calculate-I-0-1-1-1-e-x-dx-

Question Number 133964 by mathocean1 last updated on 26/Feb/21

c a l c u l a t e

I = \int_{0}^{1} \frac{1}{1 + e^{x}} d x

Answered by bobhans last updated on 26/Feb/21

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

= x - \ln (1 + e^{x}) + c

I = {[x - \ln (1 + e^{x})]}_{0}^{1} = 1 - \ln (1 + e) + \ln (2)

= 1 + \ln (\frac{2}{1 + e}) = \ln (\frac{2 e}{1 + e})

Answered by Olaf last updated on 26/Feb/21

I = \int_{0}^{1} \frac{d x}{1 + e^{x}}

I = \int_{0}^{1} (\frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1 + e^{x}}) d x

I = {[x - \ln (1 + e^{x})]}_{0}^{1}

I = 1 - \ln (1 + e) + \ln 2

I = 1 + \ln (\frac{2}{1 + e})

Answered by mathmax by abdo last updated on 26/Feb/21

I = \int_{0}^{1} \frac{dx}{e^{x} + 1} changement e^{x} = t give x = lnt \Rightarrow

I = \int_{1}^{e} \frac{dt}{t (t + 1)} = \int_{1}^{e} (\frac{1}{t} - \frac{1}{t + 1}) dt = {[\ln ∣ \frac{t}{t + 1} ∣]}_{1}^{e} = \ln (\frac{e}{e + 1}) - \ln (\frac{1}{2})

= 1 - \ln (e + 1) + \ln (2)