Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $

Question

Any suggestions how to solve it? by parts? $$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$

Thanks!

Answer 1

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

Answer 2

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