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

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

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

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

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

Solution steps

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

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

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

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

\int \frac{x + 3}{x ^{2} - 1} d x = \frac{1}{2} ln x^{2} - 1 - 3 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

\int 1 d x = x

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

Add a constant to the solution

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

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