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

Question

I have taken $(x^4+x^2+1)=u$ to differentiate it w.r.t. $x$ to get $$\int \frac{x}{2u(2x^2+1)}\ du$$

Answer 1

$$((x^2+1)+x)((x^2+1) -x) = x^4 + 2x^2 + 1 - x^2 = x^4 + x^2 + 1$$

Then use partial fraction method.

Edit: More hints,

$$((x^2+1)+x) + ((x^2+1) -x) = 2(x^2+1)$$

$$\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$$

Answer 2

$$\dfrac{2x^2}{x^4+x^2+1}=\dfrac{x^2+1}{x^4+x^2+1}+\dfrac{x^2-1}{x^4+x^2+1}$$

$$=\dfrac{1+\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}+\dfrac{1-\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}$$

For the first integral:

As $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x,$

write $x^2+1+\dfrac1{x^2}=1+\left(x-\dfrac1x\right)^2+2$

Can you handle the second integral?

Answer 3

If you are interested in the integral over $\mathbb{R}^+$, Glasser's Master Theorem allows a rapid evaluation:

$$ \int_{0}^{+\infty}\frac{x^2\,dx}{x^4+x^2+1}\stackrel{\text{parity}}{=}\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+3}\stackrel{\text{G.M.T}}{=}\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dx}{x^2+3}=\color{blue}{\frac{\pi}{2\sqrt{3}}}.$$