What is the integral of x(4x+7)^8 ?

The integral of x(4x+7)^8 is 1/1440 (9(4x+7)^{10}-70(4x+7)^9)+C

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integral of x(4x+7)^8

\int x (4 x + 7)^{8} d x

\frac{1}{1440} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9}) + C

Solution steps

\int x (4 x + 7)^{8} d x

Apply u-substitution

: \int \frac{u ^{8} ( u - 7 )}{16} d u

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

Expand

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

= \frac{1}{16} \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}{16} (\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}{16} (\frac{u ^{10}}{10} - \frac{7 u ^{9}}{9})

Substitute back

u = 4 x + 7

= \frac{1}{16} (\frac{( 4 x + 7 ) ^{10}}{10} - \frac{7 ( 4 x + 7 ) ^{9}}{9})

Simplify

\frac{1}{16} (\frac{( 4 x + 7 ) ^{10}}{10} - \frac{7 ( 4 x + 7 ) ^{9}}{9}) : \frac{1}{1440} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9})

= \frac{1}{1440} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9})

Add a constant to the solution

= \frac{1}{1440} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9}) + C

\frac{d}{d x} ((x^{2} + 2 x) ln (2 x))

(x - y) d x + (x - y + 1) d y = 0

x \to \infty + lim (x \cdot + \frac{1}{x})

\frac{1}{4} (x^{8} + 8)

\frac{\partial}{\partial y} (y + 3 x y^{2} - 18)

What is the integral of x(4x+7)^8 ?
The integral of x(4x+7)^8 is 1/1440 (9(4x+7)^{10}-70(4x+7)^9)+C