Câu hỏi 24

Đặt \(\left\{ \begin{array}{l}u = x + 1\\dv = {e^x}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = {e^x}\end{array} \right.\)

\(\begin{array}{l} \Rightarrow \int\limits_1^2 {\left( {x + 1} \right){e^x}dx}  = \left. {\left( {x + 1} \right){e^x}} \right|_1^2 - \int\limits_1^2 {{e^x}dx}  = 3{e^2} - 2e - \left. {{e^x}} \right|_1^2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3{e^2} - 2e - {e^2} + e = 3{e^2} - {e^2} - e\end{array}\)

Vậy \(a = 3,b =  - 1,c =  - 1 \Rightarrow a + b + c = 1\).

Chọn B.