Using the formula

$\displaystyle UV-\int VdU$

I come up with

$\displaystyle (\frac 1x) (\frac13 x^3) - \frac13 \int \frac{x^3}{x^2}$

Here is where I get stuck with the integration, I think I can move the

$\displaystyle x^2$

up top and make it

$\displaystyle x^-2$

So I am left with

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

Above is Supposed to be a negative 2, but couldnt find the code for it.

So if I use u sub

$\displaystyle U=x^3 \ dU=3x^2$

The 3 I can handle but not sure what to do about the fact that my exponent is negative, and my dU has a positive.

Any help?

Thanks

Rich