What is the integral of (x-5)/((x-1)(x+1)) ?

The integral of (x-5)/((x-1)(x+1)) is 1/2 ln(|x-1||x+1|)+5(1/2 ln|x+1|-1/2 ln|x-1|)+C

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integral of (x-5)/((x-1)(x+1))

\int \frac{x - 5}{( x - 1 ) ( x + 1 )} d x

\frac{1}{2} ln (∣ x - 1 ∣ ∣ x + 1 ∣) + 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

Solution steps

\int \frac{x - 5}{( x - 1 ) ( x + 1 )} d x

Expand

\frac{x - 5}{( x - 1 ) ( x + 1 )} : \frac{x}{( x - 1 ) ( x + 1 )} - \frac{5}{( x - 1 ) ( x + 1 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{( x - 1 ) ( x + 1 )} d x - \int \frac{5}{( x - 1 ) ( x + 1 )} d x

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

\int \frac{5}{( x - 1 ) ( x + 1 )} d x = - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= \frac{1}{2} ln (∣ x - 1 ∣ ∣ x + 1 ∣) - (- 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣))

Simplify

= \frac{1}{2} ln (∣ x - 1 ∣ ∣ x + 1 ∣) + 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Add a constant to the solution

= \frac{1}{2} ln (∣ x - 1 ∣ ∣ x + 1 ∣) + 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

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What is the integral of (x-5)/((x-1)(x+1)) ?
The integral of (x-5)/((x-1)(x+1)) is 1/2 ln(|x-1||x+1|)+5(1/2 ln|x+1|-1/2 ln|x-1|)+C