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Integration by parts

  1. Apr 29, 2013 #1

    trollcast

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    1. The problem statement, all variables and given/known data
    Find [itex]\int (2x^4\ln3x)\ dx[/itex]


    2. Relevant equations
    3. The attempt at a solution

    I let u = ln3x and dv/dx = x4 and I've managed to solve it and get an answer of:

    [itex]x^5(\frac{2}{5}\ln(3x) - \frac{1}{45} + c)[/itex]

    Which is really close to the answer from wolfram alpha but the last term should be 2/25 not 1/45?
     
  2. jcsd
  3. Apr 29, 2013 #2

    jedishrfu

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    Can you show your intermediate steps?

    What did you get for du?

    what did you get for v?

    and then u*v - integral(vdu) ?
     
  4. Apr 29, 2013 #3

    SammyS

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    How did you get the 1/45 ?

    The constant of integration should outside of the parentheses .

    ##\displaystyle \ x^5\left(\frac{2}{5}\ln(3x) - \frac{1}{45}\right)+C\ ##
     
  5. Apr 29, 2013 #4

    trollcast

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    [itex]\frac{du}{dx} = \frac{1}{3x}[/itex]
    [itex]v = \frac{x^5}{5}[/itex]
    [itex]uv - \int v\frac{du}{dx} =2( \ln3x * \frac{x^5}{5} - \int(\frac{x^5}{5} * \frac{1}{3x}))[/itex]
     
  6. Apr 29, 2013 #5
    What is the derivative of ln(3x)?

    Edit: You can either use the chain rule as SammyS pointed out below, or you can use the product property of logs to write ln(3x)=ln3+lnx
     
  7. Apr 29, 2013 #6

    SammyS

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    Using the chain rule:

    ##\displaystyle \frac{d}{dx}\, \ln(3x)=\frac{3}{3x}##
     
  8. Apr 29, 2013 #7

    trollcast

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    Thanks got it now.
     
  9. Apr 29, 2013 #8

    SammyS

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    Yes, ln3+lnx makes it even more obvious.
     
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