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::

$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$
$$=\int\frac{\sqrt{x+8}(\sqrt{x-3}+\sqrt{x+3})}{(\sqrt{x-3}-\sqrt{x+3})(\sqrt{x-3}+\sqrt{x+3})}dx$$
$$=\int\frac{\sqrt{x+8}\sqrt{x-3}}{6}dx+\int\frac{\sqrt{x+8}\sqrt{x+3}}{6}dx$$
$$=\int\frac{\sqrt{x^2+5x-24}}{6}dx+\int\frac{\sqrt{x^2+5x+24}}{6}dx$$
$$=\frac{I_1+I_2}{6}$$
$$I_1 = \int\sqrt{x^2+5x-24}$$
$$I_1= \int\sqrt{\left(x+\frac{5}{2}\right)^2-\frac{121}{4}}dx$$
Substitute $u = \left(x+\frac{5}{2}\right)dx$
$$I_1= \int\sqrt{u^2-\frac{121}{4}}du$$
Substitute $\frac{11\sec\theta}{2} = u;du = \frac{11\sec\theta\tan \theta}{2}d\theta$
$$I_1= \frac{121}{4}\int\tan ^2\theta  \sec \theta d\theta$$
$$I_1= \frac{121}{4}\left[\int\sec^3 \theta d\theta -  \int \sec \theta d\theta\right]$$
Using reduction formula
$$\int{sec^m \theta} = \frac{\sin \theta \sec ^{m-1} \theta}{m-1} + \frac{m-2}{m-1}\int\sec^{m-2}\theta d\theta$$

Substitute and Solve
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:
$$\int\frac{\sqrt{x^2+5x-24}+\sqrt{x^2+11x+24}}{-6}=\int\frac{\sqrt{(x-3)(x+8)}+\sqrt{(x+3)(x+8)}}{-6}=$$
$$=\frac{-1}{6}\int\sqrt{(x-3)(x+8)}-\frac{1}{6}\int\sqrt{(x+3)(x+8)}$$