What is the integral of p(p+1)^6 ?

The integral of p(p+1)^6 is (p^8)/8+(6p^7)/7+(5p^6)/2+4p^5+(15p^4)/4+2p^3+(p^2)/2+C

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integral of p(p+1)^6

\int p (p + 1)^{6} d p

\frac{p ^{8}}{8} + \frac{6 p ^{7}}{7} + \frac{5 p ^{6}}{2} + 4 p^{5} + \frac{15 p ^{4}}{4} + 2 p^{3} + \frac{p ^{2}}{2} + C

Solution steps

\int p (p + 1)^{6} d p

Expand

p (p + 1)^{6} : p^{7} + 6 p^{6} + 15 p^{5} + 20 p^{4} + 15 p^{3} + 6 p^{2} + p

= \int p^{7} + 6 p^{6} + 15 p^{5} + 20 p^{4} + 15 p^{3} + 6 p^{2} + 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^{7} d p + \int 6 p^{6} d p + \int 15 p^{5} d p + \int 20 p^{4} d p + \int 15 p^{3} d p + \int 6 p^{2} d p + \int p d p

\int p^{7} d p = \frac{p ^{8}}{8}

\int 6 p^{6} d p = \frac{6 p ^{7}}{7}

\int 15 p^{5} d p = \frac{5 p ^{6}}{2}

\int 20 p^{4} d p = 4 p^{5}

\int 15 p^{3} d p = \frac{15 p ^{4}}{4}

\int 6 p^{2} d p = 2 p^{3}

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

= \frac{p ^{8}}{8} + \frac{6 p ^{7}}{7} + \frac{5 p ^{6}}{2} + 4 p^{5} + \frac{15 p ^{4}}{4} + 2 p^{3} + \frac{p ^{2}}{2}

Add a constant to the solution

= \frac{p ^{8}}{8} + \frac{6 p ^{7}}{7} + \frac{5 p ^{6}}{2} + 4 p^{5} + \frac{15 p ^{4}}{4} + 2 p^{3} + \frac{p ^{2}}{2} + C

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What is the integral of p(p+1)^6 ?
The integral of p(p+1)^6 is (p^8)/8+(6p^7)/7+(5p^6)/2+4p^5+(15p^4)/4+2p^3+(p^2)/2+C