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

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

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

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

Solution steps

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

Expand

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

\int \frac{3 x}{x ^{2} - 1} d x = \frac{3}{2} ln x^{2} - 1

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

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

Add a constant to the solution

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

\int sin (2 x) d x

\int 2 x sec (x^{2}) d x

\int (9 x^{2} + 12) (2 x^{3} + 8 x)^{11} d x

\int (x - 2 ∣ x ∣) d x

\int \frac{8}{( y - 9 ) ^{3}} d y

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