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

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

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integral of (z^2)/y

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

z^{2} ln ∣ y ∣ + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= z^{2} ln ∣ y ∣

Add a constant to the solution

= z^{2} ln ∣ y ∣ + C

\frac{( s + 1 )}{s ^{2} + s + 10}

\frac{d}{d x} (ln (\frac{x ^{3} + 4}{2 x}))

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

\int \frac{x ^{4} + 2}{x ^{3} - x ^{2}} d x

\frac{d}{d x} (4 π^{2})