Answer
$$-\frac{1}{x}(\ln x+1)+c.$$
Work Step by Step
We do the integration by parts; we choose $u=\ln x$ and $dv=\frac{1}{x^2}dx$. Then, $du=\frac{dx}{x}$, $v=-\frac{1}{x}$, hence we have
$$\int \frac{\ln x}{x^2}dx=\int udv=uv-\int vdu\\
=-\frac{\ln x}{x}+\int x^{-2}dx\\
=-\frac{\ln x}{x}-x^{-1}+c \\
=-\frac{1}{x}(\ln x+1)+c.$$
