# Integrating a fraction/square roots when 'x' is on the bottom

• Apr 28th 2009, 05:35 PM
MikeNZ
Integrating a fraction/square roots when 'x' is on the bottom
How would I transpose this formula so I can integrate it? I'm not good with integrating fractions or square roots.

Integrate:

Q1) $\displaystyle \frac{dy}{dx}$=$\displaystyle \frac{x^3-2x}{x}$

Q2) $\displaystyle (\sqrt{x}+1)^2$

It's the fractions and square roots that are annoying. For the square root, I know I'd have to write two brackets out and possibly expand, but don't know how I would expand with a square root - So does the 'x' in the square root become: $\displaystyle x^\frac{1}{2}$? Or something along those lines..

• Apr 28th 2009, 05:52 PM
chengbin

$\displaystyle \frac{x^3-2x}{x}$

If that's right, just divide it, then it becomes $\displaystyle x^2-2$

That's easy to integrate.

For Q2, just expand it, and you'll get $\displaystyle x+2\sqrt x+1$ Integrating that is just $\displaystyle \frac {1}{2} x^2+\frac {4}{3} x\sqrt x+x$
• Apr 28th 2009, 05:59 PM
MikeNZ
Quote:

Originally Posted by chengbin

$\displaystyle \frac{x^3-2x}{x}$

If that's right, just divide it, then it becomes $\displaystyle x^2-2$

That's easy to integrate.

For Q2, just expand it, and you'll get $\displaystyle x+2\sqrt x+1$ Integrating that is just $\displaystyle \frac {1}{2} x^2+\frac {4}{3} x\sqrt x+x$

Oh ok, yep Q1 is easy to Integrate..

Just quickly, for Q2, where did you get the 4 over 3 from? and if you integrated 'x' by itself, wouldn't you get $\displaystyle \frac{x^2}{2}$?
• Apr 28th 2009, 06:28 PM
chengbin
$\displaystyle \int 2\sqrt x = \int 2x^\frac {1}{2}$

$\displaystyle \frac{2}{\frac{3}{2}}x^\frac {3}{2}$

$\displaystyle \frac {4}{3} x^\frac {3}{2} = \frac {4}{3} x(x^\frac {1}{2}) = \frac {4}{3}x\sqrt x$
• Apr 28th 2009, 06:48 PM
MikeNZ
Ohh yip, thanks :)