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

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

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

- 2 x - \frac{3}{x + 1} + \frac{x ^{2}}{x + 1} - 2 + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

= - 2 x - \frac{3}{x + 1} + \frac{x ^{2}}{x + 1} - 2

Add a constant to the solution

= - 2 x - \frac{3}{x + 1} + \frac{x ^{2}}{x + 1} - 2 + C

\int \frac{cos ( θ )}{1 + sin ( θ )} d θ

\int \frac{x ^{4} + 1}{x ^{3} + 9 x} d x

\int \frac{1}{3 x ^{2} + 2 x + 2} d x

\int (x^{2} + 5 x + 6) sin (2 x) d x

\int \frac{5}{x ^{3} + 4 x} d x

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