What is the integral of (z^{-3})/3 ?

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

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integral of (z^{-3})/3

\int \frac{z ^{- 3}}{3} d z

- \frac{1}{6 z ^{2}} + C

Solution steps

\int \frac{z ^{- 3}}{3} d z

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{2 z ^{2}}

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

Simplify

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

= - \frac{1}{6 z ^{2}}

Add a constant to the solution

= - \frac{1}{6 z ^{2}} + C

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