What is the integral of 2/((x-1)(x+1)) ?

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

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integral of 2/((x-1)(x+1))

\int \frac{2}{( x - 1 ) ( x + 1 )} d x

- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{2}{( x - 1 ) ( x + 1 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{( x - 1 ) ( x + 1 )} d x

Expand

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

= 2 \cdot \int \frac{1}{x ^{2} - 1} d x

Factor

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

= 2 \cdot \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

= 2 (- \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}

= 2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

Simplify

2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

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