What is the integral from 0 to 1 of 1/(e^x-e^{-x)} ?

The integral from 0 to 1 of 1/(e^x-e^{-x)} is diverges

integral from 0 to 1 of 1/(e^x-e^{-x)}

\int_{0}^{1} \frac{1}{e ^{x} - e ^{- x}} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{e ^{x} - e ^{- x}} d x

Rewrite using trig identities

= \int_{0}^{1} \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_{0}^{1} \frac{1}{sinh ( x )} d x

Use the Hyperbolic identity:

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

= \frac{1}{2} \cdot \int_{0}^{1} 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}))]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to 1 of 1/(e^x-e^{-x)} ?
The integral from 0 to 1 of 1/(e^x-e^{-x)} is diverges