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

f (x) = x + 8, at x = 8

y^{^{''}} + 4 y^{^{'}} = 2 e^{2 x}

\int - k

4 x + \frac{5}{x}

\frac{\partial}{\partial x} (6 - x^{2} + x y - 4 y^{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