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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

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

Add a constant to the solution

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

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