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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{ln ^{2} ( u )}{u} d u

Apply u-substitution

= \int v^{2} d v

Apply the Power Rule

= \frac{v ^{3}}{3}

Substitute back

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

Add a constant to the solution

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

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