The question:

$\displaystyle \int {\frac{1}{x^2 + 9} \ dx}$

My attempt:

It looks similar to the derivative of inverse tan, so dividing the denominator by 9:

$\displaystyle \frac{1}{9}\int{\frac{1}{(\frac{x}{3})^2 + 1} \ dx}$

= $\displaystyle \frac{1}{9}tan(\frac{x}{3})^{-1} + C$

However it doesn't seem to match what Wolfram is telling me. I'm sure I've made a silly mistake, but I'm not seeing it. Any help would be great. Thanks.