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Câu hỏi: Cho $\int\limits_{0}^{1}{\dfrac{2{{x}^{2}}+3x}{{{x}^{2}}+3x+2}dx}=a+b\ln 2+c\ln 3$ với $a,b,c$ là các số nguyên. Tổng $a+b+c$ bằng
A. 3.
B. 2.
C. 1.
D. 1.
A. 3.
B. 2.
C. 1.
D. 1.
Ta có: $\int\limits_{0}^{1}{\dfrac{2{{x}^{2}}+3x}{{{x}^{2}}+3x+2}dx}=\int\limits_{0}^{1}{\left( 2-\dfrac{3x+4}{{{x}^{2}}+3x+2} \right)dx}$
$=\int\limits_{0}^{1}{\left( 2-\dfrac{1}{x+1}-\dfrac{2}{x+2} \right)dx}=\left. \left( 2x-\ln \left| x+1 \right|-2\ln \left| x+2 \right| \right) \right|_{0}^{1}=2+\ln 2-2\ln 3$
Suy ra $a=2;b=1;c=-2$.
Vậy $a+b+c=1$.
$=\int\limits_{0}^{1}{\left( 2-\dfrac{1}{x+1}-\dfrac{2}{x+2} \right)dx}=\left. \left( 2x-\ln \left| x+1 \right|-2\ln \left| x+2 \right| \right) \right|_{0}^{1}=2+\ln 2-2\ln 3$
Suy ra $a=2;b=1;c=-2$.
Vậy $a+b+c=1$.
Đáp án C.