What is the integral of p(p+3)^8 ?

The integral of p(p+3)^8 is ((p+3)^{10}}{10}-\frac{(p+3)^9)/3+C

integral of p(p+3)^8

\int p (p + 3)^{8} d p

\frac{( p + 3 ) ^{10}}{10} - \frac{( p + 3 ) ^{9}}{3} + C

Solution steps

\int p (p + 3)^{8} d p

Apply u-substitution

= \int (u - 3) u^{8} d u

Expand

(u - 3) u^{8} : u^{9} - 3 u^{8}

= \int u^{9} - 3 u^{8} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{9} d u - \int 3 u^{8} d u

\int u^{9} d u = \frac{u ^{10}}{10}

\int 3 u^{8} d u = \frac{u ^{9}}{3}

= \frac{u ^{10}}{10} - \frac{u ^{9}}{3}

Substitute back

u = p + 3

= \frac{( p + 3 ) ^{10}}{10} - \frac{( p + 3 ) ^{9}}{3}

Add a constant to the solution

= \frac{( p + 3 ) ^{10}}{10} - \frac{( p + 3 ) ^{9}}{3} + C

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What is the integral of p(p+3)^8 ?
The integral of p(p+3)^8 is ((p+3)^{10}}{10}-\frac{(p+3)^9)/3+C