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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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    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.2012-05-01
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    @Didier: Thanks, I found the solution by rewriting 1/sqrt( x + 12 ) as (x+12)^(-0.5) and applying the partial integration rule!2012-05-01
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    Your substitution $u=\sqrt{x+2}$ works! You have $dx=2u$ $du$ and $$\begin{eqnarray*} \int \frac{x+8}{\sqrt{x+2}}dx &=&\int \frac{u^{2}-2+8}{u}\cdot 2u\; du \\ &=&\int \left( 12+2u^{2}\right)\; du \\ &=&12u+\frac{2}{3}u^{3} \\ &=&12\sqrt{x+2}+\frac{2}{3}\left( x+2\right) ^{3/2}+C. \end{eqnarray*}$$2012-05-01
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    Wolf: Yes, and likewise for $\sqrt{x+12}=(x+12)^{1/2}$, with a primitive which is a multiple of $(x+12)^{3/2}$. Now you could write down a complete solution of your question and post it here (yes, such a practice is encouraged on the site).2012-05-01
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    @Américo: $x+12$, not $x+2$.2012-05-01
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    @Didier, Wolf *Correction* $$\begin{eqnarray*} \int \frac{x+8}{\sqrt{x+12}}dx &=&\int \frac{u^{2}-12+8}{u}\cdot 2u\; du \\ &=&\int \left( 2u^{2}-8\right) du \\ &=&\frac{2}{3}u^{3}-8u \\ &=&\frac{2}{3}\left( x+12\right) ^{3/2}-8\sqrt{x+12}. \end{eqnarray*}$$2012-05-01
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    @Américo: In view of my last comment to the OP, I am not sure that posting a comment with a full solution is such a good idea. Additionally, you missed a notable simplification in your last expression.2012-05-01
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    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!2012-05-01
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    @Didier, Jay Electronica Thanks! I will not post a solution.2012-05-01
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    @AméricoTavares I had just posted an answer before I read your comment. Want me to delete it? You posted the answer first in the comments.2012-05-01
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    @JayElectronica No. That's fine.2012-05-01
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    By the way to the OP, next time you try substituting something (or any problem in general), feel 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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