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

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