Find the primitive of a real function

Question

I know that this integral should be simple but I am really struggling to understand how to get in 2arcsecx + c response. I managed to solve the problem at this point and my difficulty is in relation to this square root:

$$ \int\sqrt{\frac{4}{x^4-x^2}}dx = \int\frac{2}{\sqrt{x^4 - x^2}}dx = 2\int\frac{dx}{\sqrt{x^4 - x^2}} $$

Answer 1

Start by writing $x^4 - x^2 = x^2 (x^2 - 1)$. Take the $x^2$ outside the square root. Then use the substitution $x = \sec(u)$.

Answer 2

Let (in the Euler spirit) $$ u=\frac{x}{\sqrt{x^4-x^2}}, $$ and you will get $$ -2\int\frac{1}{1+u^2}\,du. $$ The result of course becomes $$ -2\arctan\Bigl(\frac{x}{\sqrt{x^4-x^2}}\Bigr)+C. $$ Now, that is not in the final form you were looking for, and it can be simplified a bit, but anyways...