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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{3} ( u + 1 )} d u

= \int \frac{1}{u ^{3} ( u + 1 )} d u

Take the partial fraction of

\frac{1}{u ^{3} ( u + 1 )} : \frac{1}{u} - \frac{1}{u ^{2}} + \frac{1}{u ^{3}} - \frac{1}{u + 1}

= \int \frac{1}{u} - \frac{1}{u ^{2}} + \frac{1}{u ^{3}} - \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 \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u + \int \frac{1}{u ^{3}} d u - \int \frac{1}{u + 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

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

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

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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