Câu hỏi: Cho $\int\limits_{0}^{1}{\dfrac{x}{{{\left( x+2 \right)}^{2}}}\text{d}x}=a+b\ln 2+c\ln 3$ với $a,b,c$ là các số hữu tỷ. Giá trị của biểu thức $3a+b+c$ bằng
A. $-2$.
B. $-1$.
C. $2$.
D. $1$.
A. $-2$.
B. $-1$.
C. $2$.
D. $1$.
Ta có
$\begin{aligned}
& \int\limits_{0}^{1}{\dfrac{x}{{{\left( x+2 \right)}^{2}}}\text{d}x}=\int\limits_{0}^{1}{\dfrac{x+2-2}{{{\left( x+2 \right)}^{2}}}\text{d}x}=\int\limits_{0}^{1}{\dfrac{1}{x+2}\text{d}x}+\int\limits_{0}^{1}{\dfrac{2}{{{\left( x+2 \right)}^{2}}}\text{d}x} \\
& =\left. \ln \left| x+2 \right| \right|_{0}^{1}-\left. \dfrac{2}{x+2} \right|_{0}^{1}=\dfrac{1}{3}-\ln 2+\ln 3 \\
\end{aligned}$;
Suy ra $a=\dfrac{1}{3},b=-1,c=1\Rightarrow 3a+b+c=1$.
$\begin{aligned}
& \int\limits_{0}^{1}{\dfrac{x}{{{\left( x+2 \right)}^{2}}}\text{d}x}=\int\limits_{0}^{1}{\dfrac{x+2-2}{{{\left( x+2 \right)}^{2}}}\text{d}x}=\int\limits_{0}^{1}{\dfrac{1}{x+2}\text{d}x}+\int\limits_{0}^{1}{\dfrac{2}{{{\left( x+2 \right)}^{2}}}\text{d}x} \\
& =\left. \ln \left| x+2 \right| \right|_{0}^{1}-\left. \dfrac{2}{x+2} \right|_{0}^{1}=\dfrac{1}{3}-\ln 2+\ln 3 \\
\end{aligned}$;
Suy ra $a=\dfrac{1}{3},b=-1,c=1\Rightarrow 3a+b+c=1$.
Đáp án D.