Google
Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( x \right)+f'\left( x \right)={{e}^{-x}},\forall x\in \mathbb{R}$ và $f\left( 0 \right)=2$. Tất cả các nguyên hàm của $f\left( x \right){{e}^{2x}}$ là
A. $\left( x-2 \right){{e}^{x}}+{{e}^{x}}+C.$
B. $\left( x+2 \right){{e}^{2x}}+{{e}^{x}}+C.$
C. $\left( x-1 \right){{e}^{x}}+C.$
D. $\left( x+1 \right){{e}^{x}}+C.$
A. $\left( x-2 \right){{e}^{x}}+{{e}^{x}}+C.$
B. $\left( x+2 \right){{e}^{2x}}+{{e}^{x}}+C.$
C. $\left( x-1 \right){{e}^{x}}+C.$
D. $\left( x+1 \right){{e}^{x}}+C.$
Ta có $f\left( x \right)+f'\left( x \right)={{e}^{-x}}\Leftrightarrow \left( {{e}^{x}} \right)'.f\left( x \right)+{{e}^{x}}.f'\left( x \right)=1\Leftrightarrow \left( {{e}^{x}}.f\left( x \right) \right)'=1$
$\Leftrightarrow {{e}^{x}}.f\left( x \right)=\int{dx}=x+C\Leftrightarrow f\left( x \right)=\dfrac{x+C}{{{e}^{x}}}$ mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{-x}}$
Do đó $f\left( x \right){{e}^{2x}}=\left( x+2 \right){{e}^{x}}\Rightarrow \int{f\left( x \right){{e}^{2x}}dx}=\int{\left( x+2 \right){{e}^{x}}dx}=\left( x+2 \right){{e}^{x}}-\int{{{e}^{x}}dx}=\left( x+1 \right){{e}^{x}}+C$
$\Leftrightarrow {{e}^{x}}.f\left( x \right)=\int{dx}=x+C\Leftrightarrow f\left( x \right)=\dfrac{x+C}{{{e}^{x}}}$ mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{-x}}$
Do đó $f\left( x \right){{e}^{2x}}=\left( x+2 \right){{e}^{x}}\Rightarrow \int{f\left( x \right){{e}^{2x}}dx}=\int{\left( x+2 \right){{e}^{x}}dx}=\left( x+2 \right){{e}^{x}}-\int{{{e}^{x}}dx}=\left( x+1 \right){{e}^{x}}+C$
Đáp án D.