What is the integral of 2x(4x+7)^8 ?

The integral of 2x(4x+7)^8 is 1/720 (9(4x+7)^{10}-70(4x+7)^9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2x(4x+7)^8

\int 2 x (4 x + 7)^{8} d x

\frac{1}{720} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9}) + C

Solution steps

\int 2 x (4 x + 7)^{8} d x

Take the constant out:

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

= 2 \cdot \int x (4 x + 7)^{8} d x

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{16} \cdot \int u^{8} (u - 7) d u

Expand

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

= 2 \cdot \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

= 2 \cdot \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}

= 2 \cdot \frac{1}{16} (\frac{u ^{10}}{10} - \frac{7 u ^{9}}{9})

Substitute back

u = 4 x + 7

= 2 \cdot \frac{1}{16} (\frac{( 4 x + 7 ) ^{10}}{10} - \frac{7 ( 4 x + 7 ) ^{9}}{9})

Simplify

2 \cdot \frac{1}{16} (\frac{( 4 x + 7 ) ^{10}}{10} - \frac{7 ( 4 x + 7 ) ^{9}}{9}) : \frac{1}{720} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9})

= \frac{1}{720} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9})

Add a constant to the solution

= \frac{1}{720} (9 (4 x + 7)^{10} - 70 (4 x + 7)^{9}) + C

x \to \infty lim (x!)

\int_{1}^{2} 6 x^{2} 4^{x^{3}} d x

x \to 3 lim (\frac{x - 3}{∣ x - 3 ∣})

\int \frac{4 x ^{2} + 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\frac{d}{d x} ((2 x)^{(- 3 x)})

What is the integral of 2x(4x+7)^8 ?
The integral of 2x(4x+7)^8 is 1/720 (9(4x+7)^{10}-70(4x+7)^9)+C