Help calculating this integral

Question

Prove this for every $n>1$ (belongs to $\mathbb{N}$ )

$$\displaystyle \int_{0}^{1}\left( \frac{x^{2n+3} - x^{2n+1}}{1+x} \right) \, \mathrm{d}x =\frac{1}{2n+3} - \frac{1}{2n+2}$$

I don't see how one can obtain this.

thanks answered ...

Answer 1

Let's factor the top:

$x^{2n+3} = x^{2n+1}x^2$ Then, the numerator becomes, $x^{2n+1}(x^2-1) = x^{2n+1}(x+1)(x-1)$

Now the integration should be straightforward.