given
$$r=4e^{3\theta} \space \space \space \space dr/d\theta=(3*4*e^{3\theta})$$ $$l=\int \sqrt(4e^{3\theta})^{2}+(3*4*e^{3\theta})^{2} \rightarrow $$
why does the integral $$ \int\sqrt(16e^{6\theta}+144e^{6\theta}) \rightarrow \int\sqrt(160e^{6\theta})$$ become $$ 4\sqrt(10) \int\sqrt(e^{6\theta} \rightarrow somehow \rightarrow (4/3) \sqrt(10) (e^{6 \pi}-1)$$ instead of $$ 4\sqrt(10) \int \sqrt(e^{12\theta} \rightarrow U=\sqrt(e^{12})\space ; \space DU=e^{12\theta}*12)$$ $$ \rightarrow(4 \sqrt(10)*[((2/3)e^{12 \theta}]*[e^{12 \theta}*12]\rightarrow (8/3) \sqrt(10) (e^{12 \theta}*12) du\space went \space away$$ I'm confused about the integration of e after I move the constants to the other side of the integral. I don't have the steps so I can only guess that there was a 1/3 somewhere in the integration of the first equation. this in an equation for length of a polar curve. I have omitted the original function r was equal too.
You have not included it from your post, but from your answer I believe that your questions is to evaluate the length of the curve $r = 4e^{3\theta}$ for $0 \leq \theta \leq 2\pi$.
The formula for arc length is: $$L = \int_a^{b} \sqrt{r^2 + (\frac{dr}{d\theta})^2} \,d\theta$$
Then we have $\frac{dr}{d\theta} = 12e^{3\theta}$, so substituting in we get:
$$L = \int_0^{2\pi} \sqrt{160e^{6\theta}} \,d\theta = \sqrt{160}\int_0^{2\pi} \sqrt{e^{6\theta}} \,d\theta = \sqrt{160}\int_0^{2\pi} e^{3\theta} \,d\theta$$.
Then using the substitution $u = 3\theta$, we find that $$L = \sqrt{160}\int_0^{2\pi} e^{3\theta} \,d\theta = \frac{\sqrt{160}}{3}\int_0^{6\pi} e^{u} \,du = \frac{4\sqrt{10}}{3}(e^{6\pi}-1)$$