If $$f(x) = \frac{e^x}{1 + e^x}$$, $$I_1 = \int_{f(-a)}^{f(a)} xg\{x(1 - x)\} dx$$ and $$I_2 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx$$, then the value of $$\frac{I_2}{I_1}$$ is

A

$$2$$

B

$$-3$$

C

$$-1$$

D

$$1$$