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Câu hỏi: Biết rằng $\int\limits_{1}^{4}{\dfrac{1}{2+\sqrt{x-1}}dx}=a\sqrt{3}+b\ln \dfrac{2+\sqrt{3}}{2},$ với $a,b\in \mathbb{Z}.$ Tính $S=a+b.$
A. $S=1.$
B. $S=2.$
C. $S=-1.$
D. $S=-2.$
A. $S=1.$
B. $S=2.$
C. $S=-1.$
D. $S=-2.$
Xét $I=\int\limits_{1}^{4}{\dfrac{1}{2-\sqrt{x-1}}dx}$
Đặt $\sqrt{x-1}=t\Rightarrow I=\int\limits_{0}^{\sqrt{3}}{\dfrac{1}{2+t}d\left( {{t}^{2}}+1 \right)}=\int\limits_{0}^{\sqrt{3}}{\dfrac{2t}{t+2}dt}=2\int\limits_{0}^{\sqrt{3}}{\left( 1-\dfrac{2}{t+2} \right)dt}=\left. 2\left( t-2\ln \left| t+2 \right| \right) \right|_{0}^{\sqrt{3}}$
$=2\left[ \sqrt{3}-2\ln \left( 2+\sqrt{3} \right) \right]+4\ln 2=2\sqrt{3}-4\ln \dfrac{2+\sqrt{3}}{2}\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=-4 \\
\end{aligned} \right.\Rightarrow S=-2$.
Đặt $\sqrt{x-1}=t\Rightarrow I=\int\limits_{0}^{\sqrt{3}}{\dfrac{1}{2+t}d\left( {{t}^{2}}+1 \right)}=\int\limits_{0}^{\sqrt{3}}{\dfrac{2t}{t+2}dt}=2\int\limits_{0}^{\sqrt{3}}{\left( 1-\dfrac{2}{t+2} \right)dt}=\left. 2\left( t-2\ln \left| t+2 \right| \right) \right|_{0}^{\sqrt{3}}$
$=2\left[ \sqrt{3}-2\ln \left( 2+\sqrt{3} \right) \right]+4\ln 2=2\sqrt{3}-4\ln \dfrac{2+\sqrt{3}}{2}\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=-4 \\
\end{aligned} \right.\Rightarrow S=-2$.
Đáp án D.