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Integration by Parts

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    In preparation for an exam next week, I'm solving some of the end-of-chapter questions. There are 30 questions. I've solved a few on my own, but here's one I'm getting stuck on.

    Question 2)
    [tex]\int x5^x dx[/tex], problem here is I don't know how to deal with [itex]5^x[/itex].
    The correct answer should be: [tex]\frac{x}{ln5}5^x - \frac{1}{(ln5)^2}5^x + C[/tex]

    To avoid spamming the forum with many threads, I'll just post in this one whenever I hit an integral I'm unsure of. For now, it's just the one above.

    2. Relevant equations
    [tex]\int u*dv = uv - \int v*du[/tex]
     
    Last edited: Mar 5, 2008
  2. jcsd
  3. Mar 5, 2008 #2

    Dick

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    It might look a little easier if you write 5^x=exp(ln(5)*x). Put the exp into the dv part.
     
  4. Mar 5, 2008 #3
    Did another one here:

    Question 4)
    [tex]\int x log_{10}x dx
    = \int x \frac{ln(x)}{ln(10)} dx[/tex]
    [tex]= \frac{1}{ln(10)}\int xln(x) dx[/tex]
    [tex]= \frac{1}{ln(10)}\int D(\frac{x^2}{2})ln(x) dx[/tex]
    [tex]= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} D(ln(x))dx\right)[/tex]
    [tex]= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} * \frac{1}{x}dx\right)[/tex]
    And yeah, getting too lazy to TeX it all. Question is, when I finally get rid of all integrals, the +C will be inside the 1/ln(10) bracket. Do I write C/ln(10), or do I assume that whatever C is will already have that applied, so I just write C?
     
  5. Mar 5, 2008 #4

    HallsofIvy

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    Yes, the constant C/ln(10) is the same as the constant C' - and it isn't even necessary to distinguish between "C" and "C'".
     
  6. Mar 5, 2008 #5
    So, 5^x = exp(ln(5^x)) = exp(xln(5)) = e^(xln(5))... I see that much.
    I also know the derivative/antiderivative of e^x = e^x.
    What is the derivative/antiderivative of e^(xln(5))??
     
  7. Mar 5, 2008 #6

    Dick

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    Use the chain rule. The derivative of e^(a*x)=a*e^(a*x). What's the antiderivative?
     
  8. Mar 5, 2008 #7
    e^(ax) / a, methinks... 'cause then taking the derivative of that would pull down another 'a', canceling out the division. Makes sense.

    Thanks :)
     
  9. Mar 5, 2008 #8
    Why is [tex]\int \frac{D(1+x^2)}{1+x^2}dx = ln(1+x^2)[/tex]?

    I know [itex]D(ln(x)) = 1/x[/itex], however I don't see how my book did this calculation.
     
  10. Mar 5, 2008 #9

    HallsofIvy

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    The point is that [itex]\int\frac{du}{u}= ln|u|+ C[/itex], No matrer what "u" is. In this case u= 1+ x2.
     
  11. Mar 5, 2008 #10
    It is from the more general rule that

    [tex]\int \frac{f'(x)}{f(x)}dx = ln|f(x)| + C[/tex]

    which comes from the chain rule. If you think about substituting u for 1 + x2, then what is du in terms of dx?
     
  12. Mar 5, 2008 #11
    If u = 1 + x^2, then du = 2x.
    Thanks for the ln|blah| tip! I remember being taught it, but had forgotten.

    One final question for tonight.
    I'm in the middle of integrating arcsin(x), and as a segment, I need to compute the antiderivative:
    [tex]\int \frac{x}{\sqrt{1 - x^2}}[/tex]

    From looking up the answer, I see that this integral is equivalent to [itex]\sqrt{1 - x^2}[/itex]. Why would this be? I see how it can be broken up into [itex]\frac{x}{1} * (1 - x^2)^{-\frac{1}{2}}[/itex]. Does this help me, or am I missing another rule?
     
  13. Mar 5, 2008 #12
    Yes, you are right, although the division by 1 is not really necessary. Think chain rule again, and see if you can separate one part of the integral as a derivative of another part, then substitute.
     
  14. Mar 5, 2008 #13
    Derivative of the top = 1

    Derivative of the bottom = [tex]-2x * (1 - x^2)^{-\frac{3}{2}}*-\frac{1}{2}[/tex]
    = [tex]x(1 - x^2)^{-\frac{3}{2}}[/tex]

    Sorry, but I still don't see it >.<;
    I see how the chain rule was used to find the derivative of the bottom, but otherwise...
     
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