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April 20, 2014

April 20, 2014

Posted by **keisha** on Tuesday, April 20, 2010 at 4:55pm.

ingergral[(x^2-e^3x)/(x^3-e^3x)^2]dx

- calculus -
**Reiny**, Tuesday, April 20, 2010 at 5:02pmBy 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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