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Câu hỏi: Tích phân $I=\int\limits_{0}^{1}{\dfrac{1}{2x+1}dx}$ bằng:
A. $I=\dfrac{6}{11}.$
B. $I=2\text{ln3}.$
C. $I=\dfrac{1}{2}\text{ln3}\text{.}$
D. $I=0,54.$
A. $I=\dfrac{6}{11}.$
B. $I=2\text{ln3}.$
C. $I=\dfrac{1}{2}\text{ln3}\text{.}$
D. $I=0,54.$
Đặt $t=2x+1\Rightarrow dt=2dx$
Đổi cận
Khi đó $I=\int\limits_{0}^{1}{\dfrac{1}{2x+1}dx=\int\limits_{1}^{3}{\dfrac{1}{t}.\dfrac{1}{2}dt=\dfrac{1}{2}\int\limits_{1}^{3}{\dfrac{1}{t}dt=\dfrac{1}{2}\text{ln}\left| t \right|\left| \begin{matrix}
3 \\
1 \\
\end{matrix}=\dfrac{1}{2}\left( \ln 3-\ln 1 \right)=\dfrac{1}{2}\ln 3. \right.}}}$
Vậy $I=\dfrac{1}{2}\text{ln3}\text{.}$
Đổi cận
| $x$ | $0$ $1$ |
| $t$ | $1$ $3$ |
3 \\
1 \\
\end{matrix}=\dfrac{1}{2}\left( \ln 3-\ln 1 \right)=\dfrac{1}{2}\ln 3. \right.}}}$
Vậy $I=\dfrac{1}{2}\text{ln3}\text{.}$
Đáp án C.