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calculus

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evaluate the intergral using substitution
ingergral[(x^2-e^3x)/(x^3-e^3x)^2]dx

  • calculus - ,

    By the "Just Look-at-it Theorem and Knowing your Derivatives" and knowing the derivative of ln(...)
    I noticed that the derivative of the bottom would be
    3(x^2 - e^3x) , which just happens to look awfully close to the top

    so if y' = [(x^2-e^3x)/(x^3-e^3x)^2] then

    y = (1/3)(ln(x^3 - e^3x))

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