What is the integral of (3x+9)/(x^2-4x-5) ?

The integral of (3x+9)/(x^2-4x-5) is -ln|x+1|+4ln|x-5|+C

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integral of (3x+9)/(x^2-4x-5)

\int \frac{3 x + 9}{x ^{2} - 4 x - 5} d x

- ln ∣ x + 1 ∣ + 4 ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{3 x + 9}{x ^{2} - 4 x - 5} d x

Take the partial fraction of

\frac{3 x + 9}{x ^{2} - 4 x - 5} : - \frac{1}{x + 1} + \frac{4}{x - 5}

= \int - \frac{1}{x + 1} + \frac{4}{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{1}{x + 1} d x + \int \frac{4}{x - 5} d x

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

\int \frac{4}{x - 5} d x = 4 ln ∣ x - 5 ∣

= - ln ∣ x + 1 ∣ + 4 ln ∣ x - 5 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + 4 ln ∣ x - 5 ∣ + C

\frac{d}{d x} (\frac{x}{4 x - 3 ln ( x )})

\int x (sin (x)) d x

x \to 0 lim (\frac{x}{∣ x ∣ + x})

n = 1 \sum \infty \frac{1}{10 n + 1}

\frac{d}{d x} (\frac{3 x ^{2}}{x ^{3} + 1})

What is the integral of (3x+9)/(x^2-4x-5) ?
The integral of (3x+9)/(x^2-4x-5) is -ln|x+1|+4ln|x-5|+C