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

MikeNZ · #1 $Old$ April 28th, 2009, 07:35 PM

How would I transpose this formula so I can integrate it? I'm not good with integrating fractions or square roots.

Integrate:

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

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

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: $x^\frac{1}{2}$ ? Or something along those lines..

Thanks in advance.

chengbin · #2 $Old$ April 28th, 2009, 07:52 PM

Am I reading this right?

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

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

That's easy to integrate.

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

MikeNZ · #3 $Old$ April 28th, 2009, 07:59 PM

Quote:

Originally Posted by chengbin $View Post$

Am I reading this right?

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

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

That's easy to integrate.

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

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 $\frac{x^2}{2}$ ?

chengbin · #4 $Old$ April 28th, 2009, 08:28 PM

$\int 2\sqrt x = \int 2x^\frac {1}{2}$

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

$\frac {4}{3} x^\frac {3}{2} = \frac {4}{3} x(x^\frac {1}{2}) = \frac {4}{3}x\sqrt x$

MikeNZ · #5 $Old$ April 28th, 2009, 08:48 PM

Ohh yip, thanks