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Calc 2 simple integration i'm stuck on

  1. May 3, 2013 #1
    My first post lets see if i did this typing right, if not please forgive me...

    1. The problem statement, all variables and given/known data

    [tex]\int \frac{(1+ln x)^2}{x}\,dx[/tex]

    2. Relevant equations

    Trying to attack it by using substitution..

    3. The attempt at a solution

    Using...
    u = 1 + ln(x) , du = 1/x

    [tex]
    \begin{align}
    \int \frac{(1+ln x)^2}{x}\,dx \\
    &= \int (u)^2\,du \\
    &= \frac{u^3}{3} + C \\
    &= \frac{(1+ln x)^3}{3} + C \\
    \end{align}
    [/tex]

    Where did I go wrong?
     
    Last edited: May 3, 2013
  2. jcsd
  3. May 3, 2013 #2
    What happens if you instead let u = lnx ? Work it out and see what happens. Does the result look more like the answer in your textbook ?

    Also, what makes you think your first answer is wrong ? Try expanding [itex] (1 + lnx)^3 [/itex] in your first answer What does the result look like ? Does it look like the answer in the textbook or the answer your instructor provided ?
     
    Last edited: May 3, 2013
  4. May 3, 2013 #3
    Thank you skins.
     
  5. May 3, 2013 #4

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It looks good to me.

    To check it, take the derivative with respect to x.

    Also, as skins has suggested, expand ##\ (1+\ln(x))^3\ .##

    Don't forget the constant of integration that is in one solution may not match constant of integration in another solution. In particular, if C is a constant, then C + 1 is also a constant.
     
  6. May 3, 2013 #5
    Thank you, you're welcome. As it turned out your answer was correct all along, It was just that the form of your final answer was probably different than what showed in your textbook or on the blackboard. But it was still correct nonetheless.
     
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