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Math Help - For test tomorrow- Integration by Parts

  1. #1
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    For test tomorrow- Integration by Parts

    Need some help with the following

    1. \int ln(2x+1) dx
    2. \int tan^{-1}5t dt
    3. \int (lnx)^2 dx
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  2. #2
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    Hello, c_323_h!

    If you're doing these "by parts", there isn't any choice.
    . . So what the problem?


    1)\;\;\int \ln(2x+1)\,dx

    Let: u = \ln(2x+1)\qquad\quad dv = dx
    Then: du = \frac{2}{2x+1}\,dx\qquad v = x

    And we have: . x\cdot\ln(2x+1) - \int\frac{2x}{2x+1}\,dx

    . . . =\;x\cdot\ln(2x + 1) - \int\left(1 - \frac{1}{2x + 1}\right)\,dx

    Can you finish it now?



    2)\;\;\int \tan^{-1}(5t)\,dt

    Let: u = \tan^{-1}(5x)\qquad\qquad dv = dx

    Then: du = \frac{5}{1 + 25x^2}\,dx\qquad v = x

    And we have: . x\cdot\tan^{-1}(5x) - \int\frac{5x}{1 + 25x^2}\,dx\quad\hdots\;etc.



    3)\;\;\int (\ln x)^2\,dx

    Let: u = (\ln x)^2\qquad\qquad dv = dx

    Then: du = \frac{2\ln x}{x}\,dx\qquad v = x

    And we have: . x\cdot(\ln x)^2 - \int x\cdot\frac{2\ln x}{x}\,dx

    . . . = \;x\cdot(\ln x)^2 - 2\underbrace{\int\ln x\,dx}
    . . . . . . . . . . . . . and this can be done by parts
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  3. #3
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    Quote Originally Posted by Soroban
    Hello, c_323_h!
    So what the problem?

    [/size]
    Hi, I thought since they were composition of functions they would be different. Thanks for the help...very much appreciated
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