What is the integral of (ln^4(x))/(x(ln^2(x)-1)) ?

The integral of (ln^4(x))/(x(ln^2(x)-1)) is (ln^3(x))/3+ln(x)-1/2 ln|ln(x)+1|+1/2 ln|ln(x)-1|+C

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integral of (ln^4(x))/(x(ln^2(x)-1))

\int \frac{ln ^{4} ( x )}{x ( ln ^{2} ( x ) - 1 )} d x

\frac{ln ^{3} ( x )}{3} + ln (x) - \frac{1}{2} ln ∣ ln (x) + 1 ∣ + \frac{1}{2} ln ∣ ln (x) - 1 ∣ + C

Solution steps

\int \frac{ln ^{4} ( x )}{x ( ln ^{2} ( x ) - 1 )} d x

Apply u-substitution

: \int \frac{u ^{4}}{u ^{2} - 1} d u

= \int \frac{u ^{4}}{u ^{2} - 1} d u

Long division

\frac{u ^{4}}{u ^{2} - 1} : u^{2} + 1 + \frac{1}{u ^{2} - 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

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

= \frac{u ^{3}}{3} + u - \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

Substitute back

u = ln (x)

= \frac{ln ^{3} ( x )}{3} + ln (x) - \frac{1}{2} ln ∣ ln (x) + 1 ∣ + \frac{1}{2} ln ∣ ln (x) - 1 ∣

Add a constant to the solution

= \frac{ln ^{3} ( x )}{3} + ln (x) - \frac{1}{2} ln ∣ ln (x) + 1 ∣ + \frac{1}{2} ln ∣ ln (x) - 1 ∣ + C

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What is the integral of (ln^4(x))/(x(ln^2(x)-1)) ?
The integral of (ln^4(x))/(x(ln^2(x)-1)) is (ln^3(x))/3+ln(x)-1/2 ln|ln(x)+1|+1/2 ln|ln(x)-1|+C