What is the integral of (8x-2)/(x^2-1) ?

The integral of (8x-2)/(x^2-1) is 4ln|x^2-1|+ln|x+1|-ln|x-1|+C

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integral of (8x-2)/(x^2-1)

\int \frac{8 x - 2}{x ^{2} - 1} d x

4 ln x^{2} - 1 + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{8 x - 2}{x ^{2} - 1} d x

Expand

\frac{8 x - 2}{x ^{2} - 1} : \frac{8 x}{x ^{2} - 1} - \frac{2}{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{8 x}{x ^{2} - 1} d x - \int \frac{2}{x ^{2} - 1} d x

\int \frac{8 x}{x ^{2} - 1} d x = 4 ln x^{2} - 1

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

= 4 ln x^{2} - 1 - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣)

Simplify

= 4 ln x^{2} - 1 + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= 4 ln x^{2} - 1 + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

What is the integral of (8x-2)/(x^2-1) ?
The integral of (8x-2)/(x^2-1) is 4ln|x^2-1|+ln|x+1|-ln|x-1|+C