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

$$\int \frac{\ln x}{x^2}dx=\frac{-\ln x-1}{x}+C$$
$$A=\int \frac{\ln x}{x^2}dx$$ We set $u=\ln x$ and $dv=\frac{dx}{x^2}$ That means $du=\frac{dx}{x}$ and $v=-\frac{1}{x}$ Following the formula $\int udv=uv-\int vdu$, we have $$A=-\frac{\ln x}{x}-\int-\frac{1}{x}\frac{dx}{x}$$ $$A=-\frac{\ln x}{x}+\int\frac{1}{x^2}dx$$ $$A=-\frac{\ln x}{x}-\frac{1}{x}+C$$ $$A=\frac{-\ln x-1}{x}+C$$