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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Expand

\frac{x ^{2} - 2}{( x - 1 ) ^{2}} : \frac{x ^{2}}{( x - 1 ) ^{2}} - \frac{2}{( x - 1 ) ^{2}}

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

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

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

= x - 1 + 2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - (- \frac{2}{x - 1})

Simplify

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

Add a constant to the solution

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

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