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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{4 x + 5}{2 x} + e^{x} d x

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

\int 4 + x^{2} + sin (x) d x

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

\int sin (n \cdot π \cdot x) d x

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