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I Integration by Parts without using u, v

  1. Nov 30, 2016 #1
    Hello, I'm currently taking calc 1 as an undergraduate student, and my professor just showed us a new? way of solving Integration By Parts.

    This is the example he gave"

    nht7MtzfGAU-Rje5sUvqRKRVDv0USCmVtHRQd4UQTtW3EeglrdPKleQtaVWuuXTOAycrQTG-eXPW933y_Uw=w514-h858-no.png

    Is there a name for this technique that substitutes d(___) instead of dx?

    Thank you,
     
  2. jcsd
  3. Dec 1, 2016 #2

    Mark44

    Staff: Mentor

    It doesn't substitute d(___) instead of dx. What you're calling d(___) is just dv.

    This is nothing more than integration by parts, but without stating dv explicitly. Here ##u = \ln x## and ##dv = xdx##. From the latter, we get ##v = \frac{x^2}2##.
     
  4. Dec 1, 2016 #3

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I've never used ##u, v## and have never really understood the point. I would rename integration by parts as the "reverse product rule", because it's just the product rule for differentiation in reverse. Anyway, what I do is:

    ##\int f(x)g(x) dx = f(x) \int g(x)dx - \int f'(x) (\int g(x)dx)dx## (*)

    I don't see why you need any other variables. You pick ##f## because it's easy to differentiate and/or ##g## because you know how to integrate it already. Your one I would do:

    ##\int (ln(x)) x dx = (ln(x)) \frac{x^2}{2} - \int (\frac{1}{x})(\frac{x^2}{2})dx = \frac{x^2ln(x)}{2} - \int \frac{x}{2} dx = \frac{x^2ln(x)}{2} - \frac{x^2}{4} + C##

    I knew how to differentiate ##ln(x)## and how to integrate ##x##.

    PS To prove (*), I would imagine that you already know the integral of ##g## is ##h##, which means ##h'(x) = g(x)##. So:

    ## \int f(x)g(x) + f'(x)h(x)dx = \int f(x)h'(x) + f'(x)h(x)dx = \int \frac{d}{dx} (f(x)h(x))dx = f(x)h(x) + C##

    Hence:

    ## \int f(x)g(x)dx = f(x)h(x) -\int f'(x)h(x)dx+ C##

    Where ##\int g(x)dx = h(x) + C##

    And that seems to me to take some of the "magic" out of integration by parts.
     
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