What is the integral of 1/(ln(x^x)sqrt((ln(x))^2-1)) ?

The integral of 1/(ln(x^x)sqrt((ln(x))^2-1)) is arctan(sqrt(ln^2(x)-1))+C

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integral of 1/(ln(x^x)sqrt((ln(x))^2-1))

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

arctan (ln^{2} (x) - 1) + C

Solution steps

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

Simplify

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

= \int \frac{1}{x ln ( x ) ln ^{2} ( x ) - 1} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 1} d u

= \int \frac{1}{u u ^{2} - 1} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

= \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= arctan (v)

Substitute back

= arctan (ln^{2} (x) - 1)

Add a constant to the solution

= arctan (ln^{2} (x) - 1) + C

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What is the integral of 1/(ln(x^x)sqrt((ln(x))^2-1)) ?
The integral of 1/(ln(x^x)sqrt((ln(x))^2-1)) is arctan(sqrt(ln^2(x)-1))+C