Differential equation Euler substitution I was trying to solve this differential equation but can't figure out the final integral I get by variable separable method


  The equation is 
  $$
x^3 \, y' = y^3 + y^2 \, \sqrt{y^2-x^2}
$$


I got the integral 
$$
\frac{dv}{v^3 + v^2 \sqrt{(v^2 - 1)} - v}
$$
but can't figure out how to solve it.

The Euler substitution $v= \frac{u^2 + 1}{2u}$ could work but I can't seem to proceed with it.

Any help with this problem is much appreciated.
Simply use a partial fraction decomposition to obtain:

$$ \int \frac{ dx}{x^3 + x^2 \sqrt{ x^2 -1} - x } = \int \frac{ \sqrt{x^2 -1} }{2 (x-1) } - \frac{ \sqrt{ x^2 - 1}}{ 2 (x+1)} - \frac{1}{x} dx$$

Use $u$ substitute with $x-1$ and $x+1$ since $x^2-1 = (x+1)(x-1)$. I think that should do it.
$$ x^3 \, y' = y^3 + y \, \sqrt{y^2-x^2} $$
Let (y/x)=t, then: 
y = v*x; y' = x*v' + v 

use this substitution

...and
v^3 + (v^2)sqrt[v^2 - 1] - v = v[(v^2 - 1) + sqrt(v^2 - 1)]

v^2 - 1 = p^2 => vdv = pdp

and you will get:

(dp)/[(p^2 +1)*(p + 1)] = dx/x