How to find $\int \frac{x^4-4}{x^2\sqrt{4+x^2+x^4}} \,\mathrm dx$

Question

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

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

Answer 1

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