What is the integral of (3x-4)/((x-2)(x+1)) ?

The integral of (3x-4)/((x-2)(x+1)) is 3ln|x-2|-7(-1/3 ln|x+1|+1/3 ln|x-2|)+C

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integral of (3x-4)/((x-2)(x+1))

\int \frac{3 x - 4}{( x - 2 ) ( x + 1 )} d x

3 ln ∣ x - 2 ∣ - 7 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{3 x - 4}{( x - 2 ) ( x + 1 )} d x

Apply u-substitution

: \int \frac{3 u - 7}{u ( u - 3 )} d u

= \int \frac{3 u - 7}{u ( u - 3 )} d u

Expand

\frac{3 u - 7}{u ( u - 3 )} : \frac{3 u}{u ( u - 3 )} - \frac{7}{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{3 u}{u ( u - 3 )} d u - \int \frac{7}{u ( u - 3 )} d u

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

\int \frac{7}{u ( u - 3 )} d u = 7 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

= 3 ln ∣ u - 3 ∣ - 7 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

Substitute back

u = x + 1

= 3 ln ∣ x + 1 - 3 ∣ - 7 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x + 1 - 3 ∣)

Simplify

3 ln ∣ x + 1 - 3 ∣ - 7 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x + 1 - 3 ∣) : 3 ln ∣ x - 2 ∣ - 7 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣)

= 3 ln ∣ x - 2 ∣ - 7 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣)

Add a constant to the solution

= 3 ln ∣ x - 2 ∣ - 7 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣) + C

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What is the integral of (3x-4)/((x-2)(x+1)) ?
The integral of (3x-4)/((x-2)(x+1)) is 3ln|x-2|-7(-1/3 ln|x+1|+1/3 ln|x-2|)+C