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Integration by parts

  1. Mar 22, 2005 #1
    ok i`m really struggling with the concept.
    I've been asked to find the indefinite integral of;

    [tex] \int \frac{x^2}{(2+ x^3)} dx [/tex]

    so before i beg for the answer could someone confirm that i`ve got the right rule to solve this;

    [tex] \int u(x) v'(x) = [ u(x) v(x)] - \int v(x) u'(x) [/tex]

    if this is right would you mind giving a suggestion to what u(x) to use?

    p.s. i may have to edit this if latex doesn`t come out right I've been having trouble with it and only jointed the forum a few day's ago!
     
    Last edited: Mar 22, 2005
  2. jcsd
  3. Mar 22, 2005 #2

    HallsofIvy

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    Yes, that is the correct formula (it's surely in your text book), but I don't see any good reason for using integration by parts. That's a fairly standard anti-derivative: look up "arctangent".
     
  4. Mar 22, 2005 #3
    I`ve just edited it. Now it`s integration by parts! see I`m getting all flustered
     
  5. Mar 22, 2005 #4
    Actually, you still don't need integration by parts. Substituting [itex]u = x^3[/itex] will simplify it to a standard form.

    Anyways, since differentiation is in some sense the inverse operation to integration, every differentiation rule yields an integration rule. I'm sure you remember the product rule:

    [tex] \frac{d}{dx}(f(x) g(x)) = f^\prime (x) g(x) + g^\prime (x) f(x)[/tex]

    Rearraging the equation above gives

    [tex] f(x) g^\prime (x) = \left[f(x) g(x)\right]^\prime - f^\prime(x)g(x)[/tex]

    And integrating both sides with respect to x gives

    [tex] \int f(x) g^\prime (x) \ dx = \int \left(\frac{d}{dx} (f(x) g(x)) - f^\prime(x)g(x)\right) \ dx[/tex]

    but certainly, [tex] \int \frac{d}{dx}(f(x)g(x)) \ dx = f(x)g(x)[/tex], so this just reduces to

    [tex] \int f(x)g^\prime (x) \ dx = f(x)g(x) - \int g(x) f^\prime (x) \ dx[/tex]

    which is what your teacher probably called the formula for integration by parts.
     
    Last edited: Mar 22, 2005
  6. Mar 22, 2005 #5

    dextercioby

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    [tex] x^{2}dx=\frac{1}{3}d(2+x^{3}) [/tex] so the integration is simple...

    Daniel.
     
  7. Mar 22, 2005 #6

    BobG

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    As others have said, you don't have to use integration by parts to solve this, but .....

    Pick the term whose derivative will eventually go to zero (the soonest). In this case, it's x^2. First derivative is 2x. Second derivative is 2. Third derivative is 0.
     
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