Câu hỏi 23

Ta có:  \(I = \int\limits_1^2 {\dfrac{{\ln x}}{{{x^2}}}} dx\)

Đặt \(\ln x = t \Rightarrow x = {e^t} \Rightarrow dt = \dfrac{1}{x}dx\). Đổi cận: \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 0\\x = 2 \Rightarrow t = \ln 2\end{array} \right..\)

\( \Rightarrow I = \int\limits_0^{\ln 2} {\dfrac{t}{{{e^t}}}dt = } \int\limits_0^{\ln 2} {t{e^{ - t}}dt} \)

Đặt: \(\left\{ \begin{array}{l}u = t\\dv = {e^{ - t}}dt\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v =  - {e^{ - t}}\end{array} \right.\)

\(\begin{array}{l} \Rightarrow I = \left. { - t{e^{ - t}}} \right|_0^{\ln 2} + \int\limits_0^{\ln 2} {{e^{ - t}}dt}  =  - \ln 2.{e^{ - \ln 2}} - \left. {{e^{ - t}}} \right|_0^{\ln 2} =  - \dfrac{1}{2}\ln 2 - {e^{ - \ln 2}} + 1 =  - \dfrac{1}{2}\ln 2 + \dfrac{1}{2}.\\ \Rightarrow \left\{ \begin{array}{l}a =  - \dfrac{1}{2}\\b = 1\\c = 2\end{array} \right. \Rightarrow S = 2a + 3b + c = 4.\end{array}\)

Chọn  A.