What is the integral of (z^3)/(\sqrt[3]{z^2+1)} ?

The integral of (z^3)/(\sqrt[3]{z^2+1)} is 1/2 (3/5 (z^2+1)^{5/3}-3/2 (z^2+1)^{2/3})+C

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integral of (z^3)/(\sqrt[3]{z^2+1)}

\frac{1}{2} (\frac{3}{5} (z^{2} + 1)^{\frac{5}{3}} - \frac{3}{2} (z^{2} + 1)^{\frac{2}{3}}) + C

Solution steps

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

= \frac{1}{2} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = z^{2} + 1

= \frac{1}{2} (\frac{3}{5} (z^{2} + 1)^{\frac{5}{3}} - \frac{3}{2} (z^{2} + 1)^{\frac{2}{3}})

Add a constant to the solution

= \frac{1}{2} (\frac{3}{5} (z^{2} + 1)^{\frac{5}{3}} - \frac{3}{2} (z^{2} + 1)^{\frac{2}{3}}) + C

\int x csc (x^{2}) d x

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 4

\frac{d}{d x} (x - ln (e^{x} cos (x)))

\frac{d}{d x} (\frac{1}{2} x)

What is the integral of (z^3)/(\sqrt[3]{z^2+1)} ?
The integral of (z^3)/(\sqrt[3]{z^2+1)} is 1/2 (3/5 (z^2+1)^{5/3}-3/2 (z^2+1)^{2/3})+C