What is the integral of (7x^2-x-5)/(x^2-1) ?

The integral of (7x^2-x-5)/(x^2-1) is 7x-ln|x+1|+ln|x-1|-1/2 ln|x^2-1|+C

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integral of (7x^2-x-5)/(x^2-1)

\int \frac{7 x ^{2} - x - 5}{x ^{2} - 1} d x

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

Solution steps

\int \frac{7 x ^{2} - x - 5}{x ^{2} - 1} d x

Expand

\frac{7 x ^{2} - x - 5}{x ^{2} - 1} : \frac{7 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{7 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{7 x ^{2}}{x ^{2} - 1} d x = - 7 (\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 ∣)

= - 7 (\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

- 7 (\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 ∣)) : 7 x - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} - 1

= 7 x - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} - 1

Add a constant to the solution

= 7 x - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} - 1 + C

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What is the integral of (7x^2-x-5)/(x^2-1) ?
The integral of (7x^2-x-5)/(x^2-1) is 7x-ln|x+1|+ln|x-1|-1/2 ln|x^2-1|+C