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

The integral of 1/(e^{2x)-e^{-2x}} is 1/4 ln(tanh(x))+C

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

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

\frac{1}{4} ln (tanh (x)) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( e ^{u} - e ^{- u} )} d u

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

Rewrite using trig identities

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

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} ln (tanh (\frac{2 x}{2})) : \frac{1}{4} ln (tanh (x))

= \frac{1}{4} ln (tanh (x))

Add a constant to the solution

= \frac{1}{4} ln (tanh (x)) + C

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What is the integral of 1/(e^{2x)-e^{-2x}} ?
The integral of 1/(e^{2x)-e^{-2x}} is 1/4 ln(tanh(x))+C