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Câu hỏi: Hàm số $f\left( x \right)=\left( x-1 \right){{e}^{x}}$ có một nguyên hàm $F\left( x \right)$ là kết quả nào sau đây, biết nguyên hàm này bằng 1 khi $x=0?$
A. $F\left( x \right)=\left( x-1 \right){{e}^{x}}.$
B. $F\left( x \right)=\left( x-1 \right){{e}^{x}}+1.$
C. $F\left( x \right)=\left( x-2 \right){{e}^{x}}.$
D. $F\left( x \right)=\left( x-2 \right){{e}^{x}}+3.$
A. $F\left( x \right)=\left( x-1 \right){{e}^{x}}.$
B. $F\left( x \right)=\left( x-1 \right){{e}^{x}}+1.$
C. $F\left( x \right)=\left( x-2 \right){{e}^{x}}.$
D. $F\left( x \right)=\left( x-2 \right){{e}^{x}}+3.$
Ta có: $F\left( x \right)=\int{f\left( x \right)dx}=\int{\left( x-1 \right){{e}^{x}}dx}$
Đặt: $\left\{ \begin{aligned}
& u=x-1 \\
& dv={{e}^{x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$
Do đó: $F\left( x \right)=\int{\left( x-1 \right){{e}^{x}}dx}=\left( x-1 \right){{e}^{x}}-\int{{{e}^{x}}dx}=\left( x-1 \right){{e}^{x}}-{{e}^{x}}+C$
Theo giả thiết: $F\left( 0 \right)=1\Leftrightarrow \left( 0-1 \right).{{e}^{o}}-{{e}^{o}}+C=1\Leftrightarrow C=3$
$\Rightarrow F\left( x \right)=\left( x-1 \right){{e}^{x}}-{{e}^{x}}+3=\left( x-2 \right){{e}^{x}}+3$
Đặt: $\left\{ \begin{aligned}
& u=x-1 \\
& dv={{e}^{x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$
Do đó: $F\left( x \right)=\int{\left( x-1 \right){{e}^{x}}dx}=\left( x-1 \right){{e}^{x}}-\int{{{e}^{x}}dx}=\left( x-1 \right){{e}^{x}}-{{e}^{x}}+C$
Theo giả thiết: $F\left( 0 \right)=1\Leftrightarrow \left( 0-1 \right).{{e}^{o}}-{{e}^{o}}+C=1\Leftrightarrow C=3$
$\Rightarrow F\left( x \right)=\left( x-1 \right){{e}^{x}}-{{e}^{x}}+3=\left( x-2 \right){{e}^{x}}+3$
Đáp án D.