Answer
$$\int^{8}_1\frac{\log_4\theta}{\theta}d\theta=\frac{9\ln2}{4}$$
Work Step by Step
$$A=\int^{8}_1\frac{\log_4\theta}{\theta}d\theta$$
We set $u=\log_4\theta$, which means $$du=\frac{1}{\theta\ln4}d\theta$$ $$\frac{1}{\theta}d\theta=\ln4du$$
For $\theta=1$, we have $u=\log_41=0$
For $\theta=8$, we have $$u=\log_48=\log_{2^2}2^3=\frac{3}{2}\log_22=\frac{3}{2}$$
Therefore, $$A=\ln4\int^{3/2}_{0}udu$$ $$A=\frac{\ln4}{2}u^2\Big]^{3/2}_{0}$$ $$A=\frac{\ln4}{2}\Big(\frac{3}{2}\Big)^2=\frac{\ln4}{2}\times\frac{9}{4}$$ $$A=\frac{9\ln4}{8}$$ $$A=\frac{9\ln2}{4}$$
