# Homework Help: How do you integrate (x)/(x-1)

1. Aug 29, 2011

### amanda_ou812

1. The problem statement, all variables and given/known data
How do you integrate S x/(x-1) dx

2. Relevant equations

IBP

3. 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

2. Aug 29, 2011

### micromass

You don't add +1 to the numerator. We do the substitution u=x-1. What happens if you do that substitution??

3. Aug 29, 2011

### Ivan92

Hint Hint:
$\frac{1}{x-1}$+$\frac{x-1}{x-1}$=$\frac{x}{x-1}$

4. Aug 29, 2011

### daveb

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!

5. Aug 29, 2011

### lineintegral1

Perhaps this will make it clearer,

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

$\int \frac{u+1}{u}du$

Now, it is trivial. :)

6. Aug 29, 2011

### lineintegral1

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