How to find the integral $\int\frac{x^2}{x^2+x^4-2}dx$?

Question

Find out the indefinite integral $$\int\frac{x^2}{x^2+x^4-2}dx$$

my try: $$\int\frac{x^2}{x^2+x^4-2}dx=\int\frac{x^2}{\left(x^2+\frac{1}{2}\right)^2-\frac 94}dx$$ $$=\int\frac{x^2+\frac{1}{2}-\frac 12}{\left(x^2+\frac{1}{2}\right)^2-\frac 94}dx$$ I got stuck here I don't know how to proceed. someone please suggest me idea or give the solution.

thanks

Answer 1

Write

$$ \begin{align} \left (x^2+\frac{1}{2}\right )^2-\frac{9}{4} & = \left(x^2+\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2 \\ & =\left(x^2+2\right)\left(x^2-1\right)\\ & =\left(x^2+2\right)\left(x+1\right)\left(x-1\right) \end{align} $$

Then do partial fraction decomposition and integrate.

Answer 2

HINT:

Write $x^2=y$

$$\dfrac{3y}{y^2+y-2}=\dfrac{y+2+2(y-1)}{(y+2)(y-1)}=\dfrac1{y-1}+\dfrac2{y^2+2}$$