What is the integral of x(6x+5)^8 ?

The integral of x(6x+5)^8 is 1/3240 (9(6x+5)^{10}-50(6x+5)^9)+C

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integral of x(6x+5)^8

\int x (6 x + 5)^{8} d x

\frac{1}{3240} (9 (6 x + 5)^{10} - 50 (6 x + 5)^{9}) + C

Solution steps

\int x (6 x + 5)^{8} d x

Apply u-substitution

: \int \frac{u ^{8} ( u - 5 )}{36} d u

= \int \frac{u ^{8} ( u - 5 )}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int u^{8} (u - 5) d u

Expand

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

= \frac{1}{36} \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}{36} (\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}{36} (\frac{u ^{10}}{10} - \frac{5 u ^{9}}{9})

Substitute back

u = 6 x + 5

= \frac{1}{36} (\frac{( 6 x + 5 ) ^{10}}{10} - \frac{5 ( 6 x + 5 ) ^{9}}{9})

Simplify

\frac{1}{36} (\frac{( 6 x + 5 ) ^{10}}{10} - \frac{5 ( 6 x + 5 ) ^{9}}{9}) : \frac{1}{3240} (9 (6 x + 5)^{10} - 50 (6 x + 5)^{9})

= \frac{1}{3240} (9 (6 x + 5)^{10} - 50 (6 x + 5)^{9})

Add a constant to the solution

= \frac{1}{3240} (9 (6 x + 5)^{10} - 50 (6 x + 5)^{9}) + C

f (x) = 2 e^{x} + 2 ln (x)

n = 1 \sum \infty sin (\frac{1}{n ^{2}})

\int - \frac{1}{sin ( θ )} d θ

y^{^{'}} + t^{2} y = t^{2}

\frac{d}{d x} (9 x e^{- \frac{1}{y ^{3}}})

What is the integral of x(6x+5)^8 ?
The integral of x(6x+5)^8 is 1/3240 (9(6x+5)^{10}-50(6x+5)^9)+C