What is the integral of (x-5)/((x+1)(x-2)) ?

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

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integral of (x-5)/((x+1)(x-2))

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

- ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{u - 6}{u ( u - 3 )} d u

Expand

\frac{u - 6}{u ( u - 3 )} : \frac{u}{u ( u - 3 )} - \frac{6}{u ( u - 3 )}

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{u}{u ( u - 3 )} d u - \int \frac{6}{u ( u - 3 )} d u

\int \frac{u}{u ( u - 3 )} d u = ln ∣ u - 3 ∣

\int \frac{6}{u ( u - 3 )} d u = 6 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

= ln ∣ u - 3 ∣ - 6 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

Substitute back

u = x + 1

= ln ∣ x + 1 - 3 ∣ - 6 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x + 1 - 3 ∣)

Simplify

ln ∣ x + 1 - 3 ∣ - 6 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x + 1 - 3 ∣) : - ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣

= - ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣ + C

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