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Câu hỏi: Cho $F\left( x \right)=\left( {{x}^{2}}+2x \right){{e}^{x}}$ là một nguyên hàm của $f\left( x \right).{{e}^{2x}}.$ Tìm họ nguyên hàm của hàm số $f'\left( x \right).{{e}^{2x}}.$
A. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( 2+{{x}^{2}} \right){{e}^{x}}+C$
B. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( {{x}^{2}}-2 \right){{e}^{x}}+C$
C. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( 2-{{x}^{2}} \right){{e}^{x}}+C$
D. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( -{{x}^{2}}-2 \right){{e}^{x}}+C$
A. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( 2+{{x}^{2}} \right){{e}^{x}}+C$
B. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( {{x}^{2}}-2 \right){{e}^{x}}+C$
C. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( 2-{{x}^{2}} \right){{e}^{x}}+C$
D. $\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( -{{x}^{2}}-2 \right){{e}^{x}}+C$
* Do $F\left( x \right)=\left( {{x}^{2}}+2x \right){{e}^{x}}$ là một nguyên hàm của $f\left( x \right).{{e}^{2x}}$ nên ta có:
$f\left( x \right).{{e}^{2x}}=F'\left( x \right)=\left[ \left( {{x}^{2}}+2x \right){{e}^{x}} \right]'=\left( 2x+2 \right){{e}^{x}}+\left( {{x}^{2}}+2x \right){{e}^{x}}$
$\Leftrightarrow f\left( x \right){{e}^{2x}}=\left( {{x}^{2}}+2x+2x+2 \right){{e}^{x}}\Leftrightarrow f\left( x \right){{e}^{2x}}=\left( {{x}^{2}}+4x+2 \right){{e}^{x}}.$
Tính $I=\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}.$
Đặt $\left\{ \begin{aligned}
& u={{e}^{2x}} \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2{{e}^{2x}}dx \\
& v=f\left( x \right) \\
\end{aligned} \right..$
Ta có $I=\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=f\left( x \right){{e}^{2x}}-2\int\limits_{{}}^{{}}{f\left( x \right).{{e}^{2x}}dx}=\left( {{x}^{2}}+4x+2 \right){{e}^{x}}-2\left( {{x}^{2}}+2x \right){{e}^{x}}+C$
$=\left( {{x}^{2}}+4x+2-2{{x}^{2}}-4x \right){{e}^{x}}+C=\left( 2-{{x}^{2}} \right){{e}^{x}}+C.$
Vậy $I=\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( 2-{{x}^{2}} \right){{e}^{x}}+C.$
$f\left( x \right).{{e}^{2x}}=F'\left( x \right)=\left[ \left( {{x}^{2}}+2x \right){{e}^{x}} \right]'=\left( 2x+2 \right){{e}^{x}}+\left( {{x}^{2}}+2x \right){{e}^{x}}$
$\Leftrightarrow f\left( x \right){{e}^{2x}}=\left( {{x}^{2}}+2x+2x+2 \right){{e}^{x}}\Leftrightarrow f\left( x \right){{e}^{2x}}=\left( {{x}^{2}}+4x+2 \right){{e}^{x}}.$
Tính $I=\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}.$
Đặt $\left\{ \begin{aligned}
& u={{e}^{2x}} \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2{{e}^{2x}}dx \\
& v=f\left( x \right) \\
\end{aligned} \right..$
Ta có $I=\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=f\left( x \right){{e}^{2x}}-2\int\limits_{{}}^{{}}{f\left( x \right).{{e}^{2x}}dx}=\left( {{x}^{2}}+4x+2 \right){{e}^{x}}-2\left( {{x}^{2}}+2x \right){{e}^{x}}+C$
$=\left( {{x}^{2}}+4x+2-2{{x}^{2}}-4x \right){{e}^{x}}+C=\left( 2-{{x}^{2}} \right){{e}^{x}}+C.$
Vậy $I=\int\limits_{{}}^{{}}{f'\left( x \right).{{e}^{2x}}dx}=\left( 2-{{x}^{2}} \right){{e}^{x}}+C.$
Đáp án C.