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

The integral of (x+4)/(x^2+4x-12) is 3/4 ln|x-2|+1/4 ln|x+6|+C

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

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

\frac{3}{4} ln ∣ x - 2 ∣ + \frac{1}{4} ln ∣ x + 6 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x + 4}{x ^{2} + 4 x - 12} : \frac{3}{4 ( x - 2 )} + \frac{1}{4 ( x + 6 )}

= \int \frac{3}{4 ( x - 2 )} + \frac{1}{4 ( x + 6 )} 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

= \int \frac{3}{4 ( x - 2 )} d x + \int \frac{1}{4 ( x + 6 )} d x

\int \frac{3}{4 ( x - 2 )} d x = \frac{3}{4} ln ∣ x - 2 ∣

\int \frac{1}{4 ( x + 6 )} d x = \frac{1}{4} ln ∣ x + 6 ∣

= \frac{3}{4} ln ∣ x - 2 ∣ + \frac{1}{4} ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{3}{4} ln ∣ x - 2 ∣ + \frac{1}{4} ln ∣ x + 6 ∣ + C

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What is the integral of (x+4)/(x^2+4x-12) ?
The integral of (x+4)/(x^2+4x-12) is 3/4 ln|x-2|+1/4 ln|x+6|+C