What is the integral of (x^2+3x+2)/(x^4-x^2) ?

The integral of (x^2+3x+2)/(x^4-x^2) is -3ln|x|+3ln|x-1|+2/x+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+3x+2)/(x^4-x^2)

\int \frac{x ^{2} + 3 x + 2}{x ^{4} - x ^{2}} d x

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

Solution steps

\int \frac{x ^{2} + 3 x + 2}{x ^{4} - x ^{2}} d x

Simplify

\frac{x ^{2} + 3 x + 2}{x ^{4} - x ^{2}} : \frac{x + 2}{x ^{2} ( x - 1 )}

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 15 r^{2} d r

\int - sin (x) sin (y) d x

\int \frac{d}{1} (tan (x)) d x

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

\int \frac{( ln ( x ) ) ^{2}}{4 x ^{3}} d x

What is the integral of (x^2+3x+2)/(x^4-x^2) ?
The integral of (x^2+3x+2)/(x^4-x^2) is -3ln|x|+3ln|x-1|+2/x+C