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

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

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integral of 1/(a^2-1)

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

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

Solution steps

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Add a constant to the solution

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

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