<s>[INST] Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $ Any suggestions how to solve it? by parts?
$$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$
Thanks!
 [/INST] $$\dfrac{x^4}{x^2+1}=\dfrac{(x^4+2x^2+1)-2x^2-1}{x^2+1}=\\
=\dfrac{(x^2+1)^2-2x^2-2+1}{x^2+1}=x^2+1-2+\dfrac{1}{x^2+1}=x^2-1+\dfrac{1}{x^2+1}.$$
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