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

$$\int \frac{1}{x(\ln x)^2}dx=-\frac{1}{\ln x}+C$$
$$A=\int \frac{1}{x(\ln x)^2}dx=\int\frac{1}{(\ln x)^2}\frac{dx}{x}$$ Set $a=\ln x$ We then have $$da=\frac{dx}{x}$$ Therefore, $$A=\int\frac{1}{a^2}da$$ $$A=-\frac{1}{a}+C$$ $$A=-\frac{1}{\ln x}+C$$