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 $x=\sqrt(u+\frac{1}{2})$
$\int\frac{x^2+1}{x^4-x^2+1}dx = \int\frac{u+\frac{1}{2}+1}{u^2+\frac{3}{4}}du
                             =\int\frac{u}{u^2+\frac{3}{4}} +\frac{\frac{3}{2}}{u^2+\frac{3}{4}}du$

Split it :
$\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int\left(\frac{3 x^{2}}{x^{6} + 1} + \frac{1}{x^{2} + 1}\right)dx$
Then remember that $\arctan(x)'=\frac{1}{1+x^2}$
