Giả sử tích phân \(I=\displaystyle\int\limits_{1}^{6}\dfrac{1}{2x+1}\mathrm{\,d}x=\ln M\), tìm \(M\).
\(M=13\) | |
\(M=4,33\) | |
\(M=\sqrt{\dfrac{13}{3}}\) | |
\(M=\dfrac{13}{3}\) |
Chọn phương án C.
\(\begin{aligned}
I&=\displaystyle\int\limits_{1}^{6}\dfrac{1}{2x+1}\mathrm{\,d}x=\dfrac{1}{2}\ln|2x+1|\bigg|_1^6\\
&=\dfrac{1}{2}\left(\ln13-\ln3\right)=\dfrac{1}{2}\ln\dfrac{13}{3}\\
&=\ln\left(\dfrac{13}{3}\right)^{\tfrac{1}{2}}=\ln\sqrt{\dfrac{13}{3}}.
\end{aligned}\)