1. ## Antidifferentiation

1/(2x-2)
How would I antidifferentiate this?
Exam in under two hours!

2. $\displaystyle \frac{1}{2x-2} = \frac{1}{2} \cdot \frac{1}{x-1}$

now it should be rather simple to find the antiderivative ...

3. Ah, that's what I was looking for. Thanks. It'd be just my luck that the one situation I'm not familiar with, and don't have in my notes, would show up on the exam.
Thanks
Lightning fast too!

4. You first need to know basic integration identities.

We're gonna use $\displaystyle \int\frac{du}u=\ln|u|+k.$ Note that your integral seems to be familiar to our little fact, then, how can we turn $\displaystyle \int\frac{dx}{2x-2}$ into $\displaystyle \int\frac{du}u$? We need a substitution, which is $\displaystyle u=2x-2\implies du=2\,dx$ so that $\displaystyle \frac{du}2=dx,$ and the integral becomes $\displaystyle \frac12\int\frac{du}u.$ From here, we're done, 'cause we've turned our integral into a known one, so the antiderivative of $\displaystyle \frac1{2x-2}$ is $\displaystyle \frac12\ln|2x-2|+k.$

(Note that we can't say $\displaystyle \int\frac{dx}{2x-2}=\ln|2x-2|+k,$ one has to turn the integral into a known one. You may differentiate $\displaystyle \ln|2x-2|$ but that won't yield the integrand. However, by differentiating $\displaystyle \frac12\ln|2x-2|$ will yield the integral.)

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I know I got beaten answering this, but I wanted to go in depth a bit.