Integral fraction equation I'm learning calculus and I'm trying to do the below integral (with working). 
$$\int {\sqrt x \over x^2}dx$$
So I start by making $u=\sqrt x$ and $x^2=u \cdot u^3$
$$\int {\sqrt u \over u \cdot u^3}dx$$
So I'm a bit lost with the next step. What do I do with $dx$?

Another way you could tackle this problem would be to notice that $$\frac{\sqrt{x}}{x^2} = \frac{x^{1/2}}{x^2} = x^{1/2 - 2} = x^{-3/2}$$
So that you end up integrating $$\int \frac{\sqrt{x}}{x^2} \, \mathrm{d}x = \int x^{-3/2} \, \mathrm{d}x$$
A simple application of the following (as long as $\alpha \neq -1$)$$\int x^{\alpha} \, \mathrm{d}x = \frac{x^{\alpha + 1}}{\alpha + 1}$$ with $\alpha = -3/2$ yields $$\int x^{-3/2} \, \mathrm{d}x = \frac{x^{-3/2 + 1}}{-3/1 + 1} = \frac{x^{-1/2}}{-1/2} = -\frac{2}{x^{1/2}} = -\frac{2}{\sqrt{x}} + \mathrm{C}$$
Hence you end up with $$\bbox[10px, border:solid 1px blue]{\int \frac{\sqrt{x}}{x^2} \, \mathrm{d}x = -\frac{2}{\sqrt{x}} + \mathrm{C}.}$$
