Answer
$$\int^{e}_1\frac{8\ln3\log_3\theta}{\theta}d\theta=4$$
Work Step by Step
$$A=\int^{e}_1\frac{8\ln3\log_3\theta}{\theta}d\theta$$
We set $u=\log_3\theta$, which means $$du=\frac{1}{\theta\ln3}d\theta$$ $$\frac{1}{\theta}d\theta=\ln3du$$
For $\theta=1$, we have $u=\log_31=0$
For $\theta=e$, we have $u=\log_3e$
Therefore, $$A=\int^{log_3e}_{0}8\ln3u(\ln3du)=\int^{log_3e}_{0}8(\ln3)^2udu$$ $$A=\frac{8(\ln3)^2}{2}u^2\Big]^{log_3e}_{0}=4(\ln3)^2u^2\Big]^{log_3e}_{0}$$ $$A=4(\ln3)^2(\log_3e)^2$$ $$A=4\Big(\ln3\times\frac{1}{\ln3}\Big)^2$$ $$A=4\times1^2=4$$
