What is the integral of (3x-9)/((x+4)(x-5)) ?

The integral of (3x-9)/((x+4)(x-5)) is 3ln|x+4|+6(1/9 ln|x-5|-1/9 ln|x+4|)+C

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integral of (3x-9)/((x+4)(x-5))

\int \frac{3 x - 9}{( x + 4 ) ( x - 5 )} d x

3 ln ∣ x + 4 ∣ + 6 (\frac{1}{9} ln ∣ x - 5 ∣ - \frac{1}{9} ln ∣ x + 4 ∣) + C

Solution steps

\int \frac{3 x - 9}{( x + 4 ) ( x - 5 )} d x

Apply u-substitution

: \int \frac{3 u + 6}{u ( u + 9 )} d u

= \int \frac{3 u + 6}{u ( u + 9 )} d u

Expand

\frac{3 u + 6}{u ( u + 9 )} : \frac{3 u}{u ( u + 9 )} + \frac{6}{u ( u + 9 )}

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 + 9 )} d u + \int \frac{6}{u ( u + 9 )} d u

\int \frac{3 u}{u ( u + 9 )} d u = 3 ln ∣ u + 9 ∣

\int \frac{6}{u ( u + 9 )} d u = 6 (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

= 3 ln ∣ u + 9 ∣ + 6 (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

Substitute back

u = x - 5

= 3 ln ∣ x - 5 + 9 ∣ + 6 (\frac{1}{9} ln ∣ x - 5 ∣ - \frac{1}{9} ln ∣ x - 5 + 9 ∣)

Simplify

3 ln ∣ x - 5 + 9 ∣ + 6 (\frac{1}{9} ln ∣ x - 5 ∣ - \frac{1}{9} ln ∣ x - 5 + 9 ∣) : 3 ln ∣ x + 4 ∣ + 6 (\frac{1}{9} ln ∣ x - 5 ∣ - \frac{1}{9} ln ∣ x + 4 ∣)

= 3 ln ∣ x + 4 ∣ + 6 (\frac{1}{9} ln ∣ x - 5 ∣ - \frac{1}{9} ln ∣ x + 4 ∣)

Add a constant to the solution

= 3 ln ∣ x + 4 ∣ + 6 (\frac{1}{9} ln ∣ x - 5 ∣ - \frac{1}{9} ln ∣ x + 4 ∣) + C

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What is the integral of (3x-9)/((x+4)(x-5)) ?
The integral of (3x-9)/((x+4)(x-5)) is 3ln|x+4|+6(1/9 ln|x-5|-1/9 ln|x+4|)+C