What is the integral of ((ln(z))^5)/z ?

The integral of ((ln(z))^5)/z is (ln^6(z))/6+C

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integral of ((ln(z))^5)/z

\int \frac{( ln ( z ) ) ^{5}}{z} d z

\frac{ln ^{6} ( z )}{6} + C

Solution steps

\int \frac{( ln ( z ) ) ^{5}}{z} d z

Apply u-substitution

: \int u^{5} d u

= \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{u ^{6}}{6}

Substitute back

u = ln (z)

= \frac{ln ^{6} ( z )}{6}

Add a constant to the solution

= \frac{ln ^{6} ( z )}{6} + C

\int e^{- 12 x} d x

x \to 5 lim (\frac{7 x}{x - 5})

y^{^{'}} + y = e^{2 x}

x + \frac{5}{x}

\int (x^{2} - 1)^{2} (3 x^{2}) d x

What is the integral of ((ln(z))^5)/z ?
The integral of ((ln(z))^5)/z is (ln^6(z))/6+C