Đáp án đúng: A
Giải chi tiết:\(\begin{array}{l}f\left( x \right) = \int\limits_{}^{} {f'\left( x \right)dx} = \int\limits_{}^{} {\dfrac{{dx}}{{{x^2} - 1}}} = \int\limits_{}^{} {\dfrac{{dx}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} = \dfrac{1}{2}\ln \left| {\dfrac{{x - 1}}{{x + 1}}} \right| + C\\\,\,\,\,\,\,\,\,\,\,\,\, = \left[ \begin{array}{l}\dfrac{1}{2}\ln \dfrac{{x - 1}}{{x + 1}} + {C_1}\,\,khi\,\,x \ge 1\,\, \vee \,\,x \le - 1\\\dfrac{1}{2}\ln \dfrac{{1 - x}}{{x + 1}} + {C_2}\,\,khi\,\, - 1 \le x \le 1\end{array} \right.\\\left\{ \begin{array}{l}f\left( 3 \right) + f\left( { - 3} \right) = 4\\f\left( {\dfrac{1}{3}} \right) + f\left( { - \dfrac{1}{3}} \right) = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{1}{2}\ln \dfrac{1}{2} + {C_1} + \dfrac{1}{2}\ln 2 + {C_1} = 4\\\dfrac{1}{2}\ln \dfrac{1}{2} + {C_2} + \dfrac{1}{2}\ln 2 + {C_2} = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{C_1} = 2\\{C_2} = 1\end{array} \right.\\ \Rightarrow f\left( x \right) = \left[ \begin{array}{l}\dfrac{1}{2}\ln \dfrac{{x - 1}}{{x + 1}} + 2\,\,khi\,\,x \ge 1\,\, \vee \,\,x \le - 1\\\dfrac{1}{2}\ln \dfrac{{1 - x}}{{x + 1}} + 1\,\,khi\,\, - 1 \le x \le 1\end{array} \right.\\ \Rightarrow f\left( { - 5} \right) + f\left( 0 \right) + f\left( 2 \right) = \dfrac{1}{2}\ln \dfrac{3}{2} + 2 + \dfrac{1}{2}\ln 1 + 1 + \dfrac{1}{2}\ln \dfrac{1}{3} + 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\ln \dfrac{1}{2} + 5 = 5 - \dfrac{1}{2}\ln 2\end{array}\)
Chọn A.
