What is the integral of (12)/((x-4)(x+8)) ?

The integral of (12)/((x-4)(x+8)) is ln|x-4|-ln|x+8|+C

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integral of (12)/((x-4)(x+8))

\int \frac{12}{( x - 4 ) ( x + 8 )} d x

ln ∣ x - 4 ∣ - ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{12}{( x - 4 ) ( x + 8 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int \frac{1}{( x - 4 ) ( x + 8 )} d x

Take the partial fraction of

\frac{1}{( x - 4 ) ( x + 8 )} : \frac{1}{12 ( x - 4 )} - \frac{1}{12 ( x + 8 )}

= 12 \cdot \int \frac{1}{12 ( x - 4 )} - \frac{1}{12 ( x + 8 )} d x

Apply the Sum Rule:

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

= 12 (\int \frac{1}{12 ( x - 4 )} d x - \int \frac{1}{12 ( x + 8 )} d x)

\int \frac{1}{12 ( x - 4 )} d x = \frac{1}{12} ln ∣ x - 4 ∣

\int \frac{1}{12 ( x + 8 )} d x = \frac{1}{12} ln ∣ x + 8 ∣

= 12 (\frac{1}{12} ln ∣ x - 4 ∣ - \frac{1}{12} ln ∣ x + 8 ∣)

Simplify

12 (\frac{1}{12} ln ∣ x - 4 ∣ - \frac{1}{12} ln ∣ x + 8 ∣) : ln ∣ x - 4 ∣ - ln ∣ x + 8 ∣

= ln ∣ x - 4 ∣ - ln ∣ x + 8 ∣

Add a constant to the solution

= ln ∣ x - 4 ∣ - ln ∣ x + 8 ∣ + C

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What is the integral of (12)/((x-4)(x+8)) ?
The integral of (12)/((x-4)(x+8)) is ln|x-4|-ln|x+8|+C