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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( x \right)+{f}'\left( x \right)={{e}^{-x}}$ 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 $f\left( x \right)+{f}'\left( x \right)={{e}^{-x}}\Leftrightarrow {{\left( {{e}^{x}} \right)}^{\prime }}.f\left( x \right)+{{e}^{x}}.{f}'\left( x \right)=1$
$\Leftrightarrow {{\left( {{e}^{x}}.f\left( x \right) \right)}^{\prime }}=1\Leftrightarrow {{e}^{x}}.f\left( x \right)=\int{dx}=x+C\Leftrightarrow f\left( x \right)=\dfrac{x+C}{{{e}^{x}}}$
Mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{-x}}$
Do đó $f\left( x \right).{{e}^{2x}}=\left( x+2 \right).{{e}^{x}}\Rightarrow \int{\left( x+2 \right).{{e}^{x}}dx}=\left( x+1 \right).{{e}^{x}}+C$
• ${{\left( uv \right)}^{\prime }}={u}'v+u{v}'$.
• $\int{{f}'\left( x \right)dx}=f\left( x \right)$
$\Leftrightarrow {{\left( {{e}^{x}}.f\left( x \right) \right)}^{\prime }}=1\Leftrightarrow {{e}^{x}}.f\left( x \right)=\int{dx}=x+C\Leftrightarrow f\left( x \right)=\dfrac{x+C}{{{e}^{x}}}$
Mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{-x}}$
Do đó $f\left( x \right).{{e}^{2x}}=\left( x+2 \right).{{e}^{x}}\Rightarrow \int{\left( x+2 \right).{{e}^{x}}dx}=\left( x+1 \right).{{e}^{x}}+C$
Note 16: Phương pháp chung
• ${{\left( {{e}^{x}} \right)}^{\prime }}={{e}^{x}}$. • ${{\left( uv \right)}^{\prime }}={u}'v+u{v}'$.
• $\int{{f}'\left( x \right)dx}=f\left( x \right)$
Đáp án D.