<s>[INST] Evaluating $\int \frac{x+8}{\sqrt {x+12}}dx$ How should I go about integrating the function $$\frac{x+8}{\sqrt {x+12}}$$
I have tried substituting $u = \sqrt{ x + 12 }$, but that leads me nowhere...
Could somebody possibly just tell me which steps have to be followed in order to evaluate this integral?
 [/INST] Let $u = \sqrt{x+12}$ and $du = \frac{1}{2\sqrt{x+12}} dx$
$$
\begin{align*}
\int \frac{x+8}{\sqrt{x+12}} \hspace{3pt}dx &= \int  {\hspace{3pt}\frac{u^2-12+8}{u}}\cdot2u\hspace{3pt} du\\
&= \int  {\hspace{3pt}2u^2-8}\hspace{3pt} du\\
&= 2\int  {\hspace{3pt}u^2-4}\hspace{3pt} du\\
&= 2\cdot \frac{u^3}{3} -8u  {\hspace{3pt}}\\
&= \frac{2}{3}{(x+2)^{3/2}} -8\sqrt{x+12}  {\hspace{3pt}}\\
&= \frac{2}{3}x\sqrt{x+12} +C\\
\end{align*}
$$
</s>