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

Note that $f(-a)+f(a)=\frac{e^{-a}}{1+e^{-a}}+\frac{e^a}{1+e^a}=\frac{1}{e^a+1}+\frac{e^a}{1+e^a}=1$.
Let $b=f(a)$ and $c=f(-a),$ so $b+c=1.$
$I_1={\int }_c^{b}xg\{x(1-x)\}dx$.
Using the property ${\int }_c^b\varphi (x)dx={\int }_c^b\varphi (b+c-x)dx$:
$I_1={\int }_c^b(1-x)g\{(1-x)(1-(1-x))\}dx$.
$I_1={\int }_c^b(1-x)g\{(1-x)x\}dx$.
Adding the two forms for $I_1$:
$2I_1={\int }_c^b[x+(1-x)]g\{x(1-x)\}dx={\int }_c^{b}g\{x(1-x)\}dx$.
We recognize that the right side is $I_2$.
$2I_1=I_2\Rightarrow \frac{I_2}{I_1}=2$.