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
$$\int \frac{x+1}{2\sqrt{x+1}} \mathrm dx$$
 A: $$\int \frac {x+1}{2(x+1)^{1/2}} = \int\frac{x+1}{2\sqrt{x+1}}\,dx.$$ 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: $$\int \frac12(x+1)^{1/2} \,dx \;\;=\;\; \frac12\frac{(x+1)^{3/2}}{\frac32} + C \;\;= \;\;\frac{(x+1)^{3/2}}{3} + C$$
A: 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 $$\frac{(x+1)^{3/2}}{3}+C.$$
A: $$\int\frac{x+1}{2(x+1)^\frac12}dx=\int\frac12(x+1)^\frac12dx=\frac12\times\frac{(x+1)^\frac32}{\frac32}+C=\frac13(x+1)^{\frac32}+C$$
