Evaluate the definite integral: $$ \int_{0}^{6} \frac{d x}{x^{2}+9}=\square $$

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To evaluate the definite integral:

$$
\int_{0}^{6} \frac{dx}{x^{2}+9}
$$

we can use the standard integral formula:

$$
\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C
$$

Here, $$ a^2 = 9 $$ which means $$ a = 3 $$. Applying this to the integral:

$$
\int \frac{dx}{x^2 + 9} = \frac{1}{3} \arctan\left(\frac{x}{3}\right) + C
$$

Now, evaluate the definite integral from 0 to 6:

$$
\int_{0}^{6} \frac{dx}{x^{2}+9} = \left. \frac{1}{3} \arctan\left(\frac{x}{3}\right) \right|_{0}^{6}
$$

Substitute the limits of integration:

$$
= \frac{1}{3} \left[ \arctan\left(\frac{6}{3}\right) - \arctan\left(\frac{0}{3}\right) \right]
$$
$$
= \frac{1}{3} \left[ \arctan(2) - \arctan(0) \right]
$$
$$
= \frac{1}{3} \arctan(2)
$$

Since $$ \arctan(0) = 0 $$.

**Final Answer:**

$$
\frac{1}{3} \arctan(2)
$$
The definite integral evaluates to $$ \frac{1}{3} \arctan(2) $$.

Evaluate the definite integral:

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