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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm trên mỗi khoảng $\left( -\infty ;-\dfrac{1}{2} \right)$, $\left( -\dfrac{1}{2};+\infty \right)$ đồng thời thỏa mãn ${f}'\left( x \right)=\dfrac{1}{2x+1}$ $\left( \forall x\ne -\dfrac{1}{2} \right)$, và $f\left( -1 \right)+2f\left( 0 \right)=2\ln 674$. Giá trị của biểu thức $S=f\left( -2 \right)+f\left( 1 \right)+f\left( 4 \right)$ bằng
A. $2\ln 3-\ln 674$.
B. $\ln 2022$.
C. $2\ln 2022$.
D. $3\ln 3$.
${f}'\left( x \right)=\dfrac{1}{2x+1}\Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& \dfrac{1}{2}\ln \left( 2x+1 \right)+{{C}_{1}},\ khi\ x>\dfrac{-1}{2} \\
& \dfrac{1}{2}\ln \left( -2x-1 \right)+{{C}_{2}},\ khi\ x<\dfrac{-1}{2}\ \\
\end{aligned} \right.$
$f\left( 0 \right)={{C}_{1}};\ f\left( -1 \right)={{C}_{2}}\Rightarrow 2f\left( 0 \right)+f\left( -1 \right)=2{{C}_{1}}+{{C}_{2}}\Rightarrow 2{{C}_{1}}+{{C}_{2}}=2\ln 674$.
$\begin{aligned}
& f\left( -2 \right)=\dfrac{1}{2}\ln 3+{{C}_{2}},\ f\left( 1 \right)=\dfrac{1}{2}\ln 3+{{C}_{1}};\ f\left( 4 \right)=\dfrac{1}{2}\ln 9+{{C}_{1}} \\
& \Rightarrow S=f\left( -2 \right)+f\left( 1 \right)+f\left( 4 \right)=\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 7+2{{C}_{1}}+{{C}_{2}} \\
& =\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 9+2\ln 674=2\ln 3+2\ln 674=2\ln 2002. \\
\end{aligned}$
A. $2\ln 3-\ln 674$.
B. $\ln 2022$.
C. $2\ln 2022$.
D. $3\ln 3$.
${f}'\left( x \right)=\dfrac{1}{2x+1}\Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& \dfrac{1}{2}\ln \left( 2x+1 \right)+{{C}_{1}},\ khi\ x>\dfrac{-1}{2} \\
& \dfrac{1}{2}\ln \left( -2x-1 \right)+{{C}_{2}},\ khi\ x<\dfrac{-1}{2}\ \\
\end{aligned} \right.$
$f\left( 0 \right)={{C}_{1}};\ f\left( -1 \right)={{C}_{2}}\Rightarrow 2f\left( 0 \right)+f\left( -1 \right)=2{{C}_{1}}+{{C}_{2}}\Rightarrow 2{{C}_{1}}+{{C}_{2}}=2\ln 674$.
$\begin{aligned}
& f\left( -2 \right)=\dfrac{1}{2}\ln 3+{{C}_{2}},\ f\left( 1 \right)=\dfrac{1}{2}\ln 3+{{C}_{1}};\ f\left( 4 \right)=\dfrac{1}{2}\ln 9+{{C}_{1}} \\
& \Rightarrow S=f\left( -2 \right)+f\left( 1 \right)+f\left( 4 \right)=\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 7+2{{C}_{1}}+{{C}_{2}} \\
& =\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 3+\dfrac{1}{2}\ln 9+2\ln 674=2\ln 3+2\ln 674=2\ln 2002. \\
\end{aligned}$
Đáp án C.