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

$I_2={\displaystyle \underset{f\left(-a\right)}{\overset{f\left(a\right)}{\int }}g\left(x\left(1-x\right)\right).dx}$

$I_1={\displaystyle \underset{f\left(-a\right)}{\overset{f\left(a\right)}{\int }}xg\left(x\left(1-x\right)\right)}.dx$

$\therefore {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right).dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)}}.dx$

$={\displaystyle \underset{f\left(-a\right)}{\overset{f\left(a\right)}{\int }}\left(f\left(-a\right)+f\left(a\right)-x\right)g\left(f\left(-a\right)+f\left(a\right)-x\right)}\left(1-\left(f\left(-a\right)+f\left(a\right)-x\right)\right).dx$

$f\left(a\right)+f\left(-a\right)=1$

$={\displaystyle \underset{f\left(-a\right)}{\overset{f\left(a\right)}{\int }}\left(1-x\right)g\left(1-x\right)\left(1-\left(1-x\right)\right)}.dx$

$={\displaystyle \underset{f\left(-a\right)}{\overset{f\left(a\right)}{\int }}\left(1-x\right)g\left(\left(1-x\right)x\right)}.dx$

$={\displaystyle \underset{f\left(-a\right)}{\overset{f\left(a\right)}{\int }}g\left(x\left(1-x\right)\right).dx-{\displaystyle \underset{f\left(-a\right)}{\overset{f\left(a\right)}{\int }}xg\left(x\left(1-x\right)\right)}}.dx$

$I_1=I_2-I_1$

$\Rightarrow 2I_1=I_2$

$\frac{I_2}{I_1}=2$