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Thread: integration by parts

  1. #1
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    integration by parts

    use integration by parts to find the integral of ln(x+1) dx
    any help would be appreciated
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  2. #2
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    $\displaystyle \int ln(1+x)dx=\int1.ln(1+x)dx$

    Let $\displaystyle u=ln(1+x)$ and $\displaystyle dv=1$

    Then $\displaystyle du=\frac{1}{1+x}$ and $\displaystyle v=x$

    $\displaystyle xln(1+x)- \int x \frac{1}{1+x}dx=xln(1+x)- \int 1dx=xln(1+x)-x+C$
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  3. #3
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    Don't forget the absolute value.
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  4. #4
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    how does the integral of x/(x+1) become the integral of 1??
    thanks
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  5. #5
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    because when you let $\displaystyle u=x$ and $\displaystyle dv=\frac{1}{1+x}=(1+x)^{-1}$, $\displaystyle du=1$ and $\displaystyle v=(1+x)^0=1$
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  6. #6
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    Whoah! Careful now.

    You could split

    $\displaystyle
    \frac{x}{x+1}
    $


    into


    $\displaystyle
    1 - \frac{1}{x+1}
    $


    and have

    $\displaystyle
    I = x\ \ln(x+1) - \int (\ 1 - \frac{1}{1+x}\ )\ dx
    $

    $\displaystyle
    \ \ =x\ \ln(1+x) - x + \ln(x+1)+C
    $


    which is fine.

    But here's a short cut - notice that one integral of 1 is x + 1, and choose u = x+1 for du = 1.


    I.e.


    $\displaystyle
    I = (x+1)\ \ln(x+1) - \int (x+1)\ \frac{1}{x+1}\ dx
    $

    $\displaystyle
    = (x+1)\ \ln(x+1) - \int (1)\ dx
    $

    $\displaystyle
    \ \ = (x+1)\ \ln(x+1) - x + C
    $


    Picture shortly...
    Last edited by tom@ballooncalculus; Jan 28th 2009 at 12:22 PM. Reason: + C !
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  7. #7
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    Don't integrate - balloontegrate!

    Balloon Calculus: worked examples from past papers
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