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Math Help - Evaluate the Indefinite Integral

  1. #1
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    Evaluate the Indefinite Integral

    Need some help with this indefinite integral:



    The answer I got was d/m (ln(mx+k)) by using u = mx+k.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by statelyplump View Post
    Need some help with this indefinite integral:



    The answer I got was d/m (ln(mx+k)) by using u = mx+k.
    d/m (ln(mx+k)) ???

    It should be:

    \frac{1}{m} \times ln(mx+k) = \frac{ln(mx+k)}{m}
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    Quote Originally Posted by harish21 View Post
    d/m (ln(mx+k)) ???

    It should be:

    \frac{1}{m} \times ln(mx+k) = \frac{ln(mx+k)}{m}
    So is d not to be treated as a constant? I'm confused by the notation.
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by statelyplump View Post
    So is d not to be treated as a constant? I'm confused by the notation.
    d is not a constant. "dx" means that you are integrating the function with respect to x!

    its..

    \int \frac{1}{mx+k} dx

    so when you substituted u = mx+k, you should have gotten:
    du = m dx \implies dx = \frac{1}{m} du

    then you substitute dx = \frac{1}{m} du in your question to get:

    \frac{1}{m} \int \frac{1}{u} du

    do you get it?
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  5. #5
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    Thanks. I understand everything else, I just didn't know that they were putting dx in the numerator like that. In every other problem, the dx is off to the side like it was when you reformatted it.
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by statelyplump View Post
    Thanks. I understand everything else, I just didn't know that they were putting dx in the numerator like that. In every other problem, the dx is off to the side like it was when you reformatted it.
    \frac{dx}{x} and \frac{1}{x} dx are the same!
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