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Câu hỏi: Cho hàm số $y=f\left( x \right)=\left\{ \begin{aligned}
& \dfrac{2}{x+1} khi 0\le x\le 1 \\
& 2x-1 khi 1\le x\le 3 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{3}{f\left( x \right)dx}$
A. $6+\ln 4$
B. $4+\ln 4$
C. $6+\ln 2$
D. $2+2\ln 2$
& \dfrac{2}{x+1} khi 0\le x\le 1 \\
& 2x-1 khi 1\le x\le 3 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{3}{f\left( x \right)dx}$
A. $6+\ln 4$
B. $4+\ln 4$
C. $6+\ln 2$
D. $2+2\ln 2$
$\int\limits_{0}^{3}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{1}^{3}{f\left( x \right)dx}=\int\limits_{0}^{1}{\dfrac{2}{x+1}dx+\int\limits_{1}^{3}{\left( 2x-1 \right)dx}=2\ln \left| x+1 \right|}\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.+\left( {{x}^{2}}-x \right)\left| \begin{aligned}
& ^{3} \\
& _{1} \\
\end{aligned} \right.=6+\ln 4$
& ^{1} \\
& _{0} \\
\end{aligned} \right.+\left( {{x}^{2}}-x \right)\left| \begin{aligned}
& ^{3} \\
& _{1} \\
\end{aligned} \right.=6+\ln 4$
Đáp án A.