Ta có:
$\int\limits_1^2\dfrac{\ln x}{x^3}dx$
$=-\dfrac{1}{2}\int\limits_1^2\ln xd\left(\dfrac{1}{x^2}\right)$
$=-\dfrac{\ln x}{2x^2}\left|\begin{array}{l}2\\1\end{array}\right.+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^2}d(\ln x)$
$=-\dfrac{\ln2}{8}+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^3}dx$
$=-\dfrac{\ln2}{8}-\dfrac{1}{4x^2}\left|\begin{array}{l}2\\1\end{array}\right.$
$=-\dfrac{\ln2}{8}+\dfrac{3}{16}$