What is the integral of (5x-12)/(x^2-4) ?

The integral of (5x-12)/(x^2-4) is 5/2 ln(|x+2||x-2|)-12(-1/4 ln|x+2|+1/4 ln|x-2|)+C

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integral of (5x-12)/(x^2-4)

\int \frac{5 x - 12}{x ^{2} - 4} d x

\frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - 12 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{5 x - 12}{x ^{2} - 4} d x

Factor

\frac{5 x - 12}{x ^{2} - 4}

= \int \frac{5 x - 12}{( x + 2 ) ( x - 2 )} d x

Expand

\frac{5 x - 12}{( x + 2 ) ( x - 2 )} : \frac{5 x}{( x + 2 ) ( x - 2 )} - \frac{12}{( x + 2 ) ( x - 2 )}

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{5 x}{( x + 2 ) ( x - 2 )} d x - \int \frac{12}{( x + 2 ) ( x - 2 )} d x

\int \frac{5 x}{( x + 2 ) ( x - 2 )} d x = \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣)

\int \frac{12}{( x + 2 ) ( x - 2 )} d x = 12 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣)

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - 12 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣)

Add a constant to the solution

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - 12 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣) + C

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What is the integral of (5x-12)/(x^2-4) ?
The integral of (5x-12)/(x^2-4) is 5/2 ln(|x+2||x-2|)-12(-1/4 ln|x+2|+1/4 ln|x-2|)+C