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

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Homework Help Overview

The discussion revolves around the integration of the function x/(x-1) with respect to x. Participants are exploring various methods to approach the integral, including integration by parts and substitution techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts integration by parts but finds it unproductive. They also explore substitution methods but express uncertainty about the validity of adding 1 to the numerator. Other participants suggest using the substitution u=x-1 and provide hints about rewriting the integrand to facilitate integration.

Discussion Status

Participants are actively engaging with the problem, offering alternative approaches and clarifying misconceptions. There is a collaborative atmosphere with multiple suggestions being shared, although no consensus on a single method has been reached yet.

Contextual Notes

The original poster expresses concern about the validity of certain algebraic manipulations and the use of integration tables, indicating a desire to understand the integration process without relying on external resources.

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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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You don't add +1 to the numerator. We do the substitution u=x-1. What happens if you do that substitution??
 
Hint Hint:
\frac{1}{x-1}+\frac{x-1}{x-1}=\frac{x}{x-1}
 
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 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. :)
 
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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... ;)
 

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