What is the integral of (z^4+10z^3)/(2z^2) ?

The integral of (z^4+10z^3)/(2z^2) is 1/2 ((z^3)/3+5z^2)+C

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integral of (z^4+10z^3)/(2z^2)

\int \frac{z ^{4} + 10 z ^{3}}{2 z ^{2}} d z

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

Solution steps

\int \frac{z ^{4} + 10 z ^{3}}{2 z ^{2}} d z

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{z ^{4} + 10 z ^{3}}{z ^{2}} d z

\frac{z ^{4} + 10 z ^{3}}{z ^{2}} = z^{2} + 10 z

= \frac{1}{2} \cdot \int z^{2} + 10 z 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

= \frac{1}{2} (\int z^{2} d z + \int 10 z d z)

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

\int 10 z d z = 5 z^{2}

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

Add a constant to the solution

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

(- 2, - 4), (2, - 4)

\frac{d}{d x} (x \cdot y)

\frac{d}{d x} (arctan (\frac{9}{x}) + arctan (\frac{3}{x}))

f (x) = \frac{3 x}{x ^{2} - 2}, at x = 1

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

What is the integral of (z^4+10z^3)/(2z^2) ?
The integral of (z^4+10z^3)/(2z^2) is 1/2 ((z^3)/3+5z^2)+C