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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-1)(2x^2-2x)

\int (2 x - 1) (2 x^{2} - 2 x) d x

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

Solution steps

\int (2 x - 1) (2 x^{2} - 2 x) d x

(2 x - 1) (2 x^{2} - 2 x) = 2 (2 x - 1) (x^{2} - x)

= \int 2 (2 x - 1) (x^{2} - x) d x

Take the constant out:

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

= 2 \cdot \int (2 x - 1) (x^{2} - x) d x

Apply Integration By Parts

: \frac{2 x ^{4}}{3} - x^{3} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - \int \frac{2 x ^{3}}{3} - x^{2} d x

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

\int \frac{2 x ^{3}}{3} - x^{2} d x = \frac{x ^{4}}{6} - \frac{x ^{3}}{3}

= 2 (\frac{2 x ^{4}}{3} - x^{3} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - (\frac{x ^{4}}{6} - \frac{x ^{3}}{3}))

Simplify

2 (\frac{2 x ^{4}}{3} - x^{3} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - (\frac{x ^{4}}{6} - \frac{x ^{3}}{3})) : \frac{4 x ^{4}}{3} - \frac{x ^{4}}{3} - 2 x^{3} + x^{2}

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

Add a constant to the solution

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

\int \frac{tan ( x ) sec ^{2} ( x )}{cos ( x )} d x

\int sin^{6} (\frac{3 θ}{2}) d θ

\int - \frac{e ^{2 x}}{8 + e ^{x}} d x

\int (1 + 5 x) cos (4 x) d x

\int \frac{1}{( 4 x - 1 ) ^{2} + 9} d x

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