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

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

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

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

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

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

Add a constant to the solution

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

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