Originally Posted by

**Tweety** $\displaystyle \int \frac{2x}{3-x} dx $

I have tired two methods, which both give me wrong answers, here my working , if anyone could tell me where I am going wrong, it would be greatly appreciated.

method 1: using a substitution:

$\displaystyle u = 3-x $

$\displaystyle x = 3-u $

$\displaystyle dx = -du $

$\displaystyle \int - \frac{2(3-u)}{u} du $

$\displaystyle -\int \frac{6-2u}{u} du $ = $\displaystyle -\int \frac{6}{u}-2 \\\ du $ = $\displaystyle - 6ln (3-x) + 2(3-x) + c $

so the answer is $\displaystyle 6 - 2x -6ln(3-x) +c $

This is the wrong answer, but I can't see what I have dont wrong, if you anyone could kindly point out.

Thank you!