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

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

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

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

diverges

Solution steps

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

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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