​​​​​​​Đáp án:

Đặt \( x = -t \Rightarrow dx = -dt \)

\( \begin{cases} x = -1 \Rightarrow t = 1 \\ x = 1 \Rightarrow t = -1 \end{cases} \)

\( I = \int_{-1}^{1} \frac{f(x)}{1 + 2018^x} dx = - \int_{1}^{-1} \frac{f(-t)}{1 + 2018^{-t}} (dt) = \int_{-1}^{1} \frac{f(t) \cdot 2018^t}{1 + 2018^t} dt \),

\( = \int_{-1}^{1} \frac{f(t)(2018^t + 1 - 1)}{1 + 2018^t} dt = \int_{-1}^{1} f(t) dt - I \),

\( \Rightarrow 2I = 6 \Rightarrow I = 3  \Rightarrow \boxed{B} \).