anonymous
  • anonymous
Can someone explain to me what's happening in this final step? I've seen the answer, but I don't undertstand how to get it! http://imgur.com/GSuuI0l
MIT 18.01 Single Variable Calculus (OCW)
katieb
  • katieb
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anonymous
  • anonymous
In the numerator there is derivative of the denominator so we can apply ln( ) law
phi
  • phi
if you multiply the top by 3, you get exactly the derivative of the bottom so put a ⅓ out front , and multiply the top by 3 you now have \[ \frac{1}{3} \int \frac{du}{u} = \frac{1}{3}\ln(u) \] notice we can write the above as \[ \frac{1}{3} \int \frac{du}{u} = \int \frac{du}{3u} \] it looks like they want to see 3(x^3+ 6x^2 +3x+8) or 3x^3 +18x^2 + 9x + 24 in the "box"

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