Cho hàm số f(x) xác định trên $ℝ\backslash \left\{-1;1\right\}$ thỏa mãn $f^{\text{'}}\left(x\right)=\frac{1}{x^2-1}$. Biết $f\left(3\right)+f\left(-3\right)=4$ và $f\left(\frac{1}{3}\right)+f\left(-\frac{1}{3}\right)=2$. Giá trị của biểu thức $f\left(-5\right)+f\left(0\right)+f\left(2\right)$ bằng:

A. $5-\frac{1}{2}\ln 2$

Đáp án A

$f\left(x\right)=\underset{}{\overset{}{\int }}{f}^{\text{'}}\left(x\right)dx=\underset{}{\overset{}{\int }}\frac{dx}{x^2-1}=\underset{}{\overset{}{\int }}\frac{dx}{\left(x-1\right)\left(x+1\right)}=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C$

$=\left[\begin{array}{l}\frac{1}{2}\ln \frac{x-1}{x+1}+C_{1}khix\ge 1\vee x\le -1\\ \frac{1}{2}\ln \frac{1-x}{x+1}+C_{2}khi-1\le x\le 1\end{array}\right.$

$\left\{\begin{array}{l}f\left(3\right)+f\left(-3\right)=4\\ f\left(\frac{1}{3}\right)+f\left(-\frac{1}{3}\right)=2\end{array}\right.\iff \left\{\begin{array}{l}\frac{1}{2}\ln \frac{1}{2}+C_1+\frac{1}{2}\ln 2+C_1=4\\ \frac{1}{2}\ln \frac{1}{2}+C_2+\frac{1}{2}\ln 2+C_2=2\end{array}\right.\iff \left\{\begin{array}{l}{C}_1=2\\ C_2=1\end{array}\right.$

$\Rightarrow f\left(x\right)=\left[\begin{array}{l}\frac{1}{2}\ln \frac{x-1}{x+1}+2khix\ge 1\vee x\le -1\\ \frac{1}{2}\ln \frac{1-x}{x+1}+1khi-1\le x\le 1\end{array}\right.$

$\begin{array}{l}\Rightarrow f\left(-5\right)+f\left(0\right)+f\left(2\right)=\frac{1}{2}\ln \frac{3}{2}+2+\frac{1}{2}\ln 1+1+\frac{1}{2}\ln \frac{1}{3}+2\\ =\frac{1}{2}\ln \frac{1}{2}+5=5-\frac{1}{2}\ln 2\end{array}$