\(I = \int {\frac{{\ln 2x}}{{{x^2}}}}\)
Đặt: \(\left\{ \begin{array}{l} u = \ln 2x\\ dv = \frac{1}{{{x^2}}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{x}dx\\ v = - \frac{1}{x} \end{array} \right.\)
\(I = - \frac{{\ln 2x}}{x} + \int {\frac{1}{{{x^2}}}dx} = - \frac{{\ln 2x}}{x} - \frac{1}{x} + C = \frac{1}{x}(\ln 2x + 1) + C.\)