What is the integral of p(p+9)^8 ?

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

integral of p(p+9)^8

\int p (p + 9)^{8} d p

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

Solution steps

\int p (p + 9)^{8} d p

Apply u-substitution

= \int (u - 9) u^{8} d u

Expand

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

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

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

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

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

Substitute back

u = p + 9

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

Add a constant to the solution

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

\int \frac{2 x - 3}{x} d x

x + y \frac{d y}{d x} = 0, y (16) = 4

\int \frac{- 8}{x ^{2} + 16} d x

\frac{d}{d x} (arctan (sin (7 x)))

\int o d o

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