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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{e ^{x} - 1} d x = ln ∣ e^{x} - 1 ∣ - x + C

Compute the boundaries:

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

diverges

= diverges

\frac{d}{d t} (\frac{1}{6} t^{3})

\int (x x^{2} + 3) d x

y^{^{''}} + 36 y = - 60 sec (6 t)

\int 12 (y^{4} + 4 y^{2} + 8)^{2} (y^{3} + 2 y) d y

\int x - 10 d x

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