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

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

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integral of (x^2-1)^{-1}

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

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

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \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

x^{2} + 2 y^{2} = 2

x^{2} + x - 2, [- 3, 4]

\int_{1}^{2} 3 x^{4} ln (x) d x

\frac{\partial}{\partial x} (\frac{y}{z})

\frac{d}{d x} (x^{3} - \frac{5}{x} + 4)

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