What is the integral of p(p+6)^4 ?

The integral of p(p+6)^4 is (p^6)/6+(24p^5)/5+54p^4+288p^3+648p^2+C

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integral of p(p+6)^4

\int p (p + 6)^{4} d p

\frac{p ^{6}}{6} + \frac{24 p ^{5}}{5} + 54 p^{4} + 288 p^{3} + 648 p^{2} + C

Solution steps

\int p (p + 6)^{4} d p

Expand

p (p + 6)^{4} : p^{5} + 24 p^{4} + 216 p^{3} + 864 p^{2} + 1296 p

= \int p^{5} + 24 p^{4} + 216 p^{3} + 864 p^{2} + 1296 p d p

Apply the Sum Rule:

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

= \int p^{5} d p + \int 24 p^{4} d p + \int 216 p^{3} d p + \int 864 p^{2} d p + \int 1296 p d p

\int p^{5} d p = \frac{p ^{6}}{6}

\int 24 p^{4} d p = \frac{24 p ^{5}}{5}

\int 216 p^{3} d p = 54 p^{4}

\int 864 p^{2} d p = 288 p^{3}

\int 1296 p d p = 648 p^{2}

= \frac{p ^{6}}{6} + \frac{24 p ^{5}}{5} + 54 p^{4} + 288 p^{3} + 648 p^{2}

Add a constant to the solution

= \frac{p ^{6}}{6} + \frac{24 p ^{5}}{5} + 54 p^{4} + 288 p^{3} + 648 p^{2} + C

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What is the integral of p(p+6)^4 ?
The integral of p(p+6)^4 is (p^6)/6+(24p^5)/5+54p^4+288p^3+648p^2+C