What is the integral of 1+9/(x^2-4) ?

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

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integral of 1+9/(x^2-4)

\int 1 + \frac{9}{x ^{2} - 4} d x

x - \frac{9}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C

Solution steps

\int 1 + \frac{9}{x ^{2} - 4} 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 1 d x + \int \frac{9}{x ^{2} - 4} d x

\int 1 d x = x

\int \frac{9}{x ^{2} - 4} d x = - \frac{9}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

= x - \frac{9}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

Add a constant to the solution

= x - \frac{9}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C

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