What is the integral of e^x(1+e^{-x})^{-2} ?

The integral of e^x(1+e^{-x})^{-2} is -2x+e^x-2ln|e^{-x}+1|+1/(e^{-x)+1}+C

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integral of e^x(1+e^{-x})^{-2}

\int e^{x} (1 + e^{- x})^{- 2} d x

- 2 x + e^{x} - 2 ln e^{- x} + 1 + \frac{1}{e ^{- x} + 1} + C

Solution steps

\int e^{x} (1 + e^{- x})^{- 2} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

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

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

Substitute back

u = e^{- x}

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

Simplify

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

= - 2 x + e^{x} - 2 ln e^{- x} + 1 + \frac{1}{e ^{- x} + 1}

Add a constant to the solution

= - 2 x + e^{x} - 2 ln e^{- x} + 1 + \frac{1}{e ^{- x} + 1} + C

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What is the integral of e^x(1+e^{-x})^{-2} ?
The integral of e^x(1+e^{-x})^{-2} is -2x+e^x-2ln|e^{-x}+1|+1/(e^{-x)+1}+C