$\int \sqrt{1 + \frac{x^2}{9}} dx$ is equal to

(Where C is an arbitrary constant)

Simplify the square root:

$\sqrt{1 + \frac{x^2}{9}} = \sqrt{\frac{9 + x^2}{9}}$

$= \frac{\sqrt{9 + x^2}}{\sqrt{9}}$

$= \frac{\sqrt{x^2 + 9}}{3}$

$= \frac{1}{3}\sqrt{x^2 + 9}$

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The integral becomes:

$\int \sqrt{1 + \frac{x^2}{9}} dx = \int \frac{1}{3}\sqrt{x^2 + 9} dx$

$= \frac{1}{3}\int \sqrt{x^2 + 9} dx$

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This matches the standard form $\int \sqrt{x^2 + a^2} dx$ where $a^2 = 9$, so $a = 3$.

The standard formula is:

$\int \sqrt{x^2 + a^2} dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log|x + \sqrt{x^2 + a^2}| + C$

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Substituting $a = 3$:

$\int \sqrt{x^2 + 9} dx = \frac{x}{2}\sqrt{x^2 + 9} + \frac{9}{2}\log|x + \sqrt{x^2 + 9}| + C$

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Multiplying by $\frac{1}{3}$:

$\int \sqrt{1 + \frac{x^2}{9}} dx = \frac{1}{3}\left[\frac{x}{2}\sqrt{x^2 + 9} + \frac{9}{2}\log|x + \sqrt{x^2 + 9}|\right] + C$

$= \frac{1}{3}\left[\frac{9}{2}\log|x + \sqrt{x^2 + 9}| + \frac{x}{2}\sqrt{x^2 + 9}\right] + C$

Therefore, the answer is:

$\frac{1}{3}\left[\frac{9}{2}\log|x + \sqrt{x^2 + 9}| + \frac{x}{2}\sqrt{x^2 + 9}\right] + C$