Prove $$\int\_0^1 \frac{x^\sqrt{2}}{1+x} dx= \frac{1}{\sqrt{2} + 1} - \frac{1}{\sqrt{2}+2} + \frac{1}{\sqrt{2}+3}-\frac{1}{\sqrt{2}+4}+\ldots\,.$$

I know that $\sum\_{k=0}^\infty (-1)^k x^{\sqrt{2}+k} = \frac{x^\sqrt{2}}{1+x}$ is a uniform convergence for $|x| < 1$. But the integral range includes $x = 1$. How can I approach it?