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Câu hỏi: Biết rằng tích phân $I=\int\limits_{1}^{5}{\dfrac{2\left| x-2 \right|+1}{x}dx=4+a\ln 2+b\ln 5}$, với $a,b\in \mathbb{Z}$. Tính $S=a+b?$
A. $S=5$
B. $S=9$
C. $S=-3$
D. $S=11$
A. $S=5$
B. $S=9$
C. $S=-3$
D. $S=11$
Ta có $I=\int\limits_{1}^{5}{\dfrac{2\left| x-2 \right|+1}{x}dx=\int\limits_{1}^{2}{\dfrac{2\left( 2-x \right)+1}{x}dx}+\int\limits_{2}^{5}{\dfrac{2\left( x-2 \right)+1}{x}}}dx=\int\limits_{1}^{2}{\left( \dfrac{5}{x}-2 \right)dx+\int\limits_{2}^{5}{\left( 2-\dfrac{3}{x} \right)dx}}$
$=\left. \left( 5\ln \left| x \right|-2x \right) \right|_{1}^{2}+\left. \left( 2x-3\ln \left| x \right| \right) \right|_{2}^{5}=8\ln 2+4-3\ln 5$
Suy ra $a=8,b=-3\Rightarrow S=a+b=5$
$=\left. \left( 5\ln \left| x \right|-2x \right) \right|_{1}^{2}+\left. \left( 2x-3\ln \left| x \right| \right) \right|_{2}^{5}=8\ln 2+4-3\ln 5$
Suy ra $a=8,b=-3\Rightarrow S=a+b=5$
Đáp án A.