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

Rewrite using trig identities

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

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 ( x )} d x

Use the Hyperbolic identity:

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

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

Use the common integral:

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

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

Add a constant to the solution

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

\int sin (x n) d x

\frac{d P}{d t} = \frac{9 P}{900} (9 - P), P (0) = 4

- 12 sin (2 x)

\int x^{3} \cdot 9 - x^{2} d x

\frac{d}{d x} ((\frac{3 x}{x + 1})^{2})

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