calculus
posted by keisha .
evaluate the intergral using substitution
ingergral[(x^2e^3x)/(x^3e^3x)^2]dx

By the "Just Lookatit 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^2e^3x)/(x^3e^3x)^2] then
y = (1/3)(ln(x^3  e^3x))