Can anyone please help to find an antiderivative of this function and the method used?
(Please see attatchment)
Cheers,
Use trigonometric substitution.
Let $\displaystyle x = 3\tan{\theta}$ so that $\displaystyle dx = 3\sec^2{\theta}\,d\theta$.
Also note that $\displaystyle \theta = \arctan{\frac{x}{3}}$.
So the integral becomes:
$\displaystyle \int{\frac{1}{x^2 + 9}\,dx} = \int{\frac{1}{(3\tan{\theta})^2 + 9}\,3\sec^2{\theta}\,d\theta}$
$\displaystyle = \int{\frac{3\sec^2{\theta}}{9\tan^2{\theta} + 9}\,d\theta}$
$\displaystyle = \int{\frac{3\sec^2{\theta}}{9(\tan^2{\theta} + 1)}\,d\theta}$
$\displaystyle = \int{\frac{\sec^2{\theta}}{3\sec^2{\theta}}\,d\the ta}$
$\displaystyle = \int{\frac{1}{3}\,d\theta}$
$\displaystyle = \frac{1}{3}\theta + C$
$\displaystyle = \frac{1}{3}\arctan{\frac{x}{3}} + C$.
1. Yes it does.
You should know the Pythagorean Identity
$\displaystyle \sin^2{\theta} + \cos^2{\theta} = 1$.
If you divide both sides by $\displaystyle \cos^2{\theta}$ you find
$\displaystyle \frac{\sin^2{\theta}}{\cos^2{\theta}} + \frac{\cos^2{\theta}}{\cos^2{\theta}} = \frac{1}{\cos^2{\theta}}$
$\displaystyle \tan^2{\theta} + 1 = \sec^2{\theta}$.
2. If you prefer to use a table of integrals, then you can use the formula
$\displaystyle \int{\frac{1}{a^2 + x^2}\,dx} = \frac{1}{a}\arctan{\frac{x}{a}} + C$.
As you can see, this result is found using trigonometric substitution.