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

y^{^{''}} - 36 y = 0

\frac{d}{d t} (\frac{2 t}{1 + t ^{2}})

\int_{- 3}^{3} \frac{6}{x ^{2} + 9} d x

\int x^{9} \cdot e^{x^{10} - 9} d x

y = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

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