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

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

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

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

Ei (ln (x - 1)) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{ln ( u )} d u

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

Apply u-substitution

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

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

Use the nonelementary integral:

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

= Ei (v)

Substitute back

= Ei (ln (x - 1))

Add a constant to the solution

= Ei (ln (x - 1)) + C

\int \frac{1}{x ^{2} 2 - 3 x} d x

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

\int (\frac{1 + x ^{2}}{x}) d x

\int x a + b x d x

\int (2 x - 4 x^{3} + 5) d x

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