What is the integral of (6x-12)/(x^2-4x-5) ?

The integral of (6x-12)/(x^2-4x-5) is 3ln|x+1|+3ln|x-5|+C

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integral of (6x-12)/(x^2-4x-5)

\int \frac{6 x - 12}{x ^{2} - 4 x - 5} d x

3 ln ∣ x + 1 ∣ + 3 ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{6 x - 12}{x ^{2} - 4 x - 5} d x

Take the partial fraction of

\frac{6 x - 12}{x ^{2} - 4 x - 5} : \frac{3}{x + 1} + \frac{3}{x - 5}

= \int \frac{3}{x + 1} + \frac{3}{x - 5} 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}{x + 1} d x + \int \frac{3}{x - 5} d x

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

\int \frac{3}{x - 5} d x = 3 ln ∣ x - 5 ∣

= 3 ln ∣ x + 1 ∣ + 3 ln ∣ x - 5 ∣

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ + 3 ln ∣ x - 5 ∣ + C

\int \frac{1}{u ^{\frac{1}{2}}} d u

n = 0 \sum \infty \frac{n}{3 ^{n + 1}}

\frac{\partial}{\partial x} (\frac{- y - x y}{( 1 + x ) ^{3}})

\int \frac{1}{x ^{2} + 25} d x

\int x (1 + x)^{4} d x

What is the integral of (6x-12)/(x^2-4x-5) ?
The integral of (6x-12)/(x^2-4x-5) is 3ln|x+1|+3ln|x-5|+C