What is the integral of 1/(e^{40x)-1} ?

The integral of 1/(e^{40x)-1} is 1/40 ln|e^{40x}-1|-x+C

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integral of 1/(e^{40x)-1}

\int \frac{1}{e ^{40 x} - 1} d x

\frac{1}{40} ln e^{40 x} - 1 - x + C

Solution steps

\int \frac{1}{e ^{40 x} - 1} d x

Apply u-substitution

: \int \frac{1}{40 ( e ^{u} - 1 )} d u

= \int \frac{1}{40 ( e ^{u} - 1 )} d u

Take the constant out:

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

= \frac{1}{40} \cdot \int \frac{1}{e ^{u} - 1} d u

\frac{1}{e ^{u} - 1} = \frac{e ^{u}}{e ^{u} - 1} - 1

= \frac{1}{40} \cdot \int \frac{e ^{u}}{e ^{u} - 1} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{40} (\int \frac{e ^{u}}{e ^{u} - 1} d u - \int 1 d u)

\int \frac{e ^{u}}{e ^{u} - 1} d u = ln ∣ e^{u} - 1 ∣

\int 1 d u = u

= \frac{1}{40} (ln ∣ e^{u} - 1 ∣ - u)

Substitute back

u = 40 x

= \frac{1}{40} (ln e^{40 x} - 1 - 40 x)

Simplify

\frac{1}{40} (ln e^{40 x} - 1 - 40 x) : \frac{1}{40} ln e^{40 x} - 1 - x

= \frac{1}{40} ln e^{40 x} - 1 - x

Add a constant to the solution

= \frac{1}{40} ln e^{40 x} - 1 - x + C

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What is the integral of 1/(e^{40x)-1} ?
The integral of 1/(e^{40x)-1} is 1/40 ln|e^{40x}-1|-x+C