If \(f(x) = \frac{e^x}{1+e^x}, I_1 = \int_{-a}^a xg(x(1-x)) \, dx\) and \(I_2 = \int_{-a}^a g(x(1-x)) \, dx\), then the value of \(\frac{I_2}{I_1}\) is:
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For integrals with symmetric integrands, pay attention to how the variable behaves. In cases with even symmetry, terms involving odd powers of the variable will cancel out.