Definite Integration

\(\mathrm{f}(x)=\frac{\mathrm{e}^x}{1+\mathrm{e}^x}, \mathrm{I}_1=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} x \mathrm{~g}(x(1-x)) \mathrm{d} x\) এবং \(\mathrm{I}_2=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} \mathrm{g}(x(1-x)) \mathrm{d} x\) সেক্ষেত্রে \(\frac{\mathrm{I}_2}{\mathrm{I}_1}\) এর মান হবে