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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= a \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

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

Use the Hyperbolic identity:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{x}{a}

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

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

Add a constant to the solution

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

n = 1 \sum \infty n (\frac{1}{e})^{n}

\int \frac{1}{x ( x - 7 )} d x

- tan (x)

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

\frac{1}{( s ^{2} - 6 s + 10 )}

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