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Câu hỏi: Biết tích phân $\int\limits_{-\dfrac{1}{2}}^{0}{\dfrac{2x+1}{x-1}}dx=a\ln 3+b\ln 2+c$ với $a,b,c\in \mathbb{Z}.$ Tính $S=a+b+c?$
A. $S=1.$
B. $S=-4.$
C. $S=-1.$
D. $S=-2.$
A. $S=1.$
B. $S=-4.$
C. $S=-1.$
D. $S=-2.$
Ta có $\int\limits_{-\dfrac{1}{2}}^{0}{\dfrac{2x+1}{x-1}dx}=\int\limits_{-\dfrac{1}{2}}^{0}{\dfrac{2\left( x-1 \right)+3}{x-1}dx=\int\limits_{-\dfrac{1}{2}}^{0}{\left( 2+\dfrac{3}{x-1} \right)dx}}$
$=\left. \left( 2x+3\ln \left| x-1 \right| \right) \right|_{-\dfrac{1}{2}}^{0}=0-\left( -1+3\ln \dfrac{3}{2} \right)=-3\ln 3+3\ln 2+1.$
Vậy $a=-3,b=3,c=1\Rightarrow S=a+b+c=1.$
$=\left. \left( 2x+3\ln \left| x-1 \right| \right) \right|_{-\dfrac{1}{2}}^{0}=0-\left( -1+3\ln \dfrac{3}{2} \right)=-3\ln 3+3\ln 2+1.$
Vậy $a=-3,b=3,c=1\Rightarrow S=a+b+c=1.$
Đáp án A.