What is the integral of (3z^2)/y ?

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

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integral of (3z^2)/y

\int \frac{3 z ^{2}}{y} d y

3 z^{2} ln ∣ y ∣ + C

Solution steps

\int \frac{3 z ^{2}}{y} d y

Take the constant out:

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

= 3 z^{2} \cdot \int \frac{1}{y} d y

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= 3 z^{2} ln ∣ y ∣

Add a constant to the solution

= 3 z^{2} ln ∣ y ∣ + C

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