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

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

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

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

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

Solution steps

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

Expand

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

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

Add a constant to the solution

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

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