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.
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 $\dfrac{1}{x^2}$.
So here you see it really depends how you like seeing it.
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.