$f(x)= \large\frac{e^x}{1+e^x}$ $\;I_1= \int \limits_{f(-a)}^{f(a)} x g[(x) (1-x)].dx;\;I_2= \int \limits_{f(-a)}^{f(a)} x g[(x) (1-x)].dx;$ then $\large\frac{I_1}{I_2}=?$

Question

\[\begin {array} {1 1} (a)\;\frac{1}{4} \\ (b)\;\frac{1}{3} \\ (c)\;\frac{1}{2} \\ (d)\;2 \end {array}\]