Answer
$$\int^{3}_{2}\frac{2\log_{2}(x-1)}{x-1}dx=\ln2$$
Work Step by Step
$$A=\int^{3}_{2}\frac{2\log_{2}(x-1)}{x-1}dx=2\int^{3}_{2}\frac{\frac{\ln(x-1)}{\ln2}}{x-1}dx$$ $$A=\frac{2}{\ln2}\int^{3}_{2}\frac{\ln(x-1)}{x-1}dx$$
We set $u=\ln(x-1)$, which means $$du=\frac{(x-1)'}{x-1}dx=\frac{1}{x-1}dx$$
- For $x=3$, we have $u=\ln2$
- For $x=2$, we have $u=\ln1=0$
Therefore, $$A=\frac{2}{\ln2}\int^{\ln2}_0udu=\frac{2}{\ln2}\times\frac{u^2}{2}\Big]^{\ln2}_0$$ $$A=\frac{u^2}{\ln2}\Big]^{\ln2}_0$$ $$A=\frac{(\ln2)^2-0^2}{\ln2}$$ $$A=\frac{(\ln2)^2}{\ln2}=\ln2$$
