What is the integral of x(6x+7)^8 ?

The integral of x(6x+7)^8 is 1/3240 (9(6x+7)^{10}-70(6x+7)^9)+C

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integral of x(6x+7)^8

\int x (6 x + 7)^{8} d x

\frac{1}{3240} (9 (6 x + 7)^{10} - 70 (6 x + 7)^{9}) + C

Solution steps

\int x (6 x + 7)^{8} d x

Apply u-substitution

: \int \frac{u ^{8} ( u - 7 )}{36} d u

= \int \frac{u ^{8} ( u - 7 )}{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 - 7) d u

Expand

u^{8} (u - 7) : u^{9} - 7 u^{8}

= \frac{1}{36} \cdot \int u^{9} - 7 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 7 u^{8} d u)

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

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

= \frac{1}{36} (\frac{u ^{10}}{10} - \frac{7 u ^{9}}{9})

Substitute back

u = 6 x + 7

= \frac{1}{36} (\frac{( 6 x + 7 ) ^{10}}{10} - \frac{7 ( 6 x + 7 ) ^{9}}{9})

Simplify

\frac{1}{36} (\frac{( 6 x + 7 ) ^{10}}{10} - \frac{7 ( 6 x + 7 ) ^{9}}{9}) : \frac{1}{3240} (9 (6 x + 7)^{10} - 70 (6 x + 7)^{9})

= \frac{1}{3240} (9 (6 x + 7)^{10} - 70 (6 x + 7)^{9})

Add a constant to the solution

= \frac{1}{3240} (9 (6 x + 7)^{10} - 70 (6 x + 7)^{9}) + C

\int x \cdot x^{10} d x

\frac{d y}{d x} y = x^{8 c o s (x)}

x \to 0 lim (\frac{sin ( 5 x )}{sin ( \frac{1}{3} x )})

\frac{d}{d x} (x^{3} + x^{2} y + x)

f (x) = x^{4} - 20 x^{2} + 64, at x = 1

What is the integral of x(6x+7)^8 ?
The integral of x(6x+7)^8 is 1/3240 (9(6x+7)^{10}-70(6x+7)^9)+C