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

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

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

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

- 1 - e^{- t} + ln 1 + e^{- t} + C

Solution steps

\int \frac{e ^{- 2 t}}{1 + e ^{- t}} d t

Apply u-substitution

: \int - \frac{u - 1}{u} d u

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

Take the constant out:

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

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= - \int 1 - \frac{1}{u} 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 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (u - ln ∣ u ∣)

Substitute back

u = 1 + e^{- t}

= - (1 + e^{- t} - ln 1 + e^{- t})

Simplify

= - 1 - e^{- t} + ln 1 + e^{- t}

Add a constant to the solution

= - 1 - e^{- t} + ln 1 + e^{- t} + C

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