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

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

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

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

- \frac{ln ^{4} ( 1 - x )}{4} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int v^{3} d v

= - \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= - \frac{v ^{4}}{4}

Substitute back

= - \frac{ln ^{4} ( 1 - x )}{4}

Add a constant to the solution

= - \frac{ln ^{4} ( 1 - x )}{4} + C

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