The best method would certainly be integration :)
The area is given by the integral,
[tex]\displaystyle\int_0^2\frac{\mathrm dx}{9-x^2}[/tex]
We can split up the integrand into partial fractions:
[tex]\dfrac1{9-x^2}=\dfrac16\left(\dfrac1{3-x}+\dfrac1{3+x}\right)[/tex]
Then the area is
[tex]\displaystyle\frac16\int_0^2\left(\frac1{3-x}+\frac1{3+x}\right)\,\mathrm dx[/tex]
[tex]=\dfrac16(-\ln|3-x|+\ln|3+x|)\bigg|_0^2[/tex]
[tex]=\dfrac16\left((\ln5-\ln1)-(\ln3-\ln3)\right)=\boxed{\dfrac{\ln5}6}[/tex]
