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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ -2; 1 \right\}$ thỏa mãn ${f}'\left( x \right)=\dfrac{1}{{{x}^{2}}+x-2}$, $f\left( -3 \right)-f\left( 3 \right)=0$ và $f\left( 0 \right)=\dfrac{1}{3}$. Giá trị của biểu thức $f\left( -4 \right)+f\left( -1 \right)-f\left( 4 \right)$ bằng.
A. $\dfrac{1}{3}\ln \dfrac{8}{5}+1$.
B. $\dfrac{1}{3}\ln 2+\dfrac{1}{3}$.
C. $\ln 80+1$.
D. $\dfrac{1}{3}\ln \dfrac{4}{5}+\ln 2+1$.
A. $\dfrac{1}{3}\ln \dfrac{8}{5}+1$.
B. $\dfrac{1}{3}\ln 2+\dfrac{1}{3}$.
C. $\ln 80+1$.
D. $\dfrac{1}{3}\ln \dfrac{4}{5}+\ln 2+1$.
$f\left( x \right)=\int{\dfrac{1}{{{x}^{2}}+x-2}\text{d}x}$ $=\left\{ \begin{aligned}
& \dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+{{C}_{1}}, \forall x\in \left( -\infty ;-2 \right) \\
& \dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+{{C}_{2}}, \forall x\in \left( -2;1 \right) \\
& \dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+{{C}_{3}}, \forall x\in \left( 1;+\infty \right) \\
\end{aligned} \right.$.
Ta có $f\left( -3 \right)=\dfrac{1}{3}\ln 4+{{C}_{1}} , \forall x\in \left( -\infty ;2 \right)$, $f\left( 0 \right)=\dfrac{1}{3}\ln \dfrac{1}{2}+{{C}_{1}} , \forall x\in \left( -2;1 \right)$,
$f\left( 3 \right)=\dfrac{1}{3}\ln \dfrac{2}{5}+{{C}_{3}} , \forall x\in \left( 1;+\infty \right)$,
Theo giả thiết ta có $f\left( 0 \right)=\dfrac{1}{3}$ $\Leftrightarrow {{C}_{2}}=\dfrac{1}{3}\left( 1+\ln 2 \right)$.
$\Rightarrow f\left( -1 \right)=\dfrac{2}{3}\ln 2+\dfrac{1}{3}$.
Và $f\left( -3 \right)-f\left( 3 \right)=0$ $\Leftrightarrow {{C}_{1}}-{{C}_{3}}=\dfrac{1}{3}\ln \dfrac{1}{10}$.
Vậy $f\left( -4 \right)+f\left( -1 \right)-f\left( 4 \right)$ $=\dfrac{1}{3}\ln \dfrac{5}{2}+{{C}_{1}}+\dfrac{1}{3}\ln 2+\dfrac{1}{3}+\dfrac{1}{3}\ln 2+\dfrac{1}{3}\ln 2-{{C}_{2}}$ $=\dfrac{1}{3}\ln 2+\dfrac{1}{3}$.
& \dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+{{C}_{1}}, \forall x\in \left( -\infty ;-2 \right) \\
& \dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+{{C}_{2}}, \forall x\in \left( -2;1 \right) \\
& \dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+{{C}_{3}}, \forall x\in \left( 1;+\infty \right) \\
\end{aligned} \right.$.
Ta có $f\left( -3 \right)=\dfrac{1}{3}\ln 4+{{C}_{1}} , \forall x\in \left( -\infty ;2 \right)$, $f\left( 0 \right)=\dfrac{1}{3}\ln \dfrac{1}{2}+{{C}_{1}} , \forall x\in \left( -2;1 \right)$,
$f\left( 3 \right)=\dfrac{1}{3}\ln \dfrac{2}{5}+{{C}_{3}} , \forall x\in \left( 1;+\infty \right)$,
Theo giả thiết ta có $f\left( 0 \right)=\dfrac{1}{3}$ $\Leftrightarrow {{C}_{2}}=\dfrac{1}{3}\left( 1+\ln 2 \right)$.
$\Rightarrow f\left( -1 \right)=\dfrac{2}{3}\ln 2+\dfrac{1}{3}$.
Và $f\left( -3 \right)-f\left( 3 \right)=0$ $\Leftrightarrow {{C}_{1}}-{{C}_{3}}=\dfrac{1}{3}\ln \dfrac{1}{10}$.
Vậy $f\left( -4 \right)+f\left( -1 \right)-f\left( 4 \right)$ $=\dfrac{1}{3}\ln \dfrac{5}{2}+{{C}_{1}}+\dfrac{1}{3}\ln 2+\dfrac{1}{3}+\dfrac{1}{3}\ln 2+\dfrac{1}{3}\ln 2-{{C}_{2}}$ $=\dfrac{1}{3}\ln 2+\dfrac{1}{3}$.
Đáp án B.