What is the integral of [(4x^2-8x+4)/(2x-2)] ?

The integral of [(4x^2-8x+4)/(2x-2)] is x^2-2x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of [(4x^2-8x+4)/(2x-2)]

\int [\frac{4 x ^{2} - 8 x + 4}{2 x - 2}] d x

x^{2} - 2 x + C

Solution steps

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

Long division

\frac{4 x ^{2} - 8 x + 4}{2 x - 2} : 2 x - 2

= \int 2 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 2 x d x - \int 2 d x

\int 2 x d x = x^{2}

\int 2 d x = 2 x

= x^{2} - 2 x

Add a constant to the solution

= x^{2} - 2 x + C

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