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

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

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Simplify

e^{- x} \frac{1}{- e ^{- x} + 1} : \frac{e ^{- x}}{- e ^{- x} + 1}

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = - e^{- x} + 1

= ln - e^{- x} + 1

Add a constant to the solution

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

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