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

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

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

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

diverges

Solution steps

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

Expand

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

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

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

\int_{0}^{1} \frac{1}{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 (e^x-1)/x ?
The integral from 0 to 1 of (e^x-1)/x is diverges