<s>[INST] 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.
 [/INST] 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.
</s>