What is the integral of (x^2+x-5)/(x^2-1) ?

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

\int \frac{x ^{2} + x - 5}{x ^{2} - 1} d x

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

Solution steps

\int \frac{x ^{2} + x - 5}{x ^{2} - 1} d x

Expand

\frac{x ^{2} + x - 5}{x ^{2} - 1} : \frac{x ^{2}}{x ^{2} - 1} + \frac{x}{x ^{2} - 1} - \frac{5}{x ^{2} - 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 ^{2} - 1} d x + \int \frac{x}{x ^{2} - 1} d x - \int \frac{5}{x ^{2} - 1} d x

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

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

\int \frac{5}{x ^{2} - 1} d x = - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + x + \frac{1}{2} ln x^{2} - 1 - (- 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣))

Simplify

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

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

Add a constant to the solution

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

\int \frac{5}{8 x ( x + 1 ) ^{2}} d x

\int - e^{- 2 y} + 1 d y

y = \frac{x}{x - 5}, (- 1, a)

\int_{6}^{8} \frac{48}{( x - 6 ) ^{3}} d x

\frac{\partial}{\partial x} (\frac{x}{y + 8 z})

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