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

The integral of p(p+4)^4 is (p^6)/6+(16p^5)/5+24p^4+(256p^3)/3+128p^2+C

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integral of p(p+4)^4

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

\frac{p ^{6}}{6} + \frac{16 p ^{5}}{5} + 24 p^{4} + \frac{256 p ^{3}}{3} + 128 p^{2} + C

Solution steps

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

Expand

p (p + 4)^{4} : p^{5} + 16 p^{4} + 96 p^{3} + 256 p^{2} + 256 p

= \int p^{5} + 16 p^{4} + 96 p^{3} + 256 p^{2} + 256 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 16 p^{4} d p + \int 96 p^{3} d p + \int 256 p^{2} d p + \int 256 p d p

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

\int 16 p^{4} d p = \frac{16 p ^{5}}{5}

\int 96 p^{3} d p = 24 p^{4}

\int 256 p^{2} d p = \frac{256 p ^{3}}{3}

\int 256 p d p = 128 p^{2}

= \frac{p ^{6}}{6} + \frac{16 p ^{5}}{5} + 24 p^{4} + \frac{256 p ^{3}}{3} + 128 p^{2}

Add a constant to the solution

= \frac{p ^{6}}{6} + \frac{16 p ^{5}}{5} + 24 p^{4} + \frac{256 p ^{3}}{3} + 128 p^{2} + C

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What is the integral of p(p+4)^4 ?
The integral of p(p+4)^4 is (p^6)/6+(16p^5)/5+24p^4+(256p^3)/3+128p^2+C