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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& {{x}^{2}}+x-2\text{ khi }x<2 \\
& \dfrac{1}{x}\text{ khi }x\ge 2 \\
\end{aligned} \right. $. Tích phân $ \int\limits_{-\dfrac{1}{3}}^{0}{f\left( {{e}^{3x+1}} \right){{e}^{3x}}dx}$ bằng
A. $3\left( \dfrac{17}{6}-\ln 2 \right)$.
B. $\dfrac{3}{e}\left( \dfrac{17}{6}-\ln 2 \right)$.
C. $\dfrac{3}{e}\left( \dfrac{7}{6}-\ln 2 \right)$.
D. $\dfrac{e}{3}\left( \dfrac{17}{6}+\ln 2 \right)$.
& {{x}^{2}}+x-2\text{ khi }x<2 \\
& \dfrac{1}{x}\text{ khi }x\ge 2 \\
\end{aligned} \right. $. Tích phân $ \int\limits_{-\dfrac{1}{3}}^{0}{f\left( {{e}^{3x+1}} \right){{e}^{3x}}dx}$ bằng
A. $3\left( \dfrac{17}{6}-\ln 2 \right)$.
B. $\dfrac{3}{e}\left( \dfrac{17}{6}-\ln 2 \right)$.
C. $\dfrac{3}{e}\left( \dfrac{7}{6}-\ln 2 \right)$.
D. $\dfrac{e}{3}\left( \dfrac{17}{6}+\ln 2 \right)$.
Đặt $t={{e}^{3x+1}}\Rightarrow dt=\dfrac{1}{3}{{e}^{3x+1}}dx\Leftrightarrow {{e}^{3x}}dx=\dfrac{3}{e}dt$
Đổi cận $x=-\dfrac{1}{3}\Rightarrow t=1$ ; $x=0\Rightarrow t=e$.
Suy ra $\int\limits_{-\dfrac{1}{3}}^{0}{f\left( {{e}^{3x+1}} \right){{e}^{3x}}dx}=\dfrac{3}{e}\int\limits_{1}^{e}{f\left( t \right)dt}=\dfrac{3}{e}\left( \int\limits_{1}^{2}{\left( {{x}^{2}}+x-2 \right)dx}+\int\limits_{2}^{e}{\dfrac{1}{x}dx} \right)=\dfrac{3}{e}\left( \dfrac{17}{6}-\ln 2 \right).$
Đổi cận $x=-\dfrac{1}{3}\Rightarrow t=1$ ; $x=0\Rightarrow t=e$.
Suy ra $\int\limits_{-\dfrac{1}{3}}^{0}{f\left( {{e}^{3x+1}} \right){{e}^{3x}}dx}=\dfrac{3}{e}\int\limits_{1}^{e}{f\left( t \right)dt}=\dfrac{3}{e}\left( \int\limits_{1}^{2}{\left( {{x}^{2}}+x-2 \right)dx}+\int\limits_{2}^{e}{\dfrac{1}{x}dx} \right)=\dfrac{3}{e}\left( \dfrac{17}{6}-\ln 2 \right).$
Đáp án B.