intergral of 3x/3x-2 dx
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Originally Posted by homerb intergral of 3x/3x-2 dx Note that $\displaystyle \displaystyle \frac{3x}{3x - 2} = \frac{(3x - 2) + 2}{3x - 2} = 1 + \frac{2}{3x - 2}$.
$\displaystyle \frac{3x}{3x-2}=1+\frac{2}{3x-2}$ now it's easy to integrate
So, after that do i do u= 3x-2 du= 3 dx? or what is the next step?
Originally Posted by homerb So, after that do i do u= 3x-2 du= 3 dx? [snip] Yes.
my answer comes to x+(2/3)ln[3x-2]+c, is that correct?
$\displaystyle x+\frac{2}{3}ln|3x-2|+c$ NOTE: differentiate and you can check your answer for yourself
Last edited by BAdhi; Mar 9th 2011 at 08:21 PM. Reason: code error
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