What is the integral of x(x+9)^7 ?

The integral of x(x+9)^7 is 1/72 (x+9)^8(8x-9)+C

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integral of x(x+9)^7

\int x (x + 9)^{7} d x

\frac{1}{72} (x + 9)^{8} (8 x - 9) + C

Solution steps

\int x (x + 9)^{7} d x

Apply u-substitution

: \int \frac{7 u ( 7 u - 9 )}{7} d u

= \int \frac{7 u ( 7 u - 9 )}{7} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int 7 u (7 u - 9) d u

Expand

7 u (7 u - 9) : u^{\frac{2}{7}} - 9 7 u

= \frac{1}{7} \cdot \int u^{\frac{2}{7}} - 9 7 u 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

= \frac{1}{7} (\int u^{\frac{2}{7}} d u - \int 9 7 u d u)

\int u^{\frac{2}{7}} d u = \frac{7}{9} u^{\frac{9}{7}}

\int 9 7 u d u = \frac{63}{8} u^{\frac{8}{7}}

= \frac{1}{7} (\frac{7}{9} u^{\frac{9}{7}} - \frac{63}{8} u^{\frac{8}{7}})

Substitute back

u = (x + 9)^{7}

= \frac{1}{7} (\frac{7}{9} ((x + 9)^{7})^{\frac{9}{7}} - \frac{63}{8} ((x + 9)^{7})^{\frac{8}{7}})

Simplify

\frac{1}{7} (\frac{7}{9} ((x + 9)^{7})^{\frac{9}{7}} - \frac{63}{8} ((x + 9)^{7})^{\frac{8}{7}}) : \frac{1}{72} (x + 9)^{8} (8 x - 9)

= \frac{1}{72} (x + 9)^{8} (8 x - 9)

Add a constant to the solution

= \frac{1}{72} (x + 9)^{8} (8 x - 9) + C

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What is the integral of x(x+9)^7 ?
The integral of x(x+9)^7 is 1/72 (x+9)^8(8x-9)+C