<s>[INST] Simplifying fraction with square root as denominator I'm trying to find the integral of:
$$\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}$$
but I first need to simplify it so I tried dividing by the $\sqrt{x}$ for each of the numbers on the top like so:
$$\dfrac {2\sqrt{x}}{\sqrt{x}}$$
and did the same for the others. For the one above it was easy to see that it just simplifies to $2$. But I am unsure how to do the same for the others for instance $\dfrac {-3x}{\sqrt{x}}$. I know to $-\sqrt{x}$ but i don't know what $-3x  - \sqrt{x}$ would come out with?
 [/INST] HINT :
Rewrite
$$
\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}=\dfrac {2\sqrt{x} - 3\sqrt{x}\sqrt{x} + x\cdot x}{\sqrt{x}}=\dfrac {2\sqrt{x} - 3\sqrt{x}\sqrt{x} + \sqrt{x}\sqrt{x}\cdot x}{\sqrt{x}}
$$
or
$$
\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}=\dfrac {2x^{\large\frac12} - 3x^1 + x^2}{x^{\large\frac12}},
$$
where $x=\sqrt{x}\sqrt{x}$ and $\sqrt{x}=x^{\large\frac12}$.
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