Đề bài: Tính tích phân \(I = \int\limits_1^2 {\frac{{\ln x}}{{{x^3}}}dx} .\)

Đặt \(\left\{ \begin{array}{l}u = \ln x\\dv = \frac{{dx}}{{{x^3}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{x}\\v =  – \frac{1}{{2{x^2}}}\end{array} \right.\) 

\(\Rightarrow I = \left. { – \frac{{\ln x}}{{2{x^2}}}} \right|_1^2 + \frac{1}{2}\int\limits_1^2 {\frac{{dx}}{{{x^3}}}}  = \left. { – \frac{{\ln x}}{{2{x^2}}}} \right|_1^2\left. { – \frac{1}{{4{x^2}}}} \right|_1^2 =  – \frac{{\ln 2}}{8} + \frac{3}{{16}} = \frac{{3 – \ln 2}}{{16}}.\)