What is the integral from 2 to 5 of 1/(ln(x)) ?

The integral from 2 to 5 of 1/(ln(x)) is \Ei(ln(5))-\Ei(ln(2))

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integral from 2 to 5 of 1/(ln(x))

\int_{2}^{5} \frac{1}{ln ( x )} d x

Ei (ln (5)) - Ei (ln (2))

Solution steps

\int_{2}^{5} \frac{1}{ln ( x )} d x

Apply u-substitution

: \int_{l n (2)}^{l n (5)} \frac{e ^{u}}{u} d u

= \int_{l n (2)}^{l n (5)} \frac{e ^{u}}{u} d u

Use the nonelementary integral:

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

= [Ei (u)]_{l n (2)}^{l n (5)}

Compute the boundaries:

Ei (ln (5)) - Ei (ln (2))

= Ei (ln (5)) - Ei (ln (2))

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What is the integral from 2 to 5 of 1/(ln(x)) ?
The integral from 2 to 5 of 1/(ln(x)) is \Ei(ln(5))-\Ei(ln(2))