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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?

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  • $\begingroup$ Thanks for editing LAcarguy. $\endgroup$ – joe May 28 '14 at 9:49
  • $\begingroup$ can you represent each x with a power? $\endgroup$ – Vikram May 28 '14 at 11:28
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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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Hint:

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

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$\dfrac{-3x}{\sqrt{x}} = -3\sqrt{x}$, and $\dfrac{x^2}{\sqrt{x}} = x^{\frac{3}{2}}$

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