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

Solution :
$$\eqalign{ & f\left( x \right) = \frac{{{e^x}}}{{1 + {e^x}}} \cr & \Rightarrow f\left( { - x} \right) = \frac{{{e^{ - x}}}}{{1 + {e^{ - x}}}} = \frac{1}{{{e^x} + 1}} \cr & \therefore \,f\left( x \right) + f\left( { - x} \right) = 1\,\forall \,x \cr & {\text{Now,}} \cr & {I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx} \cr & \Rightarrow {I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)g\left\{ {x\left( {1 - x} \right)} \right\}dx} \cr & \Rightarrow {I_1} = {I_2} - {I_1} \cr & \Rightarrow 2{I_1} = {I_2} \cr} $$