What is the integral of (ln(z))/(xz) ?

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

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integral of (ln(z))/(xz)

\int \frac{ln ( z )}{x z} d z

\frac{1}{2 x} ln^{2} (z) + C

Solution steps

\int \frac{ln ( z )}{x z} d z

Take the constant out:

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

= \frac{1}{x} \cdot \int \frac{ln ( z )}{z} d z

Apply u-substitution

: \int u d u

= \frac{1}{x} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{x} \cdot \frac{u ^{2}}{2}

Substitute back

u = ln (z)

= \frac{1}{x} \cdot \frac{ln ^{2} ( z )}{2}

Simplify

\frac{1}{x} \cdot \frac{ln ^{2} ( z )}{2} : \frac{1}{2 x} ln^{2} (z)

= \frac{1}{2 x} ln^{2} (z)

Add a constant to the solution

= \frac{1}{2 x} ln^{2} (z) + C

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