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Câu hỏi: Cho $\int\limits_{1}^{2}{\dfrac{x}{{{\left( x+1 \right)}^{2}}}}\text{d}x=a+b.\ln 2+c\ln 3$, với $a, b, c$ là các số hữu tỷ. Giá trị của $6a+b+c$ bằng
A. $-2$.
B. $1$.
C. $2$.
D. $-1$.
Ta có $\int\limits_{1}^{2}{\dfrac{x}{{{\left( x+1 \right)}^{2}}} }\text{d}x=\int\limits_{1}^{2}{\dfrac{1}{x+1}\text{d}x}-\int\limits_{1}^{2}{\dfrac{1}{{{\left( x+1 \right)}^{2}}}\text{d}x}$ $=\left. \ln \left| x+1 \right| \right|_{1}^{2}+\left. \dfrac{1}{x+1} \right|_{1}^{2}=\ln 3-\ln 2+\dfrac{1}{3}-\dfrac{1}{2}$
$=-\dfrac{1}{6}-\ln 2+\ln 3$. Vậy $6a+b+c=6.\dfrac{-1}{6}+\left( -1 \right)+1=-1$.
A. $-2$.
B. $1$.
C. $2$.
D. $-1$.
Ta có $\int\limits_{1}^{2}{\dfrac{x}{{{\left( x+1 \right)}^{2}}} }\text{d}x=\int\limits_{1}^{2}{\dfrac{1}{x+1}\text{d}x}-\int\limits_{1}^{2}{\dfrac{1}{{{\left( x+1 \right)}^{2}}}\text{d}x}$ $=\left. \ln \left| x+1 \right| \right|_{1}^{2}+\left. \dfrac{1}{x+1} \right|_{1}^{2}=\ln 3-\ln 2+\dfrac{1}{3}-\dfrac{1}{2}$
$=-\dfrac{1}{6}-\ln 2+\ln 3$. Vậy $6a+b+c=6.\dfrac{-1}{6}+\left( -1 \right)+1=-1$.
Đáp án D.