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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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    By the way to the OP, next time you try substituting something (or any problem in general), $f$eel free to show your work so we can specifically see where you went wrong (or even if you did!) which will benefit you greatly and prevent us from telling you something you already know. Also, just a minor note, one *evaluates* integrals - not *solve* generally.2012-05-01

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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 $

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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*} $