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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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 = 1 + z^{3}

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

Add a constant to the solution

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

\int x^{2} x^{2} + 4 d x

\frac{\partial}{\partial x} (\frac{2}{x})

\int \frac{x ^{4}}{1 - x ^{10}} d x

x \to \infty lim (e x)

\int_{1}^{2} \frac{2}{x ^{3} + 4 x} d x

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