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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=\left( 2x+1 \right){{e}^{x}}$ là
A. $2x{{e}^{x}}+C.$
B. $\left( 2x-2 \right){{e}^{x}}+C.$
C. $\left( 2x-1 \right){{e}^{x}}+C.$
D. $\left( 2x+3 \right){{e}^{x}}+C.$
A. $2x{{e}^{x}}+C.$
B. $\left( 2x-2 \right){{e}^{x}}+C.$
C. $\left( 2x-1 \right){{e}^{x}}+C.$
D. $\left( 2x+3 \right){{e}^{x}}+C.$
Đặt $\left\{ \begin{aligned}
& u=2x+1 \\
& dv={{e}^{x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$
$\Rightarrow \int{\left( 2x+1 \right){{e}^{x}}}dx=\left( 2x+1 \right){{e}^{x}}-\int{2{{e}^{x}}dx}=\left( 2x+1 \right){{e}^{x}}-2{{e}^{x}}+C=\left( 2x-1 \right){{e}^{x}}+C$.
& u=2x+1 \\
& dv={{e}^{x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$
$\Rightarrow \int{\left( 2x+1 \right){{e}^{x}}}dx=\left( 2x+1 \right){{e}^{x}}-\int{2{{e}^{x}}dx}=\left( 2x+1 \right){{e}^{x}}-2{{e}^{x}}+C=\left( 2x-1 \right){{e}^{x}}+C$.
Đáp án C.