# How do I integrate $\int \frac{x+1}{2\sqrt{x+1}} \mathrm dx$?

I know how to do the bottom part, but I can't figure it out how to get $x+1$ on top

$$\int \frac{x+1}{2\sqrt{x+1}} \mathrm dx$$

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Do you really mean $$\int\frac{x+1}{2(x+1)^{-1}}\,dx\quad?$$ If so, then you are probably thinking about it too hard. – Clayton Feb 5 '13 at 22:05
Do you realize $\frac{x+1}{(x+1)^{-1}}=(x+1)^2$? – Alex Feb 5 '13 at 22:06
oops I've made a mistake. Shoulde be ∫x+1 / 2(x+1)^1 dx – helpty Feb 5 '13 at 22:08
@helpty: Recall the elementary fact that $x/x=1$. This will help you out. – Clayton Feb 5 '13 at 22:10
If your average daily consumption of alcohol is $x$ units, then you should not post here if you have consumed more than $x/2$ units of alcohol. I speak from experience. – TonyK Feb 5 '13 at 22:15

$$\int \frac {x+1}{2(x+1)^{1/2}} = \int\frac{x+1}{2\sqrt{x+1}}\,dx.$$ Let $u=(x+1)^{1/2}$, so $du=\dfrac{1}{2(x+1)^{1/2}}\,dx.\;$ Substitution gives us $$\int u^2\,du\;\;=\;\;\frac{u^3}{3}+C.$$

Back substitute, and our answer is $$\frac {((x+1)^{1/2})^3}{3} \;=\;\frac{(x+1)^{3/2}}{3}+C.$$

This can be approached in a much more straightforward manner by noting that:

$$\frac{x+1}{2(x+1)^{1/2}}\;= =\;\frac12(x+1)^{ 1 - (1/2)}\;=\frac12(x+1)^{1/2}$$

Then since $\dfrac{d}{dx}(x+ 1) = 1$, we simply integrate: $$\int \frac12(x+1)^{1/2} \,dx \;\;=\;\; \frac12\frac{(x+1)^{3/2}}{\frac32} + C \;\;= \;\;\frac{(x+1)^{3/2}}{3} + C$$

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thank you, guys. Sorry for the oversight I had. – helpty Feb 5 '13 at 22:18
Does this make sense now? – amWhy Feb 5 '13 at 22:19
yes. I see it. thank you. – helpty Feb 5 '13 at 22:20
This answer is far more complicated than it needs to be... look at Mike's before. You just use $\displaystyle \frac{a}{\sqrt{a}}=\sqrt{a}$. – Jp McCarthy Feb 5 '13 at 23:12

If you mean $$\int\frac{x+1}{2(x+1)^{-1}}\,dx,$$ then we can reduce this to $$\frac{1}{2}\int(x+1)^2\,dx,$$ which is easy to integrate. If you mean $$\int\frac{x+1}{2(x+1)}\,dx$$ then we can cancel $x+1$ and we have $$\int\frac{1}{2}\,dx=\frac{x}{2}+C.$$ $\textbf{Update}$: The correct integral is (hopefully) $$\int\frac{x+1}{2\sqrt{x+1}}\,dx.$$ Set $u=(x+1)^{1/2}$, and $du=\frac{1}{2(x+1)^{1/2}}\,dx$. Making the substitution gives us $$\int u^2\,du=\frac{u^3}{3}+C.$$ Substitute back in, and we have as our final answer $$\frac{(x+1)^{3/2}}{3}+C.$$

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$$\int\frac{x+1}{2(x+1)^\frac12}dx=\int\frac12(x+1)^\frac12dx=\frac12\times\frac{(x+1)^\frac32}{\frac32}+C=\frac13(x+1)^{\frac32}+C$$

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This is the only sensible solution here. It's just a simple manipulation and substitution... well most people like Mike here wouldn't even bother with an explicit substitution. – Jp McCarthy Feb 5 '13 at 23:11
@JpMcCarthy In many cases, I would write out the substitution just to be safe, but in cases where $du=dx$, it's as simple as it gets. – Mike Feb 5 '13 at 23:18
Sorry my tone is a little off; please replace bother with "do". – Jp McCarthy Feb 5 '13 at 23:29