How would you find $$\int\frac{1}{\sqrt{x+1} + \sqrt x} dx$$

I used $u$-substitution and got this far:

$u = \sqrt{x+1}$ which means $(u^2)-1 = x$

$du = 1/(2\sqrt{x-1}) dx = 1/2u dx$ which means $dx = 2udu$

That means the new integral is $$\int \frac{2u}{u + \sqrt{u^2-1}}du$$

What technique do I use to solve that new integral?

Thanks