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March 26, 2017

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intergral 2x^2/(x^2+4x+8)

  • calculus(integral) - ,

    I did a long division and got

    2x^2/(x^2+4x+8) = 2 - (4x+8)/(x^2 + 4x + 8)
    = 2 - 2(2x+4)/(x^2 + 4x + 8)

    I noticed that the derivative of x^2 + 4x + 8 is 2x+4 , which I have sitting on top.
    Ahhh, logs!!!!

    ∫2x^2/(x^2+4x+8) dx
    =∫ ( 2 - 2(2x+4)/x^2 + 4x + 8) ) dx
    = 2x - 4 ln(x^2 + 4x + 8)

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