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 $x=\sqrt(u+\frac{1}{2})$
$\int\frac{x^2+1}{x^4-x^2+1}dx = \int\frac{u+\frac{1}{2}+1}{u^2+\frac{3}{4}}du
                             =\int\frac{u}{u^2+\frac{3}{4}} +\frac{\frac{3}{2}}{u^2+\frac{3}{4}}du$
 A: $$\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\dfrac1{x^2}}$$
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
A: 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 $\arctan(x)'=\frac{1}{1+x^2}$
A: As Hassan Muhammad commented, you have a mistake. Since you set $x=\sqrt{u+\frac{1}{2}}$, you then have $dx=\frac{1}{2 \sqrt{u+\frac{1}{2}}}$ and then $$I=\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int \frac{2 u+3}{\sqrt{u+\frac{1}{2}} \left(4 u^2+3\right)}du$$ which is not very nice.
If you use Hippalectryon's idea, you end with $$I=\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int\left(\frac{3 x^{2}}{x^{6} + 1} + \frac{1}{x^{2} + 1}\right)dx=\tan ^{-1}\left(x^3\right)+\tan ^{-1}\left(x\right)$$ and if you apply the formula $$\tan ^{-1}\left(a\right)+\tan ^{-1}\left(b\right)=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)$$ you have, after simplifications, $$I=\tan ^{-1}\left(\frac{x}{1-x^2}\right)$$ which is result to which lab bhattacharjee's answer is taking you to.
