Lời giải:\(I = \int\limits_1^2 {\frac{{\ln xdx}}{{{x^2}}}dx} \).
Đặt \(\left\{ \begin{array}{l}u = \ln x\\dv = \frac{1}{{{x^2}}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{x}\\v = - \frac{1}{x}\end{array} \right.\) ta có:
\(\begin{array}{l}I = \left. {\left( {\ln x.\frac{{ - 1}}{x}} \right)} \right|_1^2 + \int\limits_1^2 {\frac{{dx}}{{{x^2}}}} = - \frac{1}{2}\ln 2 - \left. {\frac{1}{x}} \right|_1^2 = - \frac{1}{2}\ln 2 - \frac{1}{2} + 1 = \frac{1}{2} - \frac{1}{2}\ln 2\\ \Rightarrow \left\{ \begin{array}{l}b = 1\\c = 2\\a = \frac{{ - 1}}{2}\end{array} \right. \Rightarrow P = 2a + 3b + c = - 1 + 3 + 2 = 4.\end{array}\)
Chọn D.
