What is the integral of (z^4-3z^2-4z)/(6z) ?

The integral of (z^4-3z^2-4z)/(6z) is 1/6 ((z^4)/4-(3z^2)/2-4z)+C

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integral of (z^4-3z^2-4z)/(6z)

\int \frac{z ^{4} - 3 z ^{2} - 4 z}{6 z} d z

\frac{1}{6} (\frac{z ^{4}}{4} - \frac{3 z ^{2}}{2} - 4 z) + C

Solution steps

\int \frac{z ^{4} - 3 z ^{2} - 4 z}{6 z} d z

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{z ^{4} - 3 z ^{2} - 4 z}{z} d z

Simplify

\frac{z ^{4} - 3 z ^{2} - 4 z}{z} : z^{3} - 3 z - 4

= \frac{1}{6} \cdot \int z^{3} - 3 z - 4 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}{6} (\int z^{3} d z - \int 3 z d z - \int 4 d z)

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

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

\int 4 d z = 4 z

= \frac{1}{6} (\frac{z ^{4}}{4} - \frac{3 z ^{2}}{2} - 4 z)

Add a constant to the solution

= \frac{1}{6} (\frac{z ^{4}}{4} - \frac{3 z ^{2}}{2} - 4 z) + C

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What is the integral of (z^4-3z^2-4z)/(6z) ?
The integral of (z^4-3z^2-4z)/(6z) is 1/6 ((z^4)/4-(3z^2)/2-4z)+C