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Câu hỏi: Tìm họ nguyên hàm của hàm số $f\left( x \right)=x{{e}^{2x}}$ ?
A. $F\left( x \right)=\dfrac{1}{3}{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
B. $~F\left( x \right)=\dfrac{1}{2}{{e}^{2x}}\left( x-2 \right)+C$ $F\left( x \right)=2{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
C. $F\left( x \right)=2{{e}^{2x}}\left( x-2 \right)+C$
D. $F\left( x \right)=2{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
A. $F\left( x \right)=\dfrac{1}{3}{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
B. $~F\left( x \right)=\dfrac{1}{2}{{e}^{2x}}\left( x-2 \right)+C$ $F\left( x \right)=2{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
C. $F\left( x \right)=2{{e}^{2x}}\left( x-2 \right)+C$
D. $F\left( x \right)=2{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
Lời giải
Ta có $F\left( x \right)=\int{x{{e}^{2x}}}dx$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
Suy ra $F\left( x \right)=\dfrac{1}{2}x{{e}^{2x}}-\dfrac{1}{2}\int{{{e}^{2x}}}dx=\dfrac{1}{2}x{{e}^{2x}}-\dfrac{1}{4}{{e}^{2x}}+C=\dfrac{1}{2}{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
Ta có $F\left( x \right)=\int{x{{e}^{2x}}}dx$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
Suy ra $F\left( x \right)=\dfrac{1}{2}x{{e}^{2x}}-\dfrac{1}{2}\int{{{e}^{2x}}}dx=\dfrac{1}{2}x{{e}^{2x}}-\dfrac{1}{4}{{e}^{2x}}+C=\dfrac{1}{2}{{e}^{2x}}\left( x-\dfrac{1}{2} \right)+C$
Đáp án A.