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Tricky Logarithmic Indefinite Integral

  1. Sep 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Our professor has only taught us these methods for Integration...thus far:

    • Direct Integration

    • Substituion Method

    So theoretically we should be able to solve this problem without using Integration by parts or partial fractions...:

    Integrate X^(2x)*ln(x+1)

    It should be solvable by substitution, direct integration...(double substitution) by algebraic manipulation etc.

    2. Relevant equations

    3. The attempt at a solution
    This is my attempt at a solution...I am concerned I am performing illegal operations, and I would appreciate some help in correcting my errors...
    thanks....(see attached)

  2. jcsd
  3. Sep 25, 2009 #2


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    There is some correct stuff in there. Did you notice that point at which ln(x)+1 turned into ln(x+1)? That's not legal. Things went downhill from there. I would suggest you write x^(2x) as e^(2x*ln(x)) and substitute u=x*ln(x). It might go easier. What's du? Uh, actually that may be the whole problem. I think ln x+1 is supposed to be ln(x)+1 not ln(x+1). They are very different.
  4. Sep 25, 2009 #3
    Thanks...I think you are right....it probably is ln(x)+1 not ln(x+1) (no WONDER it seemed so weird)...I will try again and see what happens..

    Thanks alot!
  5. Sep 27, 2009 #4

    Here is my new attempt at this problem...I follwed your suggestions and here is what I came up with...
    I appreciate any feed back or suggestions. Thank you!

  6. Sep 27, 2009 #5
    You can simplify your answer, but you also didn't integrate e2u correctly.

    I think it would have been just as easy, if not easier, to use u = x2x. Then du turns out real nice.
  7. Sep 27, 2009 #6
    Thanks for the heads up...

    Your right!...my bad.

    I re-did the last part with another round of substitution and finally got:

    e^2(x Ln(x))



    Does this seem right?
  8. Sep 27, 2009 #7
    That's right. You could always check an integral by differentiating it. :wink:
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