I need help finding integral of $\frac{1}{x^2+x^4}$

Question

I don't know how can I solve this integral rational function.

$$\int\frac{1}{x^2+x^4}\mathrm{d}x$$

Answer 1

Hint:$$\frac{1}{x^2+x^4}=\frac{1}{x^2} - \frac{1}{x^2 + 1}$$

Answer 2

\begin{align*}\int\frac{1}{x^2+x^4}dx&=\int\frac{1}{x^2(1+x^2)}dx \\ &=\int\left(\frac{1}{x^2}-\frac{1}{1+x^2}\right)dx~ \quad(\because \mbox{using partial fraction technique}) \\ &=\int\frac{1}{x^2}dx-\int\frac{1}{1+x^2}dx \\&=\frac{-1}{x}+\tan^{-1}(x)+c \quad (\mbox{c is a constant of integration})\\&\quad \left(\because \int x^{n}dx=\frac{x^{n+1}}{(n+1)}+c_1,~n\neq-1,\mbox{and} \int\frac{1}{1+x^2}dx=\tan^{-1}(x)+c_2\right)\end{align*}