Cho hàm số f(x) xác định trên R (+-1) thỏa mãn f'(x) = 1/x^2-1.

* Hướng dẫn giải

Phương pháp giải: $f\left(x\right)=\int f\text{'}\left(x\right)\text{\hspace{0.17em}d}x$

Giải chi tiết:

$\begin{array}{l}f\left(x\right)=\int f\text{'}\left(x\right)\text{\hspace{0.17em}d}x=\int \frac{1}{x^2-1}\text{\hspace{0.17em}d}x=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C\\ \Rightarrow f\left(x\right)=\left[\begin{array}{l}\frac{1}{2}\ln \frac{x-1}{x+1}+C_{1}khix\in \left(-\infty ;-1\right)\cup \left(1;+\infty \right)\\ \frac{1}{2}\ln \frac{1-x}{x+1}+C_{2}khix\in \left(-1;-1\right)\end{array}\right.\\ \Rightarrow f\left(-3\right)+f\left(3\right)=\frac{1}{2}\ln 2+C_1+\frac{1}{2}\ln \frac{1}{2}+C_1=0\iff C_1=0\\ f\left(\frac{-1}{2}\right)+f\left(3\right)=\frac{1}{2}\ln 3+C_2+\frac{1}{2}\ln \frac{1}{3}+C_2=2\iff C_2=1\\ \Rightarrow f\left(x\right)=\left[\begin{array}{l}\frac{1}{2}\ln \frac{x-1}{x+1}khix\in \left(-\infty ;-1\right)\cup \left(1;+\infty \right)\\ \frac{1}{2}\ln \frac{1-x}{x+1}+1khix\in \left(-1;-1\right)\end{array}\right.\\ \Rightarrow f\left(-2\right)+f\left(0\right)+f\left(4\right)=\frac{1}{2}\ln 3+\frac{1}{2}\ln 1+\frac{1}{2}\ln \frac{3}{5}=1+\frac{1}{2}\ln \frac{9}{5}\\ \end{array}$

Chọn D.