Post a New Question


posted by .

evaluate the intergral using substitution

  • 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))

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question