What is the integral of 1/(x(ln(-x))^3) ?

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

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integral of 1/(x(ln(-x))^3)

\int \frac{1}{x ( ln ( - x ) ) ^{3}} d x

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

Solution steps

\int \frac{1}{x ( ln ( - x ) ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{u ln ^{3} ( u )} d u

= \int \frac{1}{u ln ^{3} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{3}} d v

= \int \frac{1}{v ^{3}} d v

Apply the Power Rule

: - \frac{1}{2 v ^{2}}

= - \frac{1}{2 v ^{2}}

Substitute back

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

Add a constant to the solution

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

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