# How do you find the integral  ln(x)/(x^2)?

$- \ln \frac{x}{x} - \frac{1}{x} + C$
Integration by parts can be done in this case, $\int \frac{\ln x}{x} ^ 2 \mathrm{dx}$ =$\int \ln x \cdot \frac{d}{\mathrm{dx}} \left(- \frac{1}{x}\right) \mathrm{dx}$= $- \ln \frac{x}{x} - \int \frac{d}{\mathrm{dx}} \left(\ln x\right) \cdot \frac{- 1}{x} \mathrm{dx} + C$= $- \ln \frac{x}{x} + \int \frac{1}{x} ^ 2 \mathrm{dx} + C$= $- \ln \frac{x}{x} - \frac{1}{x} + C$