What is the integral from 1 to 9 of (e^x)/x ?

The integral from 1 to 9 of (e^x)/x is \Ei(9)-\Ei(1)

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integral from 1 to 9 of (e^x)/x

\int_{1}^{9} \frac{e ^{x}}{x} d x

Ei (9) - Ei (1)

Solution steps

\int_{1}^{9} \frac{e ^{x}}{x} d x

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= [Ei (x)]_{1}^{9}

Compute the boundaries:

Ei (9) - Ei (1)

= Ei (9) - Ei (1)

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What is the integral from 1 to 9 of (e^x)/x ?
The integral from 1 to 9 of (e^x)/x is \Ei(9)-\Ei(1)