Just having some trouble with this last problem of mine, its the derivative of the integral (u+5)/(u-8) from 9x to 3x
So the problem is
$\displaystyle \frac{d}{dx}\int_{9x}^{3x} \frac{u+5}{u-8} du$?
You should know that, by the "fundamental theorem of calculus", $\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)$
It is a little bit more complicated by the fact that the limits of integration are 9x and 3x but you can do this:
$\displaystyle \int_{9x}^{3x} \frac{u+5}{u-8}du= \int_0^{3x}\frac{u+5}{u-8}du- \int_0^{9x} \frac{u+5}{u-8}du$
And the fact that the upper limit is not just "x" can be handled by setting y= 3x in the first integral and y= 9x in the second:
The derivative of $\displaystyle \int_0^y \frac{u+5}{u-8}du$ [b]with respect to y[/tex] is just $\displaystyle \frac{y+5}{y-8}$ and the derivative with respect to x is (by the chain rule) that times the derivative of y with respect to x.