What is the integral from 2 to 9 of (e^x)/x ?

The integral from 2 to 9 of (e^x)/x is \Ei(9)-\Ei(2)

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integral from 2 to 9 of (e^x)/x

\int_{2}^{9} \frac{e ^{x}}{x} d x

Ei (9) - Ei (2)

Solution steps

\int_{2}^{9} \frac{e ^{x}}{x} d x

Use the nonelementary integral:

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

= [Ei (x)]_{2}^{9}

Compute the boundaries:

Ei (9) - Ei (2)

= Ei (9) - Ei (2)

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What is the integral from 2 to 9 of (e^x)/x ?
The integral from 2 to 9 of (e^x)/x is \Ei(9)-\Ei(2)