# How to pull out coefficient from radical in an integral

I am in an online Calculus 2 class, and before my professor gets back to me, I was wondering if you guys could help. I was reading through an example:

How was 1/27 pulled out from the coefficient next to u^2? I am probably missing something dumb. Thanks.

• The $27$ is not coming from the radical but rather from substituting the exponential in the numerator: $e^{3t} \rightarrow u^3 / 27$ Jun 19, 2016 at 1:16

Note that $e^{3t} = \frac{1}{27} u^3, 9e^{2t} = u^2, dt = \frac{du}{3e^t} = \frac{du}{u}$. Now substitute and see what you get.
There's an error in the problem-the numerator in the u-substitution should be $u^3$, not $u^2$. The numerator here is $(1/3 u)^3$ (compare to u). So a factor of 1/27 drops out into the denominator. The rest is easy to see by careful substitution.
• I don't think there's an error, a factor of $u$ gets cancelled by the $dt$ substitution