I am having trouble evaluating $\int \dfrac{1}{x\sqrt{x^4-4}} dx$
I tried making $a = 2$, $u = x^{2}$, $du = 2x dx$ and rewriting the integral as: $\dfrac{1}{2} \int \dfrac{du}{\sqrt{u^2-a^2}} $ But I believe something is not right at this step (perhaps when changing from $dx$ to $du$)?
I end up with: ${1\over 4} \operatorname{arcsec} \dfrac{1}{2}x^{2} + C$
Any help would be appreciated, I feel I am only making a simple mistake. Also, for some reason, on WA, it is showing an answer involving $\tan^{-1}$ but I do not see an $a^{2} + u^{2}$ possibility. Note that I do know how sometimes (different) inverse trig functions when integrated are equal.
Ex: $\int \dfrac{1}{\sqrt{e^{2x}-1}} dx = \arctan{\sqrt{e^{2x}-1}} + C = \operatorname{arcsec}(e^{x}) + C $