TITLE: How to find $\int \frac{x^4-4}{x^2\sqrt{4+x^2+x^4}} \,\mathrm dx$
QUESTION [9 upvotes]: Integrate $$\int \frac{(x^4-4)}{(x^2\sqrt{4+x^2+x^4})}\mathrm dx$$

My try:
$$\int \frac{(x^2-4/x^2)}{(\sqrt{4+x^2+x^4})}\mathrm dx\\
=\int \frac{ (x^2-4/x^2)}{(\sqrt{(x^2+1/2)^2+15/4})}\mathrm dx\\$$
Let $t=x^2$
$$=\int \frac{(t^2-4)}{2t\sqrt t\sqrt{(t+1/2)^2+15/4}}\mathrm dt=?$$

Answer:Spoiler

$$\frac{\sqrt{4+x^2+x^4}}{x}$$

REPLY [8 votes]: $$\int \frac{(x^4-4)dx}{(x^2\sqrt{4+x^2+x^4})}$$ 
$$ \int \frac{(x^2-4x^{-2})dx}{(\sqrt{4+x^2+x^4})} $$
$$  \int \frac{(x-4x^{-3})dx}{(\sqrt{4x^{-2}+1+x^2})} $$
Take, $$ u = 1+x^2+4x^{-2} \implies du = (2x-8x^{-3})dx \implies du/2 = (x-4x^{-3})dx$$ 
So integral = 
$$ \int \frac{du/2}{\sqrt{u}} = \sqrt{u} + C \implies \frac{\sqrt{4+x^4+x^2}}{x} +C $$