Câu hỏi 47

Đặt \(t=\sqrt{3x+1}\Leftrightarrow {{t}^{2}}=3x+1\Leftrightarrow 2tdt=3dx\)

Đổi cận \(\left\{ \begin{align}   x=0\Leftrightarrow t=1 \\   x=1\Leftrightarrow t=2 \\ \end{align} \right.\), khi đó ta có:  \(\int\limits_{0}^{1}{3{{e}^{\sqrt{3x+1}}}dx}=\int\limits_{1}^{2}{{{e}^{t}}.2tdt}=2\int\limits_{1}^{2}{t{{e}^{t}}dt}\)

Đặt \(\left\{ \begin{array}{l}u = t\\dv = {e^t}dt\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = dt\\v = {e^t}\end{array} \right. \Rightarrow 2\int\limits_1^2 {t{e^t}dt}  = 2\left( {\left. {t{e^t}} \right|_1^2 - \int\limits_1^2 {{e^t}dt} } \right) = 2\left( {\left. {t{e^t}} \right|_1^2 - \left. {{e^t}} \right|_1^2} \right) = 2\left( {2{e^2} - e - \left( {{e^2} - e} \right)} \right) = 2{e^2}\)

\(\Rightarrow \left\{ \begin{align}  a=10 \\   b=c=0 \\ \end{align} \right.\Rightarrow P=a+b+c=10\)

Chọn B.