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

The integral of (x^2-x+1)/(x+1) is -3x+3ln|x+1|+1/2 (x+1)^2-3+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-x+1)/(x+1)

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

- 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3 + C

Solution steps

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

Expand

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

Simplify

\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣ - (x + 1 - ln ∣ x + 1 ∣) + ln ∣ x + 1 ∣ : - 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3

= - 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3

Add a constant to the solution

= - 3 x + 3 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 3 + C

\frac{1}{2 ^{x}}

\int \frac{1}{( 16 + x ^{2} )} d x

\int (cos^{5} (x)) d x

\int x^{2} + x + 4 d x

\int \frac{7 x ^{2} + x + 12}{3 x ^{3} + 12 x} d x

What is the integral of (x^2-x+1)/(x+1) ?
The integral of (x^2-x+1)/(x+1) is -3x+3ln|x+1|+1/2 (x+1)^2-3+C