What is the integral of 6((ln(z))^2)/z ?

The integral of 6((ln(z))^2)/z is 2ln^3(z)+C

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integral of 6((ln(z))^2)/z

\int 6 \frac{( ln ( z ) ) ^{2}}{z} d z

2 ln^{3} (z) + C

Solution steps

\int 6 \cdot \frac{( ln ( z ) ) ^{2}}{z} d z

Take the constant out:

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

= 6 \cdot \int \frac{( ln ( z ) ) ^{2}}{z} d z

Apply u-substitution

: \int u^{2} d u

= 6 \cdot \int u^{2} d u

Apply the Power Rule

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

= 6 \cdot \frac{u ^{3}}{3}

Substitute back

u = ln (z)

= 6 \cdot \frac{ln ^{3} ( z )}{3}

Simplify

6 \cdot \frac{ln ^{3} ( z )}{3} : 2 ln^{3} (z)

= 2 ln^{3} (z)

Add a constant to the solution

= 2 ln^{3} (z) + C

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What is the integral of 6((ln(z))^2)/z ?
The integral of 6((ln(z))^2)/z is 2ln^3(z)+C