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

The integral of (z^2+z)^2 is (z^5)/5+(z^4)/2+(z^3)/3+C

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integral of (z^2+z)^2

\int (z^{2} + z)^{2} d z

\frac{z ^{5}}{5} + \frac{z ^{4}}{2} + \frac{z ^{3}}{3} + C

Solution steps

\int (z^{2} + z)^{2} d z

Expand

(z^{2} + z)^{2} : z^{4} + 2 z^{3} + z^{2}

= \int z^{4} + 2 z^{3} + z^{2} d z

Apply the Sum Rule:

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

= \int z^{4} d z + \int 2 z^{3} d z + \int z^{2} d z

\int z^{4} d z = \frac{z ^{5}}{5}

\int 2 z^{3} d z = \frac{z ^{4}}{2}

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

= \frac{z ^{5}}{5} + \frac{z ^{4}}{2} + \frac{z ^{3}}{3}

Add a constant to the solution

= \frac{z ^{5}}{5} + \frac{z ^{4}}{2} + \frac{z ^{3}}{3} + C

\frac{d}{d x} (\frac{1}{3} e^{- 3 x})

\frac{1}{( s ^{2} + 2 s + 2 ) ( s ^{2} + 9 )}

\frac{d}{d x} ((3 x + 1)^{2})

\int \frac{200}{16 x ^{2} + 256} d x

\frac{d ^{2} y}{d x ^{2}} = 4 y, y (0) = 0, y^{^{'}} (0) = 1

What is the integral of (z^2+z)^2 ?
The integral of (z^2+z)^2 is (z^5)/5+(z^4)/2+(z^3)/3+C