<s>[INST] Integral $\int \frac{x^2}{x^4+x^2+1}\ dx$ I have taken $(x^4+x^2+1)=u$ to differentiate it w.r.t. $x$ to get
$$\int \frac{x}{2u(2x^2+1)}\ du$$
 [/INST] $$((x^2+1)+x)((x^2+1) -x) = x^4 + 2x^2 + 1 - x^2 = x^4 + x^2 + 1$$
Then use partial fraction method.
Edit: More hints,
$$((x^2+1)+x) + ((x^2+1) -x) = 2(x^2+1)$$
$$\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$$
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