What is the integral of p(p+2)^6 ?

The integral of p(p+2)^6 is (p^8)/8+(12p^7)/7+10p^6+32p^5+60p^4+64p^3+32p^2+C

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integral of p(p+2)^6

\int p (p + 2)^{6} d p

\frac{p ^{8}}{8} + \frac{12 p ^{7}}{7} + 10 p^{6} + 32 p^{5} + 60 p^{4} + 64 p^{3} + 32 p^{2} + C

Solution steps

\int p (p + 2)^{6} d p

Expand

p (p + 2)^{6} : p^{7} + 12 p^{6} + 60 p^{5} + 160 p^{4} + 240 p^{3} + 192 p^{2} + 64 p

= \int p^{7} + 12 p^{6} + 60 p^{5} + 160 p^{4} + 240 p^{3} + 192 p^{2} + 64 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 12 p^{6} d p + \int 60 p^{5} d p + \int 160 p^{4} d p + \int 240 p^{3} d p + \int 192 p^{2} d p + \int 64 p d p

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

\int 12 p^{6} d p = \frac{12 p ^{7}}{7}

\int 60 p^{5} d p = 10 p^{6}

\int 160 p^{4} d p = 32 p^{5}

\int 240 p^{3} d p = 60 p^{4}

\int 192 p^{2} d p = 64 p^{3}

\int 64 p d p = 32 p^{2}

= \frac{p ^{8}}{8} + \frac{12 p ^{7}}{7} + 10 p^{6} + 32 p^{5} + 60 p^{4} + 64 p^{3} + 32 p^{2}

Add a constant to the solution

= \frac{p ^{8}}{8} + \frac{12 p ^{7}}{7} + 10 p^{6} + 32 p^{5} + 60 p^{4} + 64 p^{3} + 32 p^{2} + C

\frac{d}{d x} (\frac{1}{x ^{2}} - \frac{3}{x ^{4}})

y = x^{4}, y = x

\frac{d}{d x} (x^{2} (3 x^{4} - 7 x + 2))

\frac{d}{d x} (\frac{sec ( x )}{4 x ^{2} + 5})

\frac{s}{s ^{2} + 4 s + 8}

What is the integral of p(p+2)^6 ?
The integral of p(p+2)^6 is (p^8)/8+(12p^7)/7+10p^6+32p^5+60p^4+64p^3+32p^2+C