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Math Help - Antidifferentiation

  1. #1
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    Antidifferentiation

    1/(2x-2)
    How would I antidifferentiate this?
    Exam in under two hours!
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  2. #2
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    skeeter's Avatar
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    \frac{1}{2x-2} = \frac{1}{2} \cdot \frac{1}{x-1}

    now it should be rather simple to find the antiderivative ...
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  3. #3
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    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
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  4. #4
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    You first need to know basic integration identities.

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

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

    ----

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

    Good luck with your exam.
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  5. #5
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    I didn't have time to read it all before the exam, so I had a glance. Turns out it helped me. I also didn't know about the absolute values in there, as I've never used them in that situation, but one of the questions was multiple choice and had the exact same thing in all the answers.
    I might've been scared off or confused if I hadn't seen it before only an hour before the exam
    Thanks!
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