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

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

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

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

ln ∣ ln (x) - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = ln (x) - 1

= ln ∣ ln (x) - 1 ∣

Add a constant to the solution

= ln ∣ ln (x) - 1 ∣ + C

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

\int 2 x^{2} + 7 + \frac{9}{x ^{2} + 1} d x

\frac{d}{d x} ((ln (x))^{3})

\frac{\partial}{\partial x} (\frac{1}{x - b})

f (x) = \frac{( 10 - 3 x ) + 2}{2}

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