Calculus 2 Integral of$ \frac{1}{\sqrt{x+1} +\sqrt x}$ 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

Hint: Use that ${1 \over \sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$.
