How do you integrate (x)/(x-1)

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Homework Statement


How do you integrate S x/(x-1) dx


Homework Equations



IBP

The Attempt at a Solution


I tried using Integration by Parts with u= x/(x-1) and dv = dx and that did not work out. Then I tried using u= 1/(x-1) and dv = xdx but that did not work either.

A google search said "Just use the substitution u = x+1, then replace dx with du and you get u+1/u = 1 + 1/u which you can integrate to give u + ln u, thus = x-1 + ln(x-1)". But I am pretty sure that you can't just add +1 to the numerator like that.

I also know that it can be done using tables but is there a way not to use tables? Did I do my integration wrong?

Thanks
 
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Hint Hint:
[itex]\frac{1}{x-1}[/itex]+[itex]\frac{x-1}{x-1}[/itex]=[itex]\frac{x}{x-1}[/itex]
 
You can even do it without substitution by changing the equation to (x)/(x-1) = (x-1+1)/(x-1) = 1+1/(x-1), then integrating to get x+ln(x-1)+C.

Wow! All 3 posts within a minute!
 
Perhaps this will make it clearer,

Suppose [itex]u=x-1[/itex]. Then clearly [itex]du=dx[/itex]. But notice also that [itex]u+1=x[/itex] by the first equation. Therefore, the integral becomes,

[itex]\int \frac{u+1}{u}du[/itex]

Now, it is trivial. :)
 
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daveb said:
Wow! All 3 posts within a minute!

We're all just so stoked about integration, we can't resist. If only all students shared our passion for integration... ;)