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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\left( 0;+\infty \right)$ thỏa mãn $f\left( 1 \right)=1$ và ${{e}^{x}}f'\left( {{e}^{x}} \right)=1+{{e}^{x}}$. Khi đó $\int\limits_{1}^{e}{f\left( x \right)}dx$ bằng
A. $\dfrac{{{e}^{2}}-1}{2}$.
B. $\dfrac{3{{e}^{2}}-2}{2}$.
C. $\dfrac{{{e}^{2}}+1}{2}$.
D. $\dfrac{{{e}^{2}}}{2}$.
A. $\dfrac{{{e}^{2}}-1}{2}$.
B. $\dfrac{3{{e}^{2}}-2}{2}$.
C. $\dfrac{{{e}^{2}}+1}{2}$.
D. $\dfrac{{{e}^{2}}}{2}$.
Ta có: ${{e}^{x}}f'\left( {{e}^{x}} \right)=1+{{e}^{x}}\Rightarrow \int{{{e}^{x}}f'\left( {{e}^{x}} \right)dx=\int{\left( 1+{{e}^{x}} \right)dx}}$.
$\underrightarrow{t={{e}^{x}}\text{ }}\int{f'\left( t \right)dt}=x+{{e}^{x}}+C\Rightarrow f\left( t \right)=x+{{e}^{x}}+C$ $\Rightarrow f\left( {{e}^{x}} \right)=x+{{e}^{x}}+C$.
Vì $f\left( 1 \right)=1\Rightarrow f\left( {{e}^{0}} \right)={{e}^{0}}+C=1\Rightarrow C=0$.
Đặt $u={{e}^{x}}\Rightarrow x=\ln u\Rightarrow f\left( u \right)=\ln u+u$ hay $f\left( x \right)=\ln x+x$.
$\Rightarrow \int\limits_{1}^{e}{f\left( x \right)}dx=\int\limits_{1}^{e}{\left( \ln x+x \right)}dx=\int\limits_{1}^{e}{\ln x}dx+\int\limits_{1}^{e}{x}dx=\left. x.\ln x \right|_{1}^{e}-\left. x \right|_{1}^{e}+\left. \dfrac{1}{2}{{x}^{2}} \right|_{1}^{e}=\dfrac{{{e}^{2}}+1}{2}$.
$\underrightarrow{t={{e}^{x}}\text{ }}\int{f'\left( t \right)dt}=x+{{e}^{x}}+C\Rightarrow f\left( t \right)=x+{{e}^{x}}+C$ $\Rightarrow f\left( {{e}^{x}} \right)=x+{{e}^{x}}+C$.
Vì $f\left( 1 \right)=1\Rightarrow f\left( {{e}^{0}} \right)={{e}^{0}}+C=1\Rightarrow C=0$.
Đặt $u={{e}^{x}}\Rightarrow x=\ln u\Rightarrow f\left( u \right)=\ln u+u$ hay $f\left( x \right)=\ln x+x$.
$\Rightarrow \int\limits_{1}^{e}{f\left( x \right)}dx=\int\limits_{1}^{e}{\left( \ln x+x \right)}dx=\int\limits_{1}^{e}{\ln x}dx+\int\limits_{1}^{e}{x}dx=\left. x.\ln x \right|_{1}^{e}-\left. x \right|_{1}^{e}+\left. \dfrac{1}{2}{{x}^{2}} \right|_{1}^{e}=\dfrac{{{e}^{2}}+1}{2}$.
Đáp án C.