How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$ I have started this problem but I'm not completely sure I'm going down the right path with it.
So far I have completed the square in the denominator.
$x^4-x^2+1= (x^2-1/2)^2+\frac{3}{4}$
Then, let $u=x^2-\frac{1}{2}$   so $x=\sqrt(u+\frac{1}{2})$
$\int\frac{x^2+1}{x^4-x^2+1}dx = \int\frac{u+\frac{1}{2}+1}{u^2+\frac{3}{4}}du
                             =\int\frac{u}{u^2+\frac{3}{4}} +\frac{\frac{3}{2}}{u^2+\frac{3}{4}}du$

$$\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\dfrac1{x^2}}$$
Now $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x$
and $\displaystyle x^2-1+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2-1$
Hope you can take it from here
