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I am trying to solve $\int\frac{7}{\sqrt{x}(x+4)}~\mathrm{d}x$. So far I have $$7\int\frac{1}{\sqrt{x}(x+4)}~\mathrm{d}x$$ $$u=\sqrt{x}$$$$\mathrm{d}u=\frac{1}{2\sqrt{x}}$$ and this is where I'm not sure of what to do. This is what I tried. $$2\mathrm{d}u=\sqrt{x}$$ I am trying to solve $\int\frac{7}{\sqrt{x}(x+4)}~\mathrm{d}x$. So far I have $$14\int\frac{1}{(x+4)}~\mathrm{d}u$$ But I haven't gotten further since I'm not sure what to do. How do I go about solving this integral? Thanks in advance for all the help!

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  • $\begingroup$ In the last step put $x=u^2$. Now does the integral have any similarity to $\arctan$? $\endgroup$ – jdoicj Aug 3 '14 at 5:04
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    $\begingroup$ Yeah, so I can Just manipulate u like that and put in for that x? If so, I think I understand it now. $\endgroup$ – Kenshin Aug 3 '14 at 5:06
  • $\begingroup$ Is it $\frac{1}{\sqrt{x}(x+4)}$ or $\frac{1}{\sqrt{x(x+4)}}$? $\endgroup$ – André Nicolas Aug 3 '14 at 5:09
  • $\begingroup$ The first one is the one I'm dealing with. $\endgroup$ – Kenshin Aug 3 '14 at 5:11
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    $\begingroup$ OK. That is clear from the edited version, but was less clear before. $\endgroup$ – André Nicolas Aug 3 '14 at 5:22
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Hint: $I = \displaystyle 14\int \dfrac{d(\sqrt{x})}{(\sqrt{x})^2 + 4}$

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  • $\begingroup$ Thanks a lot! I understand now. $\endgroup$ – Kenshin Aug 3 '14 at 5:07
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With your change of variable we have

$$du=\frac{dx}{2\sqrt x}\implies dx=2udu$$ so the integral becomes

$$14\int\frac{du}{u^2+4}$$ Can you take it from here?

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  • $\begingroup$ Yes, thanks a lot! $\endgroup$ – Kenshin Aug 3 '14 at 5:07
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Aug 3 '14 at 5:07
  • $\begingroup$ @Sami Could you lend an "undelete" vote on this question? Thanks! $\endgroup$ – Namaste Aug 4 '14 at 12:14
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By substituting $x=4y^2$ you have $dx = 8y\,dy$ and: $$ I = 7\int\frac{dx}{(x+4)\sqrt{x}}=7\int\frac{dy}{y^2+1}=7\arctan(y)+C=7\arctan\sqrt{\frac{x}{4}}+C.$$

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