What is the integral of x(4x+5)^8 ?

The integral of x(4x+5)^8 is 1/1440 (9(4x+5)^{10}-50(4x+5)^9)+C

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integral of x(4x+5)^8

\int x (4 x + 5)^{8} d x

\frac{1}{1440} (9 (4 x + 5)^{10} - 50 (4 x + 5)^{9}) + C

Solution steps

\int x (4 x + 5)^{8} d x

Apply u-substitution

: \int \frac{u ^{8} ( u - 5 )}{16} d u

= \int \frac{u ^{8} ( u - 5 )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u^{8} (u - 5) d u

Expand

u^{8} (u - 5) : u^{9} - 5 u^{8}

= \frac{1}{16} \cdot \int u^{9} - 5 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

= \frac{1}{16} (\int u^{9} d u - \int 5 u^{8} d u)

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

\int 5 u^{8} d u = \frac{5 u ^{9}}{9}

= \frac{1}{16} (\frac{u ^{10}}{10} - \frac{5 u ^{9}}{9})

Substitute back

u = 4 x + 5

= \frac{1}{16} (\frac{( 4 x + 5 ) ^{10}}{10} - \frac{5 ( 4 x + 5 ) ^{9}}{9})

Simplify

\frac{1}{16} (\frac{( 4 x + 5 ) ^{10}}{10} - \frac{5 ( 4 x + 5 ) ^{9}}{9}) : \frac{1}{1440} (9 (4 x + 5)^{10} - 50 (4 x + 5)^{9})

= \frac{1}{1440} (9 (4 x + 5)^{10} - 50 (4 x + 5)^{9})

Add a constant to the solution

= \frac{1}{1440} (9 (4 x + 5)^{10} - 50 (4 x + 5)^{9}) + C

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What is the integral of x(4x+5)^8 ?
The integral of x(4x+5)^8 is 1/1440 (9(4x+5)^{10}-50(4x+5)^9)+C