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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} (x - 1)^{2} + C

Solution steps

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

Expand

\frac{x ^{2} - 2 x + 1}{x - 1} : \frac{x ^{2}}{x - 1} - \frac{2 x}{x - 1} + \frac{1}{x - 1}

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

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

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

Simplify

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

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

Add a constant to the solution

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

\int - 4 cos (4 x) d x

\int sin^{3} (x) + cos (x) d x

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

\int \frac{1}{( x + 25 ) x} d x

\int \frac{1}{6} sec (\frac{1}{9} x) tan (\frac{1}{9} x) d x

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