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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( x \right)+{f}'\left( x \right)={{\text{e}}^{-x}}, \forall x\in \mathbb{R}$ và $f\left( 0 \right)=2$. Họ nguyên hàm của hàm số $f\left( x \right).{{e}^{2x}}$ là
A. $x{{e}^{x}}+x+C$.
B. $\left( x+1 \right){{e}^{x}}+C$.
C. $x{{e}^{-x}}+x+C$.
D. $\left( x-1 \right){{e}^{x}}+C$.
A. $x{{e}^{x}}+x+C$.
B. $\left( x+1 \right){{e}^{x}}+C$.
C. $x{{e}^{-x}}+x+C$.
D. $\left( x-1 \right){{e}^{x}}+C$.
$\begin{aligned}
& f\left( x \right)+{f}'\left( x \right)={{\text{e}}^{-x}}\Leftrightarrow f\left( x \right){{\text{e}}^{x}}+{f}'\left( x \right){{\text{e}}^{x}}={{\text{e}}^{-x}}.{{\text{e}}^{x}} \Leftrightarrow f\left( x \right){{\text{e}}^{x}}+{f}'\left( x \right){{\text{e}}^{x}}=1\Leftrightarrow {{\left( f\left( x \right){{\text{e}}^{x}} \right)}^{\prime }}=1 \\
& \Leftrightarrow f\left( x \right){{\text{e}}^{x}}=x+C \\
\end{aligned}$
Với $x=0$ $\Rightarrow f\left( 0 \right){{\text{e}}^{0}}=0+C\Leftrightarrow 2.1=C\Rightarrow C=2\Rightarrow f\left( x \right){{\text{e}}^{x}}=x+2$.
$\Rightarrow f\left( x \right){{\text{e}}^{2x}}=\left( x+2 \right){{e}^{x}}$.
$\Rightarrow \int{f\left( x \right){{\text{e}}^{2x}}dx}=\int{\left( x+2 \right){{e}^{x}}dx=}\left( x+2 \right){{e}^{x}}-\int{{{e}^{x}}dx=}=\left( x+2 \right){{e}^{x}}-{{e}^{x}}+C=\left( x+1 \right){{e}^{x}}+C$.
& f\left( x \right)+{f}'\left( x \right)={{\text{e}}^{-x}}\Leftrightarrow f\left( x \right){{\text{e}}^{x}}+{f}'\left( x \right){{\text{e}}^{x}}={{\text{e}}^{-x}}.{{\text{e}}^{x}} \Leftrightarrow f\left( x \right){{\text{e}}^{x}}+{f}'\left( x \right){{\text{e}}^{x}}=1\Leftrightarrow {{\left( f\left( x \right){{\text{e}}^{x}} \right)}^{\prime }}=1 \\
& \Leftrightarrow f\left( x \right){{\text{e}}^{x}}=x+C \\
\end{aligned}$
Với $x=0$ $\Rightarrow f\left( 0 \right){{\text{e}}^{0}}=0+C\Leftrightarrow 2.1=C\Rightarrow C=2\Rightarrow f\left( x \right){{\text{e}}^{x}}=x+2$.
$\Rightarrow f\left( x \right){{\text{e}}^{2x}}=\left( x+2 \right){{e}^{x}}$.
$\Rightarrow \int{f\left( x \right){{\text{e}}^{2x}}dx}=\int{\left( x+2 \right){{e}^{x}}dx=}\left( x+2 \right){{e}^{x}}-\int{{{e}^{x}}dx=}=\left( x+2 \right){{e}^{x}}-{{e}^{x}}+C=\left( x+1 \right){{e}^{x}}+C$.
Đáp án B.