<s>[INST] 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
 [/INST] Here is a route.
$$
\begin{align}
\int\frac{1}{\sqrt{x+1} + \sqrt{x}} {\rm d}x &=\int\frac{1}{(\sqrt{x+1} + \sqrt{x})} \times \frac{(\sqrt{x+1} - \sqrt{x})}{(\sqrt{x+1} - \sqrt{x})} {\rm d}x\\\\
&=\int \left(\sqrt{x+1} - \sqrt{x}\right) {\rm d}x \\\\
& = \int(x+1)^{1/2} {\rm d}x-\int x^{1/2} {\rm d}x\\\\
& =\frac{1}{1+1/2}(x+1)^{1+1/2} -\frac{1}{1+1/2}x^{1+1/2} +C\\\\
& =\frac 23 (x+1)^{3/2}-\frac 23 x^{3/2}+C\\\\
\end{align}
$$
where $C$ is any constant.
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