What is the integral from 0 to 1 of 1/((1+e^x)^2) ?

The integral from 0 to 1 of 1/((1+e^x)^2) is -ln(1+e)+ln(2)+1/2+1/(1+e)

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integral from 0 to 1 of 1/((1+e^x)^2)

\int_{0}^{1} \frac{1}{( 1 + e ^{x} ) ^{2}} d x

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

Decimal Notation

0.14882 \dots

Solution steps

\int_{0}^{1} \frac{1}{( 1 + e ^{x} ) ^{2}} d x

Apply u-substitution

: \int_{2}^{1 + e} \frac{1}{u ^{2} ( u - 1 )} d u

= \int_{2}^{1 + e} \frac{1}{u ^{2} ( u - 1 )} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u - 1 )} : - \frac{1}{u} - \frac{1}{u ^{2}} + \frac{1}{u - 1}

= \int_{2}^{1 + e} - \frac{1}{u} - \frac{1}{u ^{2}} + \frac{1}{u - 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{2}^{1 + e} \frac{1}{u} d u - \int_{2}^{1 + e} \frac{1}{u ^{2}} d u + \int_{2}^{1 + e} \frac{1}{u - 1} d u

\int_{2}^{1 + e} \frac{1}{u} d u = ln (1 + e) - ln (2)

\int_{2}^{1 + e} \frac{1}{u ^{2}} d u = - \frac{1}{1 + e} + \frac{1}{2}

\int_{2}^{1 + e} \frac{1}{u - 1} d u = 1

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

Simplify

- (ln (1 + e) - ln (2)) - (- \frac{1}{1 + e} + \frac{1}{2}) + 1 : - ln (1 + e) + ln (2) + \frac{1}{2} + \frac{1}{1 + e}

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

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What is the integral from 0 to 1 of 1/((1+e^x)^2) ?
The integral from 0 to 1 of 1/((1+e^x)^2) is -ln(1+e)+ln(2)+1/2+1/(1+e)