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If \(I_1 =\int\frac{e^x dx}{e^x + e^{-x}}\) and \(I_2 =\int\frac{dx}{e^{2x} + 1},\)then what is I1 + I2 equal to?
Formula used:\(? x^ndx = \frac{x^{n+1}}{n+1}+C\)? dx = x + CCalculation:\(I_1 =?\frac{e^x dx}{e^x + e^{-x}}\) \(? I_1 =?\frac{e^x dx}{e^x +\frac{1}{ e^{x}}}\)\(? I_1 =?\frac{e^{2x} dx}{e^{2x }+1}\) ----(1) \(I_2 =?\frac{dx}{e^{2x} + 1},\) ----(2)Add equation (1) & (2)? I = I1 + I2? I = \(?\frac{e^{2x} dx}{e^{2x }+1}\) + \(?\frac{dx}{e^{2x} + 1}\)\(? I = ?[\frac{e^{2x} }{e^{2x }+1}+ \frac{1}{e^{2x}+1}]dx\)\(? I = ?[\frac{e^{2x}+1 }{e^{2x }+1}]dx\)? I = ?dx Using the formula given above ? I1 + I2 = x + C
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