What is the integral of ((ln(z))^2)/z ?

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

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integral of ((ln(z))^2)/z

\int \frac{( ln ( z ) ) ^{2}}{z} d z

\frac{ln ^{3} ( z )}{3} + C

Solution steps

\int \frac{( ln ( z ) ) ^{2}}{z} d z

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = ln (z)

= \frac{ln ^{3} ( z )}{3}

Add a constant to the solution

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

\int sec (5 x - 3) tan (5 x - 3) d x

f (x) = x^{3} (8 x^{2} + 2 x - 3)^{2}

\frac{d}{d x} (cos (- 4 x))

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