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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 ( t )} d t

Use the Hyperbolic identity:

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

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

Use the common integral:

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

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

Add a constant to the solution

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

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