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

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

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

\int \frac{1}{( x ^{2} - 1 )} d x

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

Solution steps

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{ln ( 14 x )}{x ^{5}} d x

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

\frac{1}{x ^{2} + y ^{2}} (x d y - y d x) = 0

cos (2 (t - \frac{5 π}{4}))

x \to 2 + lim (∣ x - 2 ∣)

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