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.
 A: 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
A: As sugested by other users:
$$\begin{align}
\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx&=\int\frac{\sqrt{x+8}(\sqrt {x-3}+\sqrt{x+3})}{(\sqrt{x-3}-\sqrt{x+3})(\sqrt{x-3}+\sqrt{x+3})}dx\\
&=\int\frac{\sqrt{x^2+5x-24}+\sqrt{x^2+11x+24}}{(x-3)-(x+3)}dx
\end{align}$$
Now let $t=\sqrt{x^2+5x-24}-x$.
Since 
$$\begin{align} 
 t=\sqrt{x^2+5x-24}-x &\Longrightarrow t+x=\sqrt{x^2+5x-24}\\
&\Longrightarrow x^2+2xt+t^2=x^2+5x-24\\
&\Longrightarrow 5x-2xt= t^2+24\\
&\Longrightarrow x(5-2t)=t^2+24\\
&\Longrightarrow x=\frac{t^2+24}{5-2t}\\
&\Longrightarrow dx=\frac{2t(5-2t)-(t^2+24)\cdot (-2)}{(5-2t)^2}dt\\
&\Longrightarrow dx=\frac{10t-4t^2+2t^2+48}{(2t-5)^2}dt\\
&\Longrightarrow dx=\frac{-t^2+5t+24}{2(2t-5)^2}dt
\end{align}$$
By sucessive modus ponens you get
$$\sqrt{x^2+5x-24}=x+t \space \wedge \space x=\frac{t^2+24}{5-2t} \space \wedge \space dx=\frac{-t^2+5t+24}{2(2t-5)^2}dt$$
Therefore 
$$\begin{align} \int \underbrace{\sqrt{x^2+5x-24}}_{\displaystyle x+t}dx&=\int \underbrace{\left(\frac{t^2+24}{5-2t}+t\right)}_{\displaystyle x+t}\cdot\underbrace{\frac{-t^2+5t+24}{2(2t-5)^2}dt}_{\displaystyle dx}
\end{align}$$
From here onwards it's a different method of integration, which you should be familiar with.
I would like to ask other users to proofread my calculations. Thanks.
A: 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:
$$\int\frac{\sqrt{x^2+5x-24}+\sqrt{x^2+11x+24}}{-6}=\int\frac{\sqrt{(x-3)(x+8)}+\sqrt{(x+3)(x+8)}}{-6}=$$
$$=\frac{-1}{6}\int\sqrt{(x-3)(x+8)}-\frac{1}{6}\int\sqrt{(x+3)(x+8)}$$
