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