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

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

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

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

- \frac{1}{2} ln (tanh (- \frac{x}{2})) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{sinh ( u )} d u

Use the Hyperbolic identity:

\frac{1}{sinh ( x )} = csch (x)

= - \frac{1}{2} \cdot \int csch (u) d u

Use the common integral:

\int csch (u) d u = ln (tanh (\frac{u}{2}))

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

Substitute back

u = - x

= - \frac{1}{2} ln (tanh (\frac{- x}{2}))

Simplify

= - \frac{1}{2} ln (tanh (- \frac{x}{2}))

Add a constant to the solution

= - \frac{1}{2} ln (tanh (- \frac{x}{2})) + C

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