Evaluate: `int_0^3 (dx)/(9+x^2)`

Answer

To evaluate the integral \( \int_0^3 \frac{dx}{9 + x^2} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more convenient form:
\[ \int_0^3 \frac{dx}{9 + x^2} = \int_0^3 \frac{dx}{3^2 + x^2} \]

### Step 2: Identify the Formula
We recognize that this integral can be solved using the formula:
\[ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
where \( a = 3 \) in our case.

### Step 3: Apply the Formula
Using the formula, we can evaluate the integral:
\[ \int \frac{dx}{9 + x^2} = \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C \]

### Step 4: Evaluate the Definite Integral
Now, we need to evaluate the definite integral from 0 to 3:
\[ \left[ \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) \right]_0^3 \]

### Step 5: Calculate the Upper Limit
First, we calculate the upper limit:
\[ \frac{1}{3} \tan^{-1} \left( \frac{3}{3} \right) = \frac{1}{3} \tan^{-1}(1) \]
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:
\[ \frac{1}{3} \cdot \frac{\pi}{4} = \frac{\pi}{12} \]

### Step 6: Calculate the Lower Limit
Now, we calculate the lower limit:
\[ \frac{1}{3} \tan^{-1} \left( \frac{0}{3} \right) = \frac{1}{3} \tan^{-1}(0) = \frac{1}{3} \cdot 0 = 0 \]

### Step 7: Subtract the Limits
Now, we subtract the lower limit from the upper limit:
\[ \frac{\pi}{12} - 0 = \frac{\pi}{12} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^3 \frac{dx}{9 + x^2} = \frac{\pi}{12} \]

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