Finding the indefinite integral $\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$ This is a homework question, I tried many subtitutions but nothing worked for me...
$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$
Any clue will help.
Thanks.

Hint: If you multiply an divide by $\sqrt{x-3}+\sqrt{x+3}$, the result will be this, getting rid of the fraction and obtainin just two integrals of a single square root. That may be easier to solve:
$$\int\frac{\sqrt{x^2+5x-24}+\sqrt{x^2+11x+24}}{-6}=\int\frac{\sqrt{(x-3)(x+8)}+\sqrt{(x+3)(x+8)}}{-6}=$$
$$=\frac{-1}{6}\int\sqrt{(x-3)(x+8)}-\frac{1}{6}\int\sqrt{(x+3)(x+8)}$$
