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    hi, how do i evaluate integral of (-x^3)/(x+1)^2 dx ?
    i would use substitution or split it in more parts but i dont know how??? can anybody show me how to do it? thanks
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    (Original post by gomer)
    hi, how do i evaluate integral of (-x^3)/(x+1)^2 dx ?
    i would use substitution or split it in more parts but i dont know how??? can anybody show me how to do it? thanks
    Integration by parts appears to be a better option in my opinion..
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    (Original post by gomer)
    hi, how do i evaluate integral of (-x^3)/(x+1)^2 dx ?
    i would use substitution or split it in more parts but i dont know how??? can anybody show me how to do it? thanks
    Try long division, to get a quotient and a remainder that isn't top-heavy, then you can integrate.
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    (Original post by XTinaA)
    Try long division, to get a quotient and a remainder that isn't top-heavy, then you can integrate.
    OK, that seems to be good idea , thank you
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    ∫ -x³/(x+1)² dx
    x³/(x+1) -3∫ x²/(x+1) dx
    x³/(x+1) -3∫ (x²-1)/(x+1) +1/(x+1) dx
    x³/(x+1) -3∫ (x-1) +1/(x+1) dx
    x³/(x+1) -3(x²/2 - x + ln(x+1)) + C
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    If you let u=x+1, the problem is reduced to integrating a simple polnomial.
 
 
 

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