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