What is the integral of p(p+3)^4 ?

The integral of p(p+3)^4 is (p^6)/6+(12p^5)/5+(27p^4)/2+36p^3+(81p^2)/2+C

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integral of p(p+3)^4

\int p (p + 3)^{4} d p

\frac{p ^{6}}{6} + \frac{12 p ^{5}}{5} + \frac{27 p ^{4}}{2} + 36 p^{3} + \frac{81 p ^{2}}{2} + C

Solution steps

\int p (p + 3)^{4} d p

Expand

p (p + 3)^{4} : p^{5} + 12 p^{4} + 54 p^{3} + 108 p^{2} + 81 p

= \int p^{5} + 12 p^{4} + 54 p^{3} + 108 p^{2} + 81 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 12 p^{4} d p + \int 54 p^{3} d p + \int 108 p^{2} d p + \int 81 p d p

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

\int 12 p^{4} d p = \frac{12 p ^{5}}{5}

\int 54 p^{3} d p = \frac{27 p ^{4}}{2}

\int 108 p^{2} d p = 36 p^{3}

\int 81 p d p = \frac{81 p ^{2}}{2}

= \frac{p ^{6}}{6} + \frac{12 p ^{5}}{5} + \frac{27 p ^{4}}{2} + 36 p^{3} + \frac{81 p ^{2}}{2}

Add a constant to the solution

= \frac{p ^{6}}{6} + \frac{12 p ^{5}}{5} + \frac{27 p ^{4}}{2} + 36 p^{3} + \frac{81 p ^{2}}{2} + C

\int_{- \infty}^{0} e^{x} d x

\int \frac{x + 3}{x ^{2} + 6 x + 10} d x

x \to 5 - lim (\frac{- 3}{x ^{2} - 25})

\frac{10 u ^{2}}{( u ^{2} + u ) ^{3}}

y^{^{''}} + y^{^{'}} - 56 y = 0

What is the integral of p(p+3)^4 ?
The integral of p(p+3)^4 is (p^6)/6+(12p^5)/5+(27p^4)/2+36p^3+(81p^2)/2+C