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Impossible evaluation NEED HELP T_T

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    evaluate the integral of 25x^2lnxdx



    2. Relevant equations



    3. The attempt at a solution
    First thing I did was set 25x^2 as D, then set lnx as I. Derivative of 25x^2 is 50x. The antiderivative of lnx is xlnx-x. Therefore it turned out to be something like.....25x^2(xlnx-x)-integral of 50x(xlnx-x). Then i proceeded by finding the antiderivative of my second integral...but soon found out that its an unending process if this keeps going up. PLEASE HELP. ADVICE IS REALLY APPRECIATED. THANK YOU SO MUCH.
     
  2. jcsd
  3. Jan 31, 2009 #2
    You need to use substitution twice. Start with In (x) as D
     
  4. Jan 31, 2009 #3
    alright Ill try that, thanks for the advice. And by the way do you perhaps know the antiderivative of e^(ln(4)*x)??? Because on this one certain problem i had to find the antiderivative of 4^x which i found to be e^(ln(4)*x) but i cant seem to know how to find the antiderivative of e^(ln(4)*x). Is it perhaps e^(ln(4)*x) all over ln(4)x???

    thank you for taking your time to help me out. Really appreciate it.
     
  5. Jan 31, 2009 #4
    Isn't this like the third thread out of 5 top ones that ask the same question?
     
  6. Jan 31, 2009 #5

    gabbagabbahey

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    There's a nice little trick here if you realize you have:

    [tex]\int 25x^2\ln x dx=25x^2(x\ln x -x)-\int50x^2\ln x dx +\int 50x^2 dx[/tex]

    [tex]\implies \int 50x^2 \ln x dx+ \int 25x^2\ln x dx=3\int 25x^2\ln x dx=25x^2(x\ln x -x)+\int 50x^2 dx[/tex]

    [tex]\implies\int 25x^2\ln x dx=\frac{1}{3}\left(25x^2(x\ln x -x)+\int 50x^2 dx\right)[/tex]
     
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