int(2x/(x-3)^2) dx ..??? not sure how to do this.
ok ya your right. makin sense but i get down to it and the real question was integral from 0 to 2 of (2x/(x-3)^2) so u=x-3 du=dx 2x=2u+6 and i get..... int from 0 to 2 ((2u +6)/u du which is int from 0 to 2 of (2u/u) du + int from 0 to 2 of (6/u) du and then that is 4 - 6 ln |x - 3 | from 0 to 2 and i get 4 - 6 ln3 ... but the back of the book says 4-ln9 which is a different answer... so did i go wrong somehwerE?
so there are two ways to go here. it is mainly a matter of personal preference. i am the kind of guy that, unless it causes way too much trouble, leaves the limits in tact in terms of x. when i integrate in terms of u i just rewrite the answer in terms of x and use the old limits
the other option is to change the limits so that you can use them with u. in that case, you use your substitution equation.
we have u = x - 3
so when x = 0, u = -3
when x = 2, u = 1
so your limits in u is from -3 up to 1, not 0 up to 2
$\displaystyle \int \frac {2x}{(x - 3)^2}~dx$
Let $\displaystyle u = x - 3 \implies x = u + 3$
$\displaystyle \Rightarrow du = dx$
So our integral becomes:
$\displaystyle \int \frac {2u + 6}{u^2}~du$
$\displaystyle = \int \left( \frac 2u + 6u^{-2} \right)~du$
$\displaystyle = 2 \ln |u| - \frac 6u + C$
the problem, i believe, is you divided by u rather than u^2