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.$$=-\dfrac{\ln2}{8}+\dfrac{3}{16}$
Ta có: $\int\limits_1^2\dfrac{\ln x}{x^3}dx$$=-\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}$
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}$