What is the integral of z^{31}+z^8 ?

The integral of z^{31}+z^8 is (z^{32}}{32}+\frac{z^9)/9+C

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integral of z^{31}+z^8

\int z^{31} + z^{8} d z

\frac{z ^{32}}{32} + \frac{z ^{9}}{9} + C

Solution steps

\int z^{31} + z^{8} 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^{31} d z + \int z^{8} d z

\int z^{31} d z = \frac{z ^{32}}{32}

\int z^{8} d z = \frac{z ^{9}}{9}

= \frac{z ^{32}}{32} + \frac{z ^{9}}{9}

Add a constant to the solution

= \frac{z ^{32}}{32} + \frac{z ^{9}}{9} + C

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What is the integral of z^{31}+z^8 ?
The integral of z^{31}+z^8 is (z^{32}}{32}+\frac{z^9)/9+C