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Math Help - integration by parts - using a previous integration

  1. #1
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    integration by parts - using a previous integration

    Hello

    I am working on a problem which involves integrating \int \frac{x^3}{1+x} dx [with limits 0, 1 (don't know how to do limits in latex)].

    So I managed to integrate using substitution as follows:

    let u = 1+x, du/dx = 1, limits: u = 1+1 = 2 and u = 1+0=1

    x = u-1.

    So get:
    \int \frac{(u - 1)^3}{u} du [with different limits 1, 2]

    I expand out:
    \int u^2 - 3u + 3 - \frac{1}{u} du

    and then:
    \frac{u^3}{3} - \frac{3u^2}{2}  + 3u - ln u

    and I get a result of 5/6 - ln2.

    I am then asked to integrate this: \int x^2 ln(1+x) with limits 0 to 1.

    So I start integrating (by parts) and notice that second integration bit - I have this:

    = ln(1+x) . \frac{x^3}{3} - \frac{1}{3} \int \frac{x^3}{1+x}

    That last bit, I already integrated above.

    So my question is, how do I combine the two. My start part of integration by parts and then the minus - and here use my previous result.

    The previous result had different limits - so I presume I cant just use the result. Do I just substitute back in 1+x for the u above like this:

     ln(1+x) . \frac{x^3}{3} - \frac{1}{3}(\frac{(1+x)^3}{3} - \frac{3(1+x)^2}{2}  + 3(1+x) - ln (1+x)) - then use my limits 0, 1 as asked for this integration?

    Angus
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  2. #2
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    Re: integration by parts - using a previous integration

    You can say that \displaystyle \frac{x^3}{1+x} = x^2-x+1-\frac{1}{1+x}
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  3. #3
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    Re: integration by parts - using a previous integration

    What you got is this: \int_0^1 x^2 \ln (1+x) \, dx = \frac{x^3}{3} \cdot \ln (1+x) \Big\vert_0^1 - \frac{1}{3} \int_0^1 \frac{x^3}{1+x} \, dx?

    You already know the value of \int_0^1 \frac{x^3}{1+x} \, dx, just divide it by 3 and subtract, yielding:

    \int_0^1 x^2 \ln (1+x) \, dx = \frac{1^3}{3} \cdot \ln 2 - \frac{1}{3} \cdot \left( \frac{5}{6} - \ln 2 \right) = \frac{2 \ln 2}{3} - \frac{5}{18}
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