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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ và $f\left( 0 \right)=1$, $F\left( x \right)=f\left( x \right)-{{e}^{x}}-x$ là một nguyên hàm của $f\left( x \right)$. Họ các nguyên hàm của $f\left( x \right)$ là
A. $\left( x+1 \right){{e}^{x}}+C$.
B. $\left( x+1 \right){{e}^{x}}-x+C$.
C. $\left( x+2 \right){{e}^{x}}-x+C$.
D. $\left( x+1 \right){{e}^{x}}+x+C$.
A. $\left( x+1 \right){{e}^{x}}+C$.
B. $\left( x+1 \right){{e}^{x}}-x+C$.
C. $\left( x+2 \right){{e}^{x}}-x+C$.
D. $\left( x+1 \right){{e}^{x}}+x+C$.
Ta có: $F\left( x \right)=f\left( x \right)-{{e}^{x}}-x$
$\Leftrightarrow {F}'\left( x \right)={f}'\left( x \right)-{{e}^{x}}-1$
$\Leftrightarrow f\left( x \right)={f}'\left( x \right)-{{e}^{x}}-1$
$\Leftrightarrow {f}'\left( x \right)-f\left( x \right)={{e}^{x}}+1$
$\Leftrightarrow {{e}^{-x}}.{f}'\left( x \right)-f\left( x \right).{{e}^{-x}}=1+{{e}^{-x}}$
$\Leftrightarrow {{\left( {{e}^{-x}}.f\left( x \right) \right)}^{\prime }}=1+{{e}^{-x}}$.
Do đó: ${{e}^{-x}}.f\left( x \right)=\int{\left( 1+{{e}^{-x}} \right)dx}=x-{{e}^{-x}}+C\Rightarrow {{e}^{0}}.f\left( 0 \right)=0-{{e}^{0}}+C\Rightarrow C=2$.
Khi đó: $f\left( x \right)=x.{{e}^{x}}-1+2{{e}^{x}}$
$\xrightarrow{{}}\int{f\left( x \right)dx}=\int{\left( x{{e}^{x}}-1+2{{e}^{x}} \right)dx}=x{{e}^{x}}-{{e}^{x}}-x+2{{e}^{x}}+C=\left( x+1 \right){{e}^{x}}-x+C$.
Vậy $\int{f\left( x \right)dx}=\left( x+1 \right){{e}^{x}}-x+C$.
$\Leftrightarrow {F}'\left( x \right)={f}'\left( x \right)-{{e}^{x}}-1$
$\Leftrightarrow f\left( x \right)={f}'\left( x \right)-{{e}^{x}}-1$
$\Leftrightarrow {f}'\left( x \right)-f\left( x \right)={{e}^{x}}+1$
$\Leftrightarrow {{e}^{-x}}.{f}'\left( x \right)-f\left( x \right).{{e}^{-x}}=1+{{e}^{-x}}$
$\Leftrightarrow {{\left( {{e}^{-x}}.f\left( x \right) \right)}^{\prime }}=1+{{e}^{-x}}$.
Do đó: ${{e}^{-x}}.f\left( x \right)=\int{\left( 1+{{e}^{-x}} \right)dx}=x-{{e}^{-x}}+C\Rightarrow {{e}^{0}}.f\left( 0 \right)=0-{{e}^{0}}+C\Rightarrow C=2$.
Khi đó: $f\left( x \right)=x.{{e}^{x}}-1+2{{e}^{x}}$
$\xrightarrow{{}}\int{f\left( x \right)dx}=\int{\left( x{{e}^{x}}-1+2{{e}^{x}} \right)dx}=x{{e}^{x}}-{{e}^{x}}-x+2{{e}^{x}}+C=\left( x+1 \right){{e}^{x}}-x+C$.
Vậy $\int{f\left( x \right)dx}=\left( x+1 \right){{e}^{x}}-x+C$.
Đáp án C.