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

The integral of (2x+3)/(x(x+1)(x-1)) is 1/2 ln|x+1|+5/2 ln|x-1|-3ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+3)/(x(x+1)(x-1))

\int \frac{2 x + 3}{x ( x + 1 ) ( x - 1 )} d x

\frac{1}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x ∣ + C

Solution steps

\int \frac{2 x + 3}{x ( x + 1 ) ( x - 1 )} d x

Expand

\frac{2 x + 3}{x ( x + 1 ) ( x - 1 )} : \frac{2 x}{x ( x + 1 ) ( x - 1 )} + \frac{3}{x ( x + 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{2 x}{x ( x + 1 ) ( x - 1 )} d x + \int \frac{3}{x ( x + 1 ) ( x - 1 )} d x

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

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

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

Simplify

- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 3 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) : \frac{1}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x ∣

= \frac{1}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x ∣ + C

\frac{\partial}{\partial x} (\frac{x + y}{z - 2 y})

\int \frac{cos ( x )}{4 - sin ( x )} d x

\int_{1}^{4} x f (x) d x

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

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

What is the integral of (2x+3)/(x(x+1)(x-1)) ?
The integral of (2x+3)/(x(x+1)(x-1)) is 1/2 ln|x+1|+5/2 ln|x-1|-3ln|x|+C