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) = x^{3} - 2 x^{2} - 4 x - 6

\int (2 x + 5) e^{2 x + 5} d x

- \frac{9}{2} x^{2} + 3 x + 3

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} + 5})

\int cos^{3} (x s i) n^{2} x d x

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