1. ## Trig integral

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.

2. u=x/3. So du=1/3 dx.

So the coefficient of arctan in the solution should be 1/3, not 1/9.

3. Thank you.

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

Make the substitution $\displaystyle \displaystyle x = 3\tan{\theta}$ so that $\displaystyle \displaystyle dx = 3\sec^2{\theta}\,d\theta$ and the integral becomes

$\displaystyle \displaystyle \int{\left[\frac{1}{(3\tan{\theta})^2 + 9}\right]\,3\sec^2{\theta}\,d\theta}$

$\displaystyle \displaystyle = \int{\frac{3\sec^2{\theta}}{9\tan^2{\theta} + 9}\,d\theta}$

$\displaystyle \displaystyle = \int{\frac{3\sec^2{\theta}}{9(\tan^2{\theta}+1)}\, d\theta}$

$\displaystyle \displaystyle = \int{\frac{\sec^2{\theta}}{3\sec^2{\theta}}\,d\the ta}$

$\displaystyle \displaystyle = \int{\frac{1}{3}\,d\theta}$

$\displaystyle \displaystyle = \frac{\theta}{3} + C$

$\displaystyle \displaystyle = \frac{\arctan{\left(\frac{x}{3}\right)}}{3} + C$.

5. Ahh, I didn't consider that. Thanks.