What is the integral of 1/(1+ln(x)) ?

The integral of 1/(1+ln(x)) is 1/e \Ei(1+ln(x))+C

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integral of 1/(1+ln(x))

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

\frac{1}{e} Ei (1 + ln (x)) + C

Solution steps

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

Apply u-substitution

= \int \frac{e ^{u - 1}}{u} d u

Simplify

e^{u - 1} : e^{u} e^{- 1}

= \int \frac{e ^{u} e ^{- 1}}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{- 1} \cdot \int \frac{e ^{u}}{u} d u

Apply exponent rule:

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

= \frac{1}{e} \cdot \int \frac{e ^{u}}{u} d u

Use the nonelementary integral:

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

= \frac{1}{e} Ei (u)

Substitute back

u = 1 + ln (x)

= \frac{1}{e} Ei (1 + ln (x))

Add a constant to the solution

= \frac{1}{e} Ei (1 + ln (x)) + C

What is the integral of 1/(1+ln(x)) ?
The integral of 1/(1+ln(x)) is 1/e \Ei(1+ln(x))+C