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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{2 x + 2}{x ^{2} - 2 x + 1} d x

2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} + C

Solution steps

\int \frac{2 x + 2}{x ^{2} - 2 x + 1} d x

Complete the square

x^{2} - 2 x + 1 : (x - 1)^{2}

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

Apply u-substitution

: \int \frac{2 u + 4}{u ^{2}} d u

= \int \frac{2 u + 4}{u ^{2}} d u

Expand

\frac{2 u + 4}{u ^{2}} : \frac{2}{u} + \frac{4}{u ^{2}}

= \int \frac{2}{u} + \frac{4}{u ^{2}} 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

= \int \frac{2}{u} d u + \int \frac{4}{u ^{2}} d u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{4}{u ^{2}} d u = - \frac{4}{u}

= 2 ln ∣ u ∣ - \frac{4}{u}

Substitute back

u = x - 1

= 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1}

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} + C

x \to 8 lim (\frac{x ^{2} - 5}{x - 8})

\int x (3 x^{2} + 1) d x

\frac{d y}{d x} = \frac{e ^{x}}{y}

\int - π cos (π x) d x

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

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