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Câu hỏi: Giá trị của tích phân $I=\int\limits_{0}^{1}{\dfrac{x}{x+1}dx}$ là
A. $I=2+\ln 2.$
B. $I=1+\ln 2.$
C. $I=1-\ln 2.$
D. $I=2-\ln 2.$
A. $I=2+\ln 2.$
B. $I=1+\ln 2.$
C. $I=1-\ln 2.$
D. $I=2-\ln 2.$
Ta có
$I=\int\limits_{0}^{1}{\dfrac{x}{x+1}dx}=\int\limits_{0}^{1}{\dfrac{x+1-1}{x+1}dx}=\int\limits_{0}^{1}{\left( 1-\dfrac{1}{x+1} \right)dx}=\int\limits_{0}^{1}{dx}-\int\limits_{0}^{1}{\dfrac{1}{x+1}dx}$
$=x\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\ln \left( x+1 \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=1-\ln 2.$
$I=\int\limits_{0}^{1}{\dfrac{x}{x+1}dx}=\int\limits_{0}^{1}{\dfrac{x+1-1}{x+1}dx}=\int\limits_{0}^{1}{\left( 1-\dfrac{1}{x+1} \right)dx}=\int\limits_{0}^{1}{dx}-\int\limits_{0}^{1}{\dfrac{1}{x+1}dx}$
$=x\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\ln \left( x+1 \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=1-\ln 2.$
Đáp án C.