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    Question: Integrate x^2 dx /((x + 1)(x^2 + x + 1))

    (in words) (x squared dx) divided by (x + 1) times (x squared + x + 1)

    I'm not really sure how I should start this, would I need to make a substitution or use "completing the square" on (x^2 + x + 1) then express in terms of partial fractions?
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    (Original post by As_Dust_Dances_)
    Question: Integrate x^2 dx /((x + 1)(x^2 + x + 1))

    (in words) (x squared dx) divided by (x + 1) times (x squared + x + 1)

    I'm not really sure how I should start this, would I need to make a substitution or use "completing the square" on (x^2 + x + 1) then express in terms of partial fractions?
    Observe that x^2 \equiv (x^2 +x + 1) - (x+1). Completing the square after a bit of manipulation from there is a good idea.
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    (Original post by Farhan.Hanif93)
    Observe that x^2 \equiv (x^2 +x + 1) - (x+1). Completing the square after a bit of manipulation from there is a good idea.
    I'm assuming then you would use partial fractions since you would get a product on the denominator?
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    (Original post by As_Dust_Dances_)
    I'm assuming then you would use partial fractions since you would get a product on the denominator?
    Nope. Note that \dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}; after cancelling down, once of the terms can be integrated directly and the other can be tackled with completing the square (namely the one with x^2+x+1 in it's denominator).
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    (Original post by Farhan.Hanif93)
    Nope. Note that \dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}; after cancelling down, once of the terms can be integrated directly and the other can be tackled with completing the square (namely the one with x^2+x+1 in it's denominator).
    Oh I understand now, I didn't think of expressing x^2 like that. Thanks so much for your help! + rep
 
 
 
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