What is the integral of (2x-1)/((x+8)(x-9)) ?

The integral of (2x-1)/((x+8)(x-9)) is ln(|x+8||x-9|)+C

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integral of (2x-1)/((x+8)(x-9))

\int \frac{2 x - 1}{( x + 8 ) ( x - 9 )} d x

ln (∣ x + 8 ∣ ∣ x - 9 ∣) + C

Solution steps

\int \frac{2 x - 1}{( x + 8 ) ( x - 9 )} d x

Apply u-substitution

: \int \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = (x + 8) (x - 9)

= ln ∣ (x + 8) (x - 9) ∣

Simplify

ln ∣ (x + 8) (x - 9) ∣ : ln (∣ x + 8 ∣ ∣ x - 9 ∣)

= ln (∣ x + 8 ∣ ∣ x - 9 ∣)

Add a constant to the solution

= ln (∣ x + 8 ∣ ∣ x - 9 ∣) + C

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What is the integral of (2x-1)/((x+8)(x-9)) ?
The integral of (2x-1)/((x+8)(x-9)) is ln(|x+8||x-9|)+C