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

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

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integral of (3x^2+1)/(x^2-1)

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

3 x - 2 ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

Simplify

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

= 3 x - 2 ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= 3 x - 2 ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

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