What is the integral of (x^2-2x)^2-(6x-x^2)^2 ?

The integral of (x^2-2x)^2-(6x-x^2)^2 is 2x^4-(32x^3)/3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-2x)^2-(6x-x^2)^2

\int (x^{2} - 2 x)^{2} - (6 x - x^{2})^{2} d x

2 x^{4} - \frac{32 x ^{3}}{3} + C

Solution steps

\int (x^{2} - 2 x)^{2} - (6 x - x^{2})^{2} d x

Simplify

(x^{2} - 2 x)^{2} - (6 x - x^{2})^{2} : 8 x^{3} - 32 x^{2}

= \int 8 x^{3} - 32 x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 8 x^{3} d x - \int 32 x^{2} d x

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

\int 32 x^{2} d x = \frac{32 x ^{3}}{3}

= 2 x^{4} - \frac{32 x ^{3}}{3}

Add a constant to the solution

= 2 x^{4} - \frac{32 x ^{3}}{3} + C

What is the integral of (x^2-2x)^2-(6x-x^2)^2 ?
The integral of (x^2-2x)^2-(6x-x^2)^2 is 2x^4-(32x^3)/3+C