Câu hỏi: Tính tích phân $I=\int\limits_{0}^{1}{{{\left( x+1 \right)}^{2}}{{e}^{x}}}dx$
A. $I=2e+1$.
B. $I=e+1$.
C. $I=2e-1$.
D. $I=e-1$.
A. $I=2e+1$.
B. $I=e+1$.
C. $I=2e-1$.
D. $I=e-1$.
Đặt: $u={{\left( x+1 \right)}^{2}} \Rightarrow du=2\left( x+1 \right)dx$
$dv={{e}^{x}}dx \Rightarrow v={{e}^{x}}$
$\Rightarrow $ $I=\left. \left[ {{\left( x+1 \right)}^{2}}.{{e}^{x}} \right] \right|_{0}^{1}-2\int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}}dx=4e-1-2{{I}_{1}}$
Với ${{I}_{1}}=\int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}}dx$
Đặt $u=x+1 \Rightarrow du=dx$
$dv={{e}^{x}}dx \Rightarrow v={{e}^{x}}$
$\Rightarrow $ ${{I}_{1}}=\left. \left[ \left( x+1 \right).{{e}^{x}} \right] \right|_{0}^{1}-\int\limits_{0}^{1}{{{e}^{x}}}dx=2e-1-\left. {{e}^{x}} \right|_{0}^{1}=e$
Vậy $I=4e-1-2e=2e-1$
$dv={{e}^{x}}dx \Rightarrow v={{e}^{x}}$
$\Rightarrow $ $I=\left. \left[ {{\left( x+1 \right)}^{2}}.{{e}^{x}} \right] \right|_{0}^{1}-2\int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}}dx=4e-1-2{{I}_{1}}$
Với ${{I}_{1}}=\int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}}dx$
Đặt $u=x+1 \Rightarrow du=dx$
$dv={{e}^{x}}dx \Rightarrow v={{e}^{x}}$
$\Rightarrow $ ${{I}_{1}}=\left. \left[ \left( x+1 \right).{{e}^{x}} \right] \right|_{0}^{1}-\int\limits_{0}^{1}{{{e}^{x}}}dx=2e-1-\left. {{e}^{x}} \right|_{0}^{1}=e$
Vậy $I=4e-1-2e=2e-1$
Đáp án C.