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!

$$\text{Note that, we have }\dfrac{x^4}{x^2+1} = \dfrac{x^4-1}{x^2+1}+\dfrac1{x^2+1} = x^2-1 + \dfrac1{x^2+1}$$
$$\text{Hence, }\int \dfrac{x^4}{x^2+1} dx = \int (x^2-1) + \int \dfrac1{x^2+1} = \dfrac{x^3}3 - x + \arctan(x) + \text{ constant}$$
