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

The integral of (x^2-9)/(x^2+9) is -6arctan(x/3)+x+C

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

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

- 6 arctan (\frac{x}{3}) + x + C

Solution steps

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

\frac{x ^{2} - 9}{x ^{2} + 9} = - \frac{18}{x ^{2} + 9} + 1

= \int - \frac{18}{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{18}{x ^{2} + 9} d x + \int 1 d x

\int \frac{18}{x ^{2} + 9} d x = 6 arctan (\frac{x}{3})

\int 1 d x = x

= - 6 arctan (\frac{x}{3}) + x

Add a constant to the solution

= - 6 arctan (\frac{x}{3}) + x + C

What is the integral of (x^2-9)/(x^2+9) ?
The integral of (x^2-9)/(x^2+9) is -6arctan(x/3)+x+C