Cho hàm số 

C

- Hướng dẫn giải

Đáp án C

Ta có:  

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

Với

$\begin{array}{l}-1<x<1\\ \Rightarrow f\left(x\right)=\frac{1}{2}\ln \left|\frac{1-x}{x+1}\right|+C_1\end{array}$

Với

$$\begin{gathered} \left[\begin{array}{l}x>1\\ x<-1\end{array}\right. \\ \Rightarrow f\left(x\right)=\frac{1}{2}\ln \frac{x-1}{x+1}+C_2 \end{gathered}$$

Do $f\left(-3\right)+f\left(3\right)=0$ và

$$\begin{gathered} f\left(-\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=2 \\ \Rightarrow \left\{\begin{array}{l}\frac{1}{2}\ln 2+C_2+\frac{1}{2}\ln \frac{1}{2}+C_2=0\\ \frac{1}{2}\ln 3+C_1+\frac{1}{2}\ln \frac{1}{3}+C_1=2\end{array}\right. \\ \iff \left\{\begin{array}{l}{C}_2=0\\ C_1=1\end{array}\right. \end{gathered}$$

Do đó  

$\begin{array}{l}P=f\left(0\right)+f\left(4\right)\\ =1+\frac{1}{2}\ln \frac{3}{5}\end{array}$