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Câu hỏi: Cho $F\left( x \right)=\left( x-1 \right){{e}^{x}}$ là một nguyên hàm của hàm số $f\left( x \right){{e}^{2x}}$. Tìm nguyên hàm của hàm số ${f}'\left( x \right){{e}^{2x}}$.
A. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\left( x-2 \right){{e}^{x}}+C$
B. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\dfrac{2-x}{2}{{e}^{x}}+C$
C. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\left( 2-x \right){{e}^{x}}+C$
D. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\left( 4-2x \right){{e}^{x}}+C$
A. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\left( x-2 \right){{e}^{x}}+C$
B. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\dfrac{2-x}{2}{{e}^{x}}+C$
C. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\left( 2-x \right){{e}^{x}}+C$
D. $\int{{f}'\left( x \right){{e}^{2x}}dx}=\left( 4-2x \right){{e}^{x}}+C$
Theo đề bài ta có $\int{f\left( x \right).{{e}^{2x}}dx}=\left( x-1 \right){{e}^{x}}+C$
Suy ra $f\left( x \right).{{e}^{2x}}={{\left[ \left( x-1 \right){{e}^{x}} \right]}^{\prime }}={{e}^{x}}+\left( x-1 \right){{e}^{x}}\Rightarrow f\left( x \right)={{e}^{-x}}+\left( x-1 \right){{e}^{-x}}\Rightarrow {f}'\left( x \right)=\left( 1-x \right){{e}^{-x}}$
$\int{{f}'\left( x \right){{e}^{2x}}dx}=\int{\left( 1-x \right){{e}^{x}}dx}=\int{\left( 1-x \right)d\left( {{e}^{x}} \right)}={{e}^{x}}\left( 1-x \right)+\int{{{e}^{x}}dx}={{e}^{x}}\left( 2-x \right)+C$.
Suy ra $f\left( x \right).{{e}^{2x}}={{\left[ \left( x-1 \right){{e}^{x}} \right]}^{\prime }}={{e}^{x}}+\left( x-1 \right){{e}^{x}}\Rightarrow f\left( x \right)={{e}^{-x}}+\left( x-1 \right){{e}^{-x}}\Rightarrow {f}'\left( x \right)=\left( 1-x \right){{e}^{-x}}$
$\int{{f}'\left( x \right){{e}^{2x}}dx}=\int{\left( 1-x \right){{e}^{x}}dx}=\int{\left( 1-x \right)d\left( {{e}^{x}} \right)}={{e}^{x}}\left( 1-x \right)+\int{{{e}^{x}}dx}={{e}^{x}}\left( 2-x \right)+C$.
Đáp án C.