What is the integral of 4x(3x-1)^5 ?

The integral of 4x(3x-1)^5 is 2/189 (6(3x-1)^7+7(3x-1)^6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4x(3x-1)^5

\int 4 x (3 x - 1)^{5} d x

\frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6}) + C

Solution steps

\int 4 x (3 x - 1)^{5} d x

Take the constant out:

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

= 4 \cdot \int x (3 x - 1)^{5} d x

Apply u-substitution

: \int \frac{u ^{5} ( u + 1 )}{9} d u

= 4 \cdot \int \frac{u ^{5} ( u + 1 )}{9} d u

Take the constant out:

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

= 4 \cdot \frac{1}{9} \cdot \int u^{5} (u + 1) d u

Expand

u^{5} (u + 1) : u^{6} + u^{5}

= 4 \cdot \frac{1}{9} \cdot \int u^{6} + u^{5} 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

= 4 \cdot \frac{1}{9} (\int u^{6} d u + \int u^{5} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{5} d u = \frac{u ^{6}}{6}

= 4 \cdot \frac{1}{9} (\frac{u ^{7}}{7} + \frac{u ^{6}}{6})

Substitute back

u = 3 x - 1

= 4 \cdot \frac{1}{9} (\frac{( 3 x - 1 ) ^{7}}{7} + \frac{( 3 x - 1 ) ^{6}}{6})

Simplify

4 \cdot \frac{1}{9} (\frac{( 3 x - 1 ) ^{7}}{7} + \frac{( 3 x - 1 ) ^{6}}{6}) : \frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6})

= \frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6})

Add a constant to the solution

= \frac{2}{189} (6 (3 x - 1)^{7} + 7 (3 x - 1)^{6}) + C

x^{2} + 2 x y - y^{2} + x = 20, (3, 4)

(3 x + 6)^{\frac{1}{5}}, at x = 2

\frac{\partial}{\partial x} (x^{2} y^{2} e^{4 x y})

\frac{d}{d x} ((x \cdot sinh (x) + cos (x))^{5})

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 8})

What is the integral of 4x(3x-1)^5 ?
The integral of 4x(3x-1)^5 is 2/189 (6(3x-1)^7+7(3x-1)^6)+C