Evaluate the integral: integral (dx)/(sqrt(9+x2))

The integral is of the form $\int \frac{du}{\sqrt{a^2+u^2}}$. Here, $a^2=9 \implies a=3$ and $u^2=x^2 \implies u=x$.
Let $x = 3 \tan \theta$.
Then, $dx = 3 \sec^2 \theta \, d\theta$.

First, simplify the term under the square root:

Now substitute $dx$ and $\sqrt{9+x^2}$ into the integral:

The integral of $\sec \theta$ is a standard integral:

From our substitution $x = 3 \tan \theta$, we have $\tan \theta = \frac{x}{3}$.
To find $\sec \theta$, we can construct a right triangle where the opposite side is $x$ and the adjacent side is 3. The hypotenuse is $\sqrt{x^2+3^2} = \sqrt{x^2+9}$.
Thus, $\sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{x^2+9}}{3}$.
Substitute these expressions back into the result from Step 3:

Using the logarithm property $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$:

Since $-\ln 3$ is a constant, we can absorb it into the arbitrary constant $C$.

The final answer is $\boxed{\ln|x+\sqrt{x^2+9}| + C}$.

The integral is of the form $\int \frac{k\,du}{\sqrt{a^2+u^2}}$. Here, $a^2=1 \implies a=1$. For $9x^2$, let $u^2 = 9x^2 \implies u = 3x$.
Let $3x = 1 \tan \theta = \tan \theta$.
Then $x = \frac{1}{3} \tan \theta$.
And $dx = \frac{1}{3} \sec^2 \theta \, d\theta$.

First, simplify the term under the square root:

Now substitute $dx$ and $\sqrt{1+9x^2}$ into the integral:

The integral of $\sec \theta$ is a standard integral:

From our substitution $3x = \tan \theta$, we have $\tan \theta = 3x$.
To find $\sec \theta$, we can construct a right triangle where the opposite side is $3x$ and the adjacent side is 1. The hypotenuse is $\sqrt{(3x)^2+1^2} = \sqrt{9x^2+1}$.
Thus, $\sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{9x^2+1}}{1} = \sqrt{9x^2+1}$.
Substitute these expressions back into the result from Step 3:

The final answer is $\boxed{\ln|3x+\sqrt{9x^2+1}| + C}$.