I am trying to solve the following integral but I am not sure where to begin.

$$\int{\frac{1}{\sqrt{x+y+z} + 1}}dx$$

I tried substituting $u = \sqrt{x+y+z}$ but I keep getting stuck.

How do I proceed?

  • $\begingroup$ Does your integral make sense? $\endgroup$ – Naive Dec 10 '17 at 17:02
  • $\begingroup$ @Naive when I do the substitution? $\endgroup$ – Omari Celestine Dec 10 '17 at 17:04
  • $\begingroup$ I made a mistake but I updated the question. The original equation is a triple integral but I am not trying to determine the indefinite integral of the equation with respects to $x$. $\endgroup$ – Omari Celestine Dec 10 '17 at 17:07
  • $\begingroup$ What does not work with your substitutution? It is a good sub $\endgroup$ – Shashi Dec 10 '17 at 17:15
  • $\begingroup$ $\frac{2u}{u+1}=2-\frac 2{u+1}$ and you integrate with a $\ln$. $\endgroup$ – zwim Dec 10 '17 at 17:19

$$\int\frac{1}{\sqrt{x+y+z}+1}\rm dx$$ Let $u={x+y+z}$

And $\rm du=\rm dx$

Your integral becomes $$\int \frac{1}{\sqrt{u}+1}\rm du$$

Let $w=\sqrt{u}+1,\;\;\;\rm dw=\frac{1}{2\sqrt{u}}\rm du=\frac{1}{2(w-1)}\rm du$

$$\int \frac{1}{\sqrt{u}+1}\rm du=2\int\frac{w-1}{w}dw=2(w-\ln|w|)=2(\sqrt{u}+1-\ln|\sqrt{u}+1|)=2(\sqrt{x+y+z}+1-\ln|\sqrt{x+y+z}+1|)$$


So I figured where I went wrong and thanks to those who commented I did the following:

$$ u = \sqrt{x+y+z} \\ du = \frac{1}{2}(x+y+z)^{-\frac{1}{2}}(1)dx = \frac{1}{2}\frac{1}{\sqrt{(x+y+z)}}dx = \frac{1}{2}\frac{1}{u}dx = \frac{1}{2u}dx \\ dx = 2u\ du $$


$$ \int\frac{1}{\sqrt{x+y+z}+1}dx = \int\frac{2u}{u+1} du = \int(2-\frac{2}{u+1})du \\ = \int2\ du - 2\int\frac{1}{u+1}du \\ = 2u-2\ \ln{(u+1)} \\ = 2\sqrt{x+y+z}-2\ln(\sqrt{x+y+z}+1) $$


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