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

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

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

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

- e^{x} - 1 + ln ∣ e^{x} + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - \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 = e^{x} + 1

= - (e^{x} + 1 - ln ∣ e^{x} + 1 ∣)

Simplify

= - e^{x} - 1 + ln ∣ e^{x} + 1 ∣

Add a constant to the solution

= - e^{x} - 1 + ln ∣ e^{x} + 1 ∣ + C

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