What is the integral of ((ln(z))^6)/z ?

The integral of ((ln(z))^6)/z is (ln^7(z))/7+C

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integral of ((ln(z))^6)/z

\int \frac{( ln ( z ) ) ^{6}}{z} d z

\frac{ln ^{7} ( z )}{7} + C

Solution steps

\int \frac{( ln ( z ) ) ^{6}}{z} d z

Apply u-substitution

= \int u^{6} d u

Apply the Power Rule

= \frac{u ^{7}}{7}

Substitute back

u = ln (z)

= \frac{ln ^{7} ( z )}{7}

Add a constant to the solution

= \frac{ln ^{7} ( z )}{7} + C

What is the integral of ((ln(z))^6)/z ?
The integral of ((ln(z))^6)/z is (ln^7(z))/7+C