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Câu hỏi: Tính tích phân $I=\int\limits_{0}^{1}{\dfrac{dx}{3-2x}}$
A. $-\dfrac{1}{2}\ln 3.$
B. $-\ln 3.$
C. $\dfrac{1}{2}\ln 3.$
D. $\dfrac{1}{2}\log 3.$
A. $-\dfrac{1}{2}\ln 3.$
B. $-\ln 3.$
C. $\dfrac{1}{2}\ln 3.$
D. $\dfrac{1}{2}\log 3.$
$I=\int\limits_{0}^{1}{\dfrac{dx}{3-2x}}=-\dfrac{1}{2}\ln \left| 3-2x \right|\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=-\dfrac{1}{2}\ln 1+\dfrac{1}{2}\ln 3=\dfrac{1}{2}\ln 3$
& 1 \\
& 0 \\
\end{aligned} \right.=-\dfrac{1}{2}\ln 1+\dfrac{1}{2}\ln 3=\dfrac{1}{2}\ln 3$
Đáp án C.