What is the integral of p(p+7)^4 ?

The integral of p(p+7)^4 is (p^6)/6+(28p^5)/5+(147p^4)/2+(1372p^3)/3+(2401p^2)/2+C

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integral of p(p+7)^4

\int p (p + 7)^{4} d p

\frac{p ^{6}}{6} + \frac{28 p ^{5}}{5} + \frac{147 p ^{4}}{2} + \frac{1372 p ^{3}}{3} + \frac{2401 p ^{2}}{2} + C

Solution steps

\int p (p + 7)^{4} d p

Expand

p (p + 7)^{4} : p^{5} + 28 p^{4} + 294 p^{3} + 1372 p^{2} + 2401 p

= \int p^{5} + 28 p^{4} + 294 p^{3} + 1372 p^{2} + 2401 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 28 p^{4} d p + \int 294 p^{3} d p + \int 1372 p^{2} d p + \int 2401 p d p

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

\int 28 p^{4} d p = \frac{28 p ^{5}}{5}

\int 294 p^{3} d p = \frac{147 p ^{4}}{2}

\int 1372 p^{2} d p = \frac{1372 p ^{3}}{3}

\int 2401 p d p = \frac{2401 p ^{2}}{2}

= \frac{p ^{6}}{6} + \frac{28 p ^{5}}{5} + \frac{147 p ^{4}}{2} + \frac{1372 p ^{3}}{3} + \frac{2401 p ^{2}}{2}

Add a constant to the solution

= \frac{p ^{6}}{6} + \frac{28 p ^{5}}{5} + \frac{147 p ^{4}}{2} + \frac{1372 p ^{3}}{3} + \frac{2401 p ^{2}}{2} + C

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What is the integral of p(p+7)^4 ?
The integral of p(p+7)^4 is (p^6)/6+(28p^5)/5+(147p^4)/2+(1372p^3)/3+(2401p^2)/2+C