int frac{e^x-1}{e^x+1} dx = | vedclass

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(B) Let $I = \int \frac{e^x-1}{e^x+1} dx$.We can rewrite the integrand as: $I = \int \frac{(e^x+1)-2}{e^x+1} dx$.This simplifies to: $I = \int (1 - \f

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$2 \log_e(1+e^x) - x + c$

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(B) Let $I = \int \frac{e^x-1}{e^x+1} dx$.We can rewrite the integrand as: $I = \int \frac{(e^x+1)-2}{e^x+1} dx$.This simplifies to: $I = \int (1 - \frac{2}{e^x+1}) dx = \int 1 dx - 2 \int \frac{1}{e^x+1} dx$.To integrate $\int \frac{1}{e^x+1} dx$,multiply the numerator and denominator by $e^{-x}$:$I = x - 2 \int \frac{e^{-x}}{1+e^{-x}} dx$.Let $t = 1+e^{-x}$,then $dt = -e^{-x} dx$,which implies $e^{-x} dx = -dt$.Substituting these into the integral:$I = x - 2 \int \frac{-dt}{t} = x + 2 \int \frac{1}{t} dt$.$I = x + 2 \log_e|t| + c = x + 2 \log_e(1+e^{-x}) + c$.Since $1+e^{-x} = \frac{e^x+1}{e^x}$,we have:$I = x + 2 \log_e(\frac{e^x+1}{e^x}) + c = x + 2 \log_e(e^x+1) - 2 \log_e(e^x) + c$.$I = x + 2 \log_e(e^x+1) - 2x + c = 2 \log_e(e^x+1) - x + c$.

$\int \frac{e^x-1}{e^x+1} dx =$

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