 skynugurq7

2022-07-13

Simplifying fraction with square root as denominator
I'm trying to find the integral of:
$\frac{2\sqrt{x}-3x+{x}^{2}}{\sqrt{x}}$
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:
$\frac{2\sqrt{x}}{\sqrt{x}}$
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 $\frac{-3x}{\sqrt{x}}$ I know to $-\sqrt{x}$ but i don't know what $-3x-\sqrt{x}$ would come out with? Janiyah Patton

Expert

HINT :
Rewrite
$\frac{2\sqrt{x}-3x+{x}^{2}}{\sqrt{x}}=\frac{2\sqrt{x}-3\sqrt{x}\sqrt{x}+x\cdot x}{\sqrt{x}}=\frac{2\sqrt{x}-3\sqrt{x}\sqrt{x}+\sqrt{x}\sqrt{x}\cdot x}{\sqrt{x}}$
or
$\frac{2\sqrt{x}-3x+{x}^{2}}{\sqrt{x}}=\frac{2{x}^{\frac{1}{2}}-3{x}^{1}+{x}^{2}}{{x}^{\frac{1}{2}}},$
where $x=\sqrt{x}\sqrt{x}$ and $\sqrt{x}={x}^{\frac{1}{2}}$ Ayaan Barr

Expert

Hint:
$\frac{2\sqrt{x}-3x+{x}^{2}}{\sqrt{x}}=2-3\sqrt{x}+{x}^{3/2}$

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