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 $dx$?
 A: 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 $$\bbox[10px, border:solid 1px blue]{\int \frac{\sqrt{x}}{x^2} \, \mathrm{d}x = -\frac{2}{\sqrt{x}} + \mathrm{C}.}$$
A: Hint
If you do not want to work with fractional exponents :
$$I=\int {\sqrt x \over x^2}dx$$ Change variable $u=\sqrt x$, $x=u^2$, $dx=2u \,du$. All of that makes $$I=\int \frac{u}{u^4}2u\,du$$ I am sure that you can take from here.
A: $$ \int \frac{\sqrt x}{x^2}dx$$
You've already made $u= \sqrt x$, Hence $u^2 = x$
Now differentiate $u$ wrt. $x$.
$$ \frac{du}{dx} = \frac{1}{2 \sqrt x}$$
$$ dx = 2 \sqrt x \ du = 2u \ du$$
$$ \int \frac{\sqrt x}{x^2}dx = \int \frac{u}{u^4}2u \ du = \int 2 u^{-2} du \\ =   -2u^{-1} + c = -2x^{-\frac{1}{2}} + c $$
Alternatively: 
$$ \frac{\sqrt x}{x^2} = x^{-\frac{3}{2}}$$
$$\int  \frac{\sqrt x}{x^2} dx = \int x^{-\frac{3}{2}}dx =  -2x^{-\frac{1}{2}} + c
$$
I think the second method is a lot easier in my opinion (c is a constant of integration).
