Answer
$$\int^2_0\frac{\log_2(x+2)}{x+2}dx=\frac{\ln8}{2}$$
Work Step by Step
$$A=\int^2_0\frac{\log_2(x+2)}{x+2}dx=\int^2_0\frac{\frac{\ln(x+2)}{\ln2}}{x+2}dx$$ $$A=\frac{1}{\ln2}\int^2_0\frac{\ln(x+2)}{x+2}dx$$
We set $u=\ln(x+2)$, which means $$du=\frac{(x+2)'}{x+2}dx=\frac{1}{x+2}dx$$
- For $x=2$, we have $u=\ln4$
- For $x=0$, we have $u=\ln2$
Therefore, $$A=\frac{1}{\ln2}\int^{\ln4}_{\ln2}udu=\frac{1}{\ln2}\times\frac{u^2}{2}\Big]^{\ln4}_{\ln2}$$ $$A=\frac{(\ln4)^2-(\ln2)^2}{2\ln2}$$ $$A=\frac{(\ln4-\ln2)(\ln4+\ln2)}{2\ln2}$$ $$A=\frac{(\ln\frac{4}{2})(\ln(4\times2))}{2\ln2}$$ $$A=\frac{\ln2\ln8}{2\ln2}=\frac{\ln8}{2}$$
