Need Help Solving this Indefinite Integral

Question

I was viewing past questions of the MIT Integration Bee and came across the problem:

$$ \int \frac{x^2+1}{x^4-x^2+1}\,dx$$

According to MIT, first semester calculus should be sufficient enough to solve this problem.

I have tried many conventional methods, unsuccessfully, and would appreciate any and all help.

Edit: Changed LaTeX formatting

Answer 1

Maybe not as elegant but more intuitive, there's always partial fractions.

$$\int\frac{x^2+1}{x^4-x^2+1}dx=\int\frac{x^2+1}{x^4+2x^2+1-3x^2}dx=\int\frac{x^2+1}{(x^2+1+x\sqrt3)(x^2+1-x\sqrt3)}dx=$$ $$\frac12\int\frac{dx}{(x+\frac{\sqrt3}2)^2+\frac14}+\frac12\int\frac{dx}{(x-\frac{\sqrt3}2)^2+\frac14}=$$ $$\int\frac{2dx}{(2x+\sqrt3)^2+1}+\int\frac{2dx}{(2x-\sqrt3)^2+1}=$$ $$\tan^{-1}(2x+\sqrt3)+\tan^{-1}(2x-\sqrt3)$$

Answer 2

$$F=\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\frac1{x^2}}$$

Observe that $\displaystyle\int\left(1+\dfrac1{x^2}\right)=x-\frac1x$

So,$\displaystyle F=\frac{1+\dfrac1{x^2}}{\left(x-\frac1x\right)^2+2-1}$