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

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

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

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

diverges

Solution steps

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

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

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

Apply the Sum Rule:

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{1}^{e} (18 + ln (x))^{2} d x

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\int_{- 4}^{4} (x - 2) d x

\int_{2}^{3} x^{2} + 3 d x

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