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

Question

( A ) 2
( B ) -3
( C ) 1
( D ) -1