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Câu hỏi: Tìm một nguyên hàm ${F\left( x \right)}$ của hàm số ${f\left( x \right)g\left( x \right)}$ biết ${\int{f\left( x \right)}\text{d}x={{e}^{x}}+C}$, ${\int{g\left( x \right)}\text{d}x=\dfrac{{{x}^{2}}}{2}+C}$ và ${F\left( 0 \right)=1}$.
A. ${F\left( x \right)=-x{{e}^{x}}+{{e}^{x}}+1}$.
B. ${F\left( x \right)=x{{e}^{x}}+2{{e}^{x}}-1}$.
C. ${F\left( x \right)=x{{e}^{x}}-{{e}^{x}}}$.
D. ${F\left( x \right)=x{{e}^{x}}-{{e}^{x}}+2}$.
A. ${F\left( x \right)=-x{{e}^{x}}+{{e}^{x}}+1}$.
B. ${F\left( x \right)=x{{e}^{x}}+2{{e}^{x}}-1}$.
C. ${F\left( x \right)=x{{e}^{x}}-{{e}^{x}}}$.
D. ${F\left( x \right)=x{{e}^{x}}-{{e}^{x}}+2}$.
Ta có $\int{f}(x)\text{d}x={{e}^{x}}+C\Rightarrow f(x)={{e}^{x}}$
$\int{g}(x)\text{d}x=\dfrac{{{x}^{2}}}{2}+C\Rightarrow g(x)=x$
Ta có $I=\int{f}(x).g(x)\text{d}x=\int{x}{{e}^{x}}\text{d}x$
Đặt$\left\{ \begin{array}{*{35}{l}}
u=x \\
\text{d}v={{e}^{x}}\text{d}x \\
\end{array}\Rightarrow \left\{ \begin{array}{*{35}{l}}
\text{d}u=\text{d}x \\
v={{e}^{x}} \\
\end{array} \right. \right.$
$\Rightarrow I=x{{e}^{x}}-\int{{{e}^{x}}}dx=x{{e}^{x}}-{{e}^{x}}+C$
Mà $F\left( 0 \right)=1\Rightarrow 0-1+C=1\Rightarrow C=2$
Vậy $F(x)=x{{e}^{x}}-{{e}^{x}}+2$
$\int{g}(x)\text{d}x=\dfrac{{{x}^{2}}}{2}+C\Rightarrow g(x)=x$
Ta có $I=\int{f}(x).g(x)\text{d}x=\int{x}{{e}^{x}}\text{d}x$
Đặt$\left\{ \begin{array}{*{35}{l}}
u=x \\
\text{d}v={{e}^{x}}\text{d}x \\
\end{array}\Rightarrow \left\{ \begin{array}{*{35}{l}}
\text{d}u=\text{d}x \\
v={{e}^{x}} \\
\end{array} \right. \right.$
$\Rightarrow I=x{{e}^{x}}-\int{{{e}^{x}}}dx=x{{e}^{x}}-{{e}^{x}}+C$
Mà $F\left( 0 \right)=1\Rightarrow 0-1+C=1\Rightarrow C=2$
Vậy $F(x)=x{{e}^{x}}-{{e}^{x}}+2$
Đáp án D.