# 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?

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Do you know a primitive of the function $x\mapsto1/\sqrt{x+12}$? And of the function $x\mapsto\sqrt{x+12}$? If so, you are done. If not, you could start with these. – Did May 1 '12 at 20:27
@Didier: Thanks, I found the solution by rewriting 1/sqrt( x + 12 ) as (x+12)^(-0.5) and applying the partial integration rule! – Wolf May 1 '12 at 20:42
Your substitution $u=\sqrt{x+2}$ works! You have $dx=2u$ $du$ and $$\begin{eqnarray*} \int \frac{x+8}{\sqrt{x+2}}dx &=&\int \frac{u^{2}-2+8}{u}\cdot 2u\; du \\ &=&\int \left( 12+2u^{2}\right)\; du \\ &=&12u+\frac{2}{3}u^{3} \\ &=&12\sqrt{x+2}+\frac{2}{3}\left( x+2\right) ^{3/2}+C. \end{eqnarray*}$$ – Américo Tavares May 1 '12 at 21:00
@Didier, Wolf Correction $$\begin{eqnarray*} \int \frac{x+8}{\sqrt{x+12}}dx &=&\int \frac{u^{2}-12+8}{u}\cdot 2u\; du \\ &=&\int \left( 2u^{2}-8\right) du \\ &=&\frac{2}{3}u^{3}-8u \\ &=&\frac{2}{3}\left( x+12\right) ^{3/2}-8\sqrt{x+12}. \end{eqnarray*}$$ – Américo Tavares May 1 '12 at 21:16
To expand on Didier's comment, Americo's last expression is equal to $$\frac{2}{3}x\sqrt{x+12} +C$$ Can't forget the constant of integration! – Joe May 1 '12 at 21:41

The way I approach these things is like this: with $u=x+12$, $$\int\frac{x+8}{\sqrt{x+12}}\,dx=\int\frac{x+12}{\sqrt{x+12}}-\frac4{\sqrt{x+12}}\,dx=\int{\sqrt{x+12}}-\frac4{\sqrt{x+12}}\,dx=\int{\sqrt{u}}-\frac4{\sqrt{u}}\,du=\frac{2u^{3/2}}3-8\sqrt u+C=\frac{2(x+12)^{3/2}}3-8\sqrt{x+12}+C =\frac23\,x\sqrt{x+12}+C$$
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*}