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

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

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integral of 1/(s^2-1)

\int \frac{1}{s ^{2} - 1} d s

- \frac{1}{2} ln ∣ s + 1 ∣ + \frac{1}{2} ln ∣ s - 1 ∣ + C

Solution steps

\int \frac{1}{s ^{2} - 1} d s

Factor

s^{2} - 1 : - (- s^{2} + 1)

= \int \frac{1}{- ( - s ^{2} + 1 )} d s

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

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

Simplify

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

= - \frac{1}{2} ln ∣ s + 1 ∣ + \frac{1}{2} ln ∣ s - 1 ∣

Add a constant to the solution

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

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