What is the integral of (z^2)/(z^3+1) ?

The integral of (z^2)/(z^3+1) is 1/3 ln|z^3+1|+C

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integral of (z^2)/(z^3+1)

\int \frac{z ^{2}}{z ^{3} + 1} d z

\frac{1}{3} ln z^{3} + 1 + C

Solution steps

\int \frac{z ^{2}}{z ^{3} + 1} d z

Apply u-substitution

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = z^{3} + 1

= \frac{1}{3} ln z^{3} + 1

Add a constant to the solution

= \frac{1}{3} ln z^{3} + 1 + C

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What is the integral of (z^2)/(z^3+1) ?
The integral of (z^2)/(z^3+1) is 1/3 ln|z^3+1|+C