If $f\left(x\right)=\frac{e^x}{1+e^x},$ $I_1=$$\underset{f(-a)}{\overset{f\left(a\right)}{\int }}xg\left(x\right(1-x\left)\right)dx$ and $I_2=$$\underset{f(-a)}{\overset{f\left(a\right)}{\int }}g\left(x\left(1-x\right)\right)dx$ then $\frac{I_2}{I_1}$ is 

$f\left(x\right)=\frac{e^x}{1+e^x}$

$\begin{array}{l}Nowf\left(a\right)+f(-a)=\frac{e^a}{1+e^a}+\frac{e^{-a}}{1+e^{-a}}\\ =\frac{e^a}{1+e^a}+\frac{1}{e^a+1}=\frac{e^a+1}{1+e^a}=1\\ \therefore f\left(a\right)+f(-a)=1\\ I_1=\underset{f(-a)}{\overset{f\left(a\right)}{\int }}xg\{x(1-x)\}dx\\ =\underset{f(-a)}{\overset{f\left(a\right)}{\int }}(1-x)g\left(\right(1-x\left)\right(1-(1-x))dx\\ (\therefore \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}\begin{array}{l}f(a+b-x)herea+bmeans\\ f(-a)+f\left(a\right)itsvalue=1\end{array})\\ =\underset{f(-a)}{\overset{f\left(a\right)}{\int }}(1-x)g\left(x\right(1-x\left)\right)dx\\ I_1=\underset{f(-a)}{\overset{f\left(a\right)}{\int }}g\left(x\right(1-x\left)\right)dx-I_1\\ =I_2-I_1\\ 2I_1=I_2\Rightarrow \frac{I_2}{I_1}=2\end{array}$