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: $$\int \dfrac{1}{\sqrt{e^{2x}-1}} dx = \arctan{\sqrt{e^{2x}-1}} + C = \operatorname{arcsec}(e^{x}) + C $$
Best Answer
You did not make the substitution correctly (your substitution would work as you wrote it if $x$ were originally upstairs).
But the choice you made for $u$ will work:
You have $u=x^2$ and $du=2x\,dx$.
From $du=2x\,dx$, you have, dividing both sides by $2x^2$ $$\tag{1}{du\over 2x^2}={x\,dx\over x^2}.$$ Substituting $u=x^2$ on the left hand side of $(1)$ and simplifying the right hand side, we have $$ \color{maroon}{{du\over 2 u}}=\color{maroon}{{dx\over x}}.$$ Substituting into the integral gives $$\int {\color{maroon}{dx}\over\color{maroon} x \sqrt{ x^4-4}}= \int {\color{maroon}{du}\over\color{maroon}{ 2u}\sqrt {u^2-4}} $$ which is an $\rm arcsec$ form.