## University Calculus: Early Transcendentals (3rd Edition)

$$\int^4_1\frac{\ln2\log_{2}x}{x}dx=\frac{(\ln4)^2}{2}$$
$$A=\int^4_1\frac{\ln2\log_{2}x}{x}dx=\int^4_1\frac{\ln2\times\frac{\ln x}{\ln2}}{x}dx$$ $$A=\int^4_1\frac{\ln x}{x}dx$$ We set $u=\ln x$, which means $$du=\frac{1}{x}dx$$ - For $x=4$, we have $u=\ln4$ - for $x=1$, we have $u=\ln1=0$ Therefore, $$A=\int^{\ln4}_0 udu$$ $$A=\frac{u^2}{2}\Big]^{\ln4}_0$$ $$A=\frac{1}{2}((\ln4)^2-0^2)$$ $$A=\frac{(\ln4)^2}{2}$$