int frac{1}{1+e^{x}} d x is equal to | vedclass

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(C) To evaluate the integral $ I = \int \frac{1}{1+e^{x}} d x $:Multiply the numerator and denominator by $ e^{-x} $:$ I = \int \frac{e^{-x}}{e^{-x}(1

Vedclass · Accepted Answer

$ \log _{e}\left(\frac{e^{x}}{e^{x}+1}\right)+C $

Vedclass · Answer

(C) To evaluate the integral $ I = \int \frac{1}{1+e^{x}} d x $:Multiply the numerator and denominator by $ e^{-x} $:$ I = \int \frac{e^{-x}}{e^{-x}(1+e^{x})} d x $$ I = \int \frac{e^{-x}}{e^{-x}+1} d x $Let $ u = e^{-x} + 1 $. Then $ du = -e^{-x} d x $,which implies $ e^{-x} d x = -du $.Substituting these into the integral:$ I = \int \frac{-du}{u} $$ I = -\ln |u| + C $$ I = -\ln |e^{-x} + 1| + C $Simplify the expression inside the logarithm:$ I = -\ln \left| \frac{1}{e^{x}} + 1 \right| + C $$ I = -\ln \left| \frac{1+e^{x}}{e^{x}} \right| + C $Using the property $ -\ln(a/b) = \ln(b/a) $:$ I = \ln \left| \frac{e^{x}}{1+e^{x}} \right| + C $Thus,the correct option is $ C $.

$ \int \frac{1}{1+e^{x}} d x $ is equal to

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