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Câu hỏi: Cho hàm số $y=f\left( x \right)=\left\{ \begin{array}{*{35}{l}}
3{{x}^{2}} khi 0\le x\le 1 \\
4-x khi 1\le x\le 2 \\
\end{array} \right. $. Tính $ \int\limits_{0}^{{{e}^{2}}-1}{\dfrac{\ln \left( x+1 \right)}{x+1}dx}$
A. $\dfrac{7}{2}$
B. 1
C. $\dfrac{5}{2}$
D. $\dfrac{3}{2}$
3{{x}^{2}} khi 0\le x\le 1 \\
4-x khi 1\le x\le 2 \\
\end{array} \right. $. Tính $ \int\limits_{0}^{{{e}^{2}}-1}{\dfrac{\ln \left( x+1 \right)}{x+1}dx}$
A. $\dfrac{7}{2}$
B. 1
C. $\dfrac{5}{2}$
D. $\dfrac{3}{2}$
Đặt $t=\ln \left( x+1 \right)\Rightarrow dt=\dfrac{1}{x+1}dx$
Đổi cận $\left\{ \begin{array}{*{35}{l}}
{{x}_{2}}={{e}^{2}}-1\Rightarrow {{t}_{2}}=\ln \left( {{e}^{2}}-1+1 \right)=2 \\
{{x}_{1}}=0\Rightarrow {{t}_{1}}=\ln \left( 0+1 \right)=0 \\
\end{array} \right.$
Ta có: $\int\limits_{0}^{2}{f\left( t \right)dt}=\int\limits_{0}^{1}{f\left( t \right)dt}+\int\limits_{1}^{2}{f\left( t \right)}=\int\limits_{0}^{1}{3{{x}^{2}}+\int\limits_{1}^{2}{4-x}=\dfrac{7}{2}}$
Đổi cận $\left\{ \begin{array}{*{35}{l}}
{{x}_{2}}={{e}^{2}}-1\Rightarrow {{t}_{2}}=\ln \left( {{e}^{2}}-1+1 \right)=2 \\
{{x}_{1}}=0\Rightarrow {{t}_{1}}=\ln \left( 0+1 \right)=0 \\
\end{array} \right.$
Ta có: $\int\limits_{0}^{2}{f\left( t \right)dt}=\int\limits_{0}^{1}{f\left( t \right)dt}+\int\limits_{1}^{2}{f\left( t \right)}=\int\limits_{0}^{1}{3{{x}^{2}}+\int\limits_{1}^{2}{4-x}=\dfrac{7}{2}}$
Đáp án A.