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

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

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

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

ln ∣ ln (x) ∣ + C

Solution steps

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

Simplify

ln (x^{x}) : x ln (x)

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

= ln ∣ ln (x) ∣

Add a constant to the solution

= ln ∣ ln (x) ∣ + C

\int \frac{x}{12} d x

x \to - 2 lim (\frac{x ^{2}}{x - 1})

\frac{sin ( x ^{2} )}{x ^{2}}

\int \frac{1 + x + x}{x} d x

\frac{d y}{d x} = \frac{13 x}{y}

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