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

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

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

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

x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 + C

Solution steps

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

Expand

\frac{x ^{2} - 5 x - 6}{x ^{2} - 4} : \frac{x ^{2}}{x ^{2} - 4} - \frac{5 x}{x ^{2} - 4} - \frac{6}{x ^{2} - 4}

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

\int \frac{x ^{2}}{x ^{2} - 4} d x = - ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x

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

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

= - ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x - \frac{5}{2} ln x^{2} - 4 - (- 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1))

Simplify

- ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x - \frac{5}{2} ln x^{2} - 4 - (- 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)) : x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1

= x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1

Add a constant to the solution

= x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 + C

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