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Unwanted constants after integration

  1. Mar 30, 2009 #1

    Cyosis

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    Homework Helper

    While integrating a rational function I stumbled upon the following problem (In the calculation of the integral the substitution [tex]u=x+1, du=d(x+1)=dx[/tex] was used).

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

    The problem now is that if I substitute x back into the integral during step (2) I get [tex]x - \frac{1}{1+x}-2 \log(1+x)[/tex].

    Obviously taking the derivative of both primitives yields the same integrand.

    My problem with this is that instead of getting an unknown constant I get this unwanted extra 1. Secondly if I plug this integral into mathematica it gives the result without the constant 1.

    So my question is why do I get different functions without having specified the integration constant?
     
  2. jcsd
  3. Mar 30, 2009 #2

    CompuChip

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    You didn't specify the integration boundaries, so you will get an integration constant. So instead of (3), you get u - 1/u - 2 log(u) + C and instead of (4) you get 1 + x - 1/(1 + x) - 2 log(1 + x) + C.

    The extra 1 you have can simply be absorbed in C, i.e. define C' = C - 1 or something like that.

    Once you specify the integration boundaries (x from a to b) the extra constant term will be absorbed by the change of boundaries in the substitution (i.e. the u-integral will run from a + 1 to b + 1).
     
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