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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

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