What is the integral of (ln(x))^{-1} ?

The integral of (ln(x))^{-1} is \Ei(ln(x))+C

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integral of (ln(x))^{-1}

\int (ln (x))^{- 1} d x

Ei (ln (x)) + C

Solution steps

\int (ln (x))^{- 1} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \int \frac{1}{ln ( x )} d x

Apply u-substitution

: \int \frac{e ^{u}}{u} d u

= \int \frac{e ^{u}}{u} d u

Use the nonelementary integral:

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

= Ei (u)

Substitute back

u = ln (x)

= Ei (ln (x))

Add a constant to the solution

= Ei (ln (x)) + C

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What is the integral of (ln(x))^{-1} ?
The integral of (ln(x))^{-1} is \Ei(ln(x))+C