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

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

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

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

- 4 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1) + x + C

Solution steps

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

\frac{x ^{2} + 3}{x ^{2} - 9} = \frac{12}{x ^{2} - 9} + 1

= \int \frac{12}{x ^{2} - 9} + 1 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{12}{x ^{2} - 9} d x + \int 1 d x

\int \frac{12}{x ^{2} - 9} d x = - 4 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1)

\int 1 d x = x

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

Add a constant to the solution

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

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