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Câu hỏi: Cho biết tích phân $\int\limits_{0}^{1}{\dfrac{5+\left( x-4 \right){{e}^{x}}}{x{{e}^{x}}+1}}dx=a+b\ln \left( e+1 \right)$ trong đó a, b là các số nguyên. Khẳng định nào sau đây là sai?
A. $a-b=9.$
B. $a.b<-18.$
C. $a+b=5.$
D. $a-2b=13.$
A. $a-b=9.$
B. $a.b<-18.$
C. $a+b=5.$
D. $a-2b=13.$
Cách 1:
$\begin{aligned}
& I=\int\limits_{0}^{1}{\dfrac{5+\left( x-4 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\dfrac{5\left( x{{e}^{x}}+1 \right)-4\left( x+1 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\left[ 5-4\dfrac{\left( x+1 \right){{e}^{x}}}{x{{e}^{x}}+1} \right]dx} \\
& I=\int\limits_{0}^{1}{5dx-4\int\limits_{0}^{1}{\dfrac{\left( x+1 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=5-4\int\limits_{0}^{1}{\dfrac{d\left( x{{e}^{x}}+1 \right)}{x{{e}^{x}}+1}=5-4\ln \left. \left( x{{e}^{x}}+1 \right) \right|_{0}^{1}}} \\
& I=5-4\ln \left( e+1 \right)\equiv a+b\ln \left( e+1 \right). \\
\end{aligned}$
Như vậy ta có: $a=5$ và $b=-4\Rightarrow $ A, B, D đúng.
$\Rightarrow a+b=1\Rightarrow $ Sai.
Cách 2:
$\begin{aligned}
& I=\int\limits_{0}^{1}{\dfrac{5+\left( x-4 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\dfrac{\left( x{{e}^{x}}+1 \right)+4\left( 1-{{e}^{x}} \right)}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\left[ 1+4\dfrac{1-{{e}^{x}}}{x{{e}^{x}}+1} \right]dx} \\
& I=\int\limits_{0}^{1}{dx}+4\int\limits_{0}^{1}{\dfrac{1-{{e}^{x}}}{x{{e}^{x}}+1}}dx=\int\limits_{0}^{1}{dx}+4\int\limits_{0}^{1}{\dfrac{\dfrac{1-{{e}^{x}}}{{{e}^{x}}}}{\dfrac{x{{e}^{x}}+1}{{{e}^{x}}}}dx}=1+4\int\limits_{0}^{1}{\dfrac{{{e}^{-x}}-1}{x+{{e}^{-x}}}dx} \\
& I=1-4\int\limits_{0}^{1}{\dfrac{1-{{e}^{-x}}}{x+{{e}^{-x}}}}dx=1-4\int\limits_{0}^{1}{\dfrac{d\left( x+{{e}^{-x}} \right)}{x+{{e}^{-x}}}}=1-\left. 4\ln \left( x+{{e}^{-x}} \right) \right|_{0}^{1} \\
\end{aligned}$
$I=5-4\ln \left( e+1 \right)\equiv a+b\ln \left( e+1 \right).$
Như vậy ta có: $a=5$ và $b=-4\Rightarrow $ A, B, D đúng.
$\Rightarrow a+b=1\Rightarrow $ C sai
$\begin{aligned}
& I=\int\limits_{0}^{1}{\dfrac{5+\left( x-4 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\dfrac{5\left( x{{e}^{x}}+1 \right)-4\left( x+1 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\left[ 5-4\dfrac{\left( x+1 \right){{e}^{x}}}{x{{e}^{x}}+1} \right]dx} \\
& I=\int\limits_{0}^{1}{5dx-4\int\limits_{0}^{1}{\dfrac{\left( x+1 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=5-4\int\limits_{0}^{1}{\dfrac{d\left( x{{e}^{x}}+1 \right)}{x{{e}^{x}}+1}=5-4\ln \left. \left( x{{e}^{x}}+1 \right) \right|_{0}^{1}}} \\
& I=5-4\ln \left( e+1 \right)\equiv a+b\ln \left( e+1 \right). \\
\end{aligned}$
Như vậy ta có: $a=5$ và $b=-4\Rightarrow $ A, B, D đúng.
$\Rightarrow a+b=1\Rightarrow $ Sai.
Cách 2:
$\begin{aligned}
& I=\int\limits_{0}^{1}{\dfrac{5+\left( x-4 \right){{e}^{x}}}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\dfrac{\left( x{{e}^{x}}+1 \right)+4\left( 1-{{e}^{x}} \right)}{x{{e}^{x}}+1}dx}=\int\limits_{0}^{1}{\left[ 1+4\dfrac{1-{{e}^{x}}}{x{{e}^{x}}+1} \right]dx} \\
& I=\int\limits_{0}^{1}{dx}+4\int\limits_{0}^{1}{\dfrac{1-{{e}^{x}}}{x{{e}^{x}}+1}}dx=\int\limits_{0}^{1}{dx}+4\int\limits_{0}^{1}{\dfrac{\dfrac{1-{{e}^{x}}}{{{e}^{x}}}}{\dfrac{x{{e}^{x}}+1}{{{e}^{x}}}}dx}=1+4\int\limits_{0}^{1}{\dfrac{{{e}^{-x}}-1}{x+{{e}^{-x}}}dx} \\
& I=1-4\int\limits_{0}^{1}{\dfrac{1-{{e}^{-x}}}{x+{{e}^{-x}}}}dx=1-4\int\limits_{0}^{1}{\dfrac{d\left( x+{{e}^{-x}} \right)}{x+{{e}^{-x}}}}=1-\left. 4\ln \left( x+{{e}^{-x}} \right) \right|_{0}^{1} \\
\end{aligned}$
$I=5-4\ln \left( e+1 \right)\equiv a+b\ln \left( e+1 \right).$
Như vậy ta có: $a=5$ và $b=-4\Rightarrow $ A, B, D đúng.
$\Rightarrow a+b=1\Rightarrow $ C sai
Đáp án C.