What is the integral of (p^2-9p)/(\sqrt[5]{p)} ?

The integral of (p^2-9p)/(\sqrt[5]{p)} is 5/14 p^{14/5}-5p^{9/5}+C

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integral of (p^2-9p)/(\sqrt[5]{p)}

\frac{5}{14} p^{\frac{14}{5}} - 5 p^{\frac{9}{5}} + C

Solution steps

Expand

Apply the Sum Rule:

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

= \frac{5}{14} p^{\frac{14}{5}} - 5 p^{\frac{9}{5}}

Add a constant to the solution

= \frac{5}{14} p^{\frac{14}{5}} - 5 p^{\frac{9}{5}} + C

What is the integral of (p^2-9p)/(\sqrt[5]{p)} ?
The integral of (p^2-9p)/(\sqrt[5]{p)} is 5/14 p^{14/5}-5p^{9/5}+C