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Let $I=\int \sqrt{1+\frac{x^{2}}{9}} d x=\frac{1}{3} \int \sqrt{9+x^{2}} d x=\frac{1}{3} \int \sqrt{(3)^{2}+x^{2}} d x$

It is known that, $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{C}$

$\begin{aligned} \therefore I &=\frac{1}{3}\left[\frac{x}{2} \sqrt{x^{2}+9}+\frac{9}{2} \log \left|x+\sqrt{x^{2}+9}\right|\right]+\mathrm{C} \\ &=\frac{x}{6} \sqrt{x^{2}+9}+\frac{3}{2} \log \left|x+\sqrt{x^{2}+9}\right|+\mathrm{C} \end{aligned}$