Calculate $$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx $$
My try:
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx = \left| {x + 1 = {u^2}} \right| = 2\int {\frac{{(u - \sqrt { - 2 + {u^2}} )}}{{u + \sqrt { - 2 + {u^2}} }}du} $$
I tried to do first Euler substitute:
$$\sqrt { - 2 + {u^2}} = {u_1} - u $$
But it did not lead me to the goal. Any thoughts will be appriciated.
Try
$$\left (\sqrt{x+1}+\sqrt{x-1} \right )\left (\sqrt{x+1}-\sqrt{x-1} \right ) = 2$$
Then the integral is
$$\frac12 \int dx \, \left (\sqrt{x+1}-\sqrt{x-1} \right )^2$$
which is
$$\int dx \left (x - \sqrt{x^2-1}\right ) $$
Can you take it from here?
$$\begin{eqnarray*}\int\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\,dx &=& \int\frac{2x-2\sqrt{x^2-1}}{2}\,dx\\ &=&\; C+\frac{x^2}{2}-\frac{x}{2}\sqrt{x^2-1}+\frac{1}{2}\log\left(x+\sqrt{x^2-1}\right).\end{eqnarray*}$$
