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

Question

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

Answer 1

Hint: instead of using a substitution, just rationalise the denominator in the original integral.

If you would really like to use a substitution, try $x=\sinh^2\theta$ and then reduce everything to exponentials.

If you actually want to evaluate your new integral, probably the easiest way is to go back to the original one. But $u=\cosh\theta$ should also work.

Answer 2

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

Answer 3

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.