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

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

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

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

ln (∣ x + 1 ∣ ∣ x - 1 ∣) + C

Solution steps

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

Apply u-substitution

= 2 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = (x + 1) (x - 1)

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

Simplify

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

= ln (∣ x + 1 ∣ ∣ x - 1 ∣)

Add a constant to the solution

= ln (∣ x + 1 ∣ ∣ x - 1 ∣) + C

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