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Math Help - One integration by parts problem

  1. #1
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    One integration by parts problem

    Just, the Integral of (x^3)/(x^3+1) dx

    I know you can turn the denominator into (X+1)(X^2-x+1). What do you do from there? thanks =)
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  2. #2
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    I doublechecked this but I'm a bit rusty lately so perhaps someone else can verify it,

    int dx/(x+1)(x^2-x+1)

    You have an irreducible quadratic factor so the equation is

    a/(x+1) + (bx+c)/(x^2-x+1) = 1

    x=0 -> a+c=1
    x=1 -> a+2b+2c=1
    x=-1 -> 3a=1

    Solving that system, a=1/3, b=-1/3, c=2/3

    Now all you have to do is to solve the last part:

    int 1/(3(x+1)) + ((-1/3)x + (2/3))/(x^2-x+1)

    Hope this helps
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    dax918 is right for the part he did, but he left out an important detail. your integral is actually the integral of 1 minus all he said.

    Note that \frac {x^3}{x^3 + 1} = \frac {x^3 + 1 - 1}{x^3 + 1} = \frac {x^3 + 1}{x^3 + 1} - \frac 1{x^3 + 1} =  1 - \frac 1{x^3 + 1}

    as dax918 said, \frac 1{x^3 + 1} = \frac 13 \bigg( \frac 1{x + 1} - \frac {x - 2}{x^2 - x + 1} \bigg)

    thus, your integral is \int \Bigg[ 1 - \frac 13 \bigg( \frac 1{x + 1} - \frac {x - 2}{x^2 - x + 1} \bigg) \Bigg]~dx
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  4. #4
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    Thank you Jhevon for pointing that out, I forgot the fact that 3deltat already did the first step
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