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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

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

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

Add a constant to the solution

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

y = 6 x, y = x^{2} - 7

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\int sin (x + 3) cos (1 - 2 x) d x

n = 2 \sum \infty \frac{7}{1 0 ^{n - 1}}

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