Integrate the function: $\sqrt{1+\frac{x^{2}}{9}}$

Let $I = \int \sqrt{1+\frac{x^{2}}{9}} \, dx$.
We can rewrite the integrand as:
$I = \int \sqrt{\frac{9+x^{2}}{9}} \, dx = \frac{1}{3} \int \sqrt{9+x^{2}} \, dx = \frac{1}{3} \int \sqrt{3^{2}+x^{2}} \, dx$.
Using the standard integration formula $\int \sqrt{x^{2}+a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2} \ln |x + \sqrt{x^{2}+a^{2}}| + C$,where $a = 3$:
$I = \frac{1}{3} \left[ \frac{x}{2} \sqrt{x^{2}+3^{2}} + \frac{3^{2}}{2} \ln |x + \sqrt{x^{2}+3^{2}}| \right] + C$.
$I = \frac{1}{3} \left[ \frac{x}{2} \sqrt{x^{2}+9} + \frac{9}{2} \ln |x + \sqrt{x^{2}+9}| \right] + C$.
$I = \frac{x}{6} \sqrt{x^{2}+9} + \frac{3}{2} \ln |x + \sqrt{x^{2}+9}| + C$,where $C$ is an arbitrary constant.