Integration by substitution

jeng

New member
Joined
Apr 13, 2006
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6
How do you integrate from 5 to 9: x/(x-3) dx ??

So far I've done this:

Let u=x-3 x=u+3
du=dx

=integral of 1/u (u+3) du

I can't go any further than that..help please..

thanks
 
Because you've chosen the u-substitution route (hitherto quite perfectly), Jeng, the next step could be to change the x-limits (5 to 9) to u-limits(? to ?), and integrate (1/u)(u + 3) (expand the parantheses and integrate term-by-term), agreed?


Once you've evaluated that, you might like to reflect on what you've done. x/(x - 3) can be written ((x - 3) + 3)/(x - 3), right?
 
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