish finding $y(x)$ explicitly after all that.
