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

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

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integral of 1/((x+1)(x-1))

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

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

Solution steps

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

Expand

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

= \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) = (2 x - 4) (2 x^{3} - x^{2} + 1)

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

h (y, w) \to 0 lim ((y - h (y, w))^{w (y)})

x \to 0 lim (\frac{( x ^{2} )}{∣ x ∣})

\frac{d y}{d x} - y = 8

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