# Getting rid of a denominator

this is a bit of a dumb question and I haven't really found anything that tells me the general rule or answer for this.

So when you have a fraction say $1/x^2.$ If I wanted to write this with a denominator of $1,$ (basically I want to bring $x^2$ to the numerator when I do simple integrating questions). I thought it would be $x^{-2}$, but that's incorrect. Why?I thought you just make the denominator with a negative power and multiply it by the numerator.That's what I did below and its correct.

Conversely If I have $(x+1)/ x^{1/2}$ it would be $(x+1)\cdot x^{-1/2}.$ How is this different from the one above? Is it because it's a function on the top?

Edit: Could it be that i'm right and i'm integrating wrong. I was trying to integrate $1/x^2.$ SO I thought thats equal to $x^-2$. which is $x^{-1}/-1$?

Edit: So sorry I got what I was doing wrong, I was comparing my answer to the book, which put the integrand back in fraction form, which confused me as to why my answer was wrong.

• Both are correct. I don't understand your confusion. $1/x^2 = x^{-2}$, – B. Goddard Apr 10 '17 at 20:10

It really depends how you see it. Your answer of $x^{-2}$ is correct and incorrect. For example, $x^{-2}$ can be wrriten as $\dfrac{x^{-2}}{1}$, which is what you want.
It is incorrect too because $x^{-2}$ is just another way of writing $\dfrac{1}{x^2}$.
The general rule is that $everything$ is a fraction, that has a denominator $=1$, but we just don't put it. So $\dfrac{1}{x^2}$ is actually $\dfrac{1}{\dfrac{x^2}{1}}$, so it's really the same thing. Remember a $negative$ exponent is another way to get your variable to the $top$.
For example, $\dfrac{x+1}{x^{1/2}}$, we can rewrite the $x^{1/2}$ as $x^{-1/2}$, since $x^{-1/2}=\dfrac{1}{x^{1/2}}$. Hope this clears things up for you.