What is the integral of 4x(2x-1)(4x+1) ?

The integral of 4x(2x-1)(4x+1) is 4((8x^4)/3+x^3-(4x^3)/3-(x^2)/2-(2x^4)/3-(x^3)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 4x(2x-1)(4x+1)

\int 4 x (2 x - 1) (4 x + 1) d x

4 (\frac{8 x ^{4}}{3} + x^{3} - \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2} - \frac{2 x ^{4}}{3} - \frac{x ^{3}}{3}) + C

Solution steps

\int 4 x (2 x - 1) (4 x + 1) d x

Take the constant out:

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

= 4 \cdot \int x (2 x - 1) (4 x + 1) d x

Apply Integration By Parts

= 4 (\frac{8 x ^{4}}{3} + x^{3} - \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2} - \int \frac{8 x ^{3}}{3} + x^{2} d x)

\int \frac{8 x ^{3}}{3} + x^{2} d x = \frac{2 x ^{4}}{3} + \frac{x ^{3}}{3}

= 4 (\frac{8 x ^{4}}{3} + x^{3} - \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2} - (\frac{2 x ^{4}}{3} + \frac{x ^{3}}{3}))

Simplify

= 4 (\frac{8 x ^{4}}{3} + x^{3} - \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2} - \frac{2 x ^{4}}{3} - \frac{x ^{3}}{3})

Add a constant to the solution

= 4 (\frac{8 x ^{4}}{3} + x^{3} - \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2} - \frac{2 x ^{4}}{3} - \frac{x ^{3}}{3}) + C

What is the integral of 4x(2x-1)(4x+1) ?
The integral of 4x(2x-1)(4x+1) is 4((8x^4)/3+x^3-(4x^3)/3-(x^2)/2-(2x^4)/3-(x^3)/3)+C