We want to compute, for any $n \in \mathbb{N}$ the following integral: $$\int_{-1}^{1} \frac{x^n}{\sqrt[n]{1+x}+\sqrt[n]{1-x}}dx$$
My attempt: if $n$ is odd, the integral is trivially equal to $0$ since $x^n/(\sqrt[n]{1+x}+\sqrt[n]{1-x})$ is itself an odd function. If $n$ is even, so does $x^n/(\sqrt[n]{1+x}+\sqrt[n]{1-x})$ and therefore $$\int_{-1}^{1} \frac{x^n}{\sqrt[n]{1+x}+\sqrt[n]{1-x}}dx=2\int_{0}^{1} \frac{x^n}{\sqrt[n]{1+x}+\sqrt[n]{1-x}}dx$$ but this is not really helping. Any ideas? Thanks