What is the integral of p(p+6)^8 ?

The integral of p(p+6)^8 is ((p+6)^{10}}{10}-\frac{2(p+6)^9)/3+C

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integral of p(p+6)^8

\int p (p + 6)^{8} d p

\frac{( p + 6 ) ^{10}}{10} - \frac{2 ( p + 6 ) ^{9}}{3} + C

Solution steps

\int p (p + 6)^{8} d p

Apply u-substitution

: \int (u - 6) u^{8} d u

= \int (u - 6) u^{8} d u

Expand

(u - 6) u^{8} : u^{9} - 6 u^{8}

= \int u^{9} - 6 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 6 u^{8} d u

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

\int 6 u^{8} d u = \frac{2 u ^{9}}{3}

= \frac{u ^{10}}{10} - \frac{2 u ^{9}}{3}

Substitute back

u = p + 6

= \frac{( p + 6 ) ^{10}}{10} - \frac{2 ( p + 6 ) ^{9}}{3}

Add a constant to the solution

= \frac{( p + 6 ) ^{10}}{10} - \frac{2 ( p + 6 ) ^{9}}{3} + C

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What is the integral of p(p+6)^8 ?
The integral of p(p+6)^8 is ((p+6)^{10}}{10}-\frac{2(p+6)^9)/3+C