indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx}  {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx}  {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
 A: Hint: $x^4+x^2-2$ can be factored as $(x-1)(x+1)(x^2+2)$. then find $A,B,C$, and $D$ which are constants, such that
$$
\frac{x^2}{x^4+x^2-2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+2}
$$
and integrate termwise.
A: $$I= \int \frac { x^2dx}  {x^4 + x^2 -2}$$
Divide by $x^4$.
$$I=\int \frac{\frac{1}{x^2}dx}{1+\frac{1}{x^2}-\frac{2}{x^4}}$$
Let $\frac1x=u$. Then, $du=-\frac{1}{x^2}dx$.
$$I=\int\frac{du}{2u^4-u^2-1}=\int\frac{du}{2u^4-2u^2+u^2-1}=\int\frac{du}{(2u^2+1)(u^2-1)}=\int\frac{du}{(2u^2+1)(u+1)(u-1)}$$
Now this can be solved by using partial fractions.
A: You are better off using partial fractions, which you should know if you have been given this integral to evaluate.  You have the following factors for the denominator:
$x^4+x^2-2
= (x^2-1)(x^2+2)
= (x+1)(x-1)(x^2+2)$
And then
$x^2/(x^4+x^2-2) = a/(x+1)+b/(x-1)+(cx+d)/(x^2+2)$
You are to find the condtants a, b, c, d that fit this equation.  Then you integrate the resulting partial fractions using logatlrithms and inverse tangents, and add those partial integrals up. 
A: HINT:
Setting $x^2=y,$
$$A=\frac{x^2}{x^4+x^2-2}=\dfrac y{y^2+y-2}=\dfrac y{(y+2)(y-1)}$$
As $y+2+2(y-1)=3y$
$$3A=\dfrac{y+2+2(y-1)}{(y+2)(y-1)}=\dfrac1{y-1}+\dfrac2{y+2}$$
Replace $y$ with $x^2$
