`int(dx)/(9x^(2) +1)`

Options

A. ` (1)/(3) tan^(-1) (2x) + c`
B. `(1)/(3) tan^(-1) x + c `
C. `(1)/(3) tan^(-1) (3x) + c`
D. `(1)/(3) tan^(-1) (6x) + c `

Explanation

To solve the integral \(\int \frac{dx}{9x^2 + 1}\), we can use the formula for the integration of the form \(\int \frac{dx}{a x^2 + b^2}\).

### Step-by-Step Solution:

1. **Identify the constants**:
We can rewrite the denominator \(9x^2 + 1\) in the form \(a x^2 + b^2\). Here, we have:
- \(a = 9\) (which is \(3^2\))
- \(b = 1\) (which is \(1^2\))

2. **Use the integration formula**:
The formula for the integral is:
\[ \int \frac{dx}{a x^2 + b^2} = \frac{1}{b} \cdot \frac{1}{\sqrt{a}} \tan^{-1} \left( \frac{\sqrt{a}}{b} x \right) + C \]
In our case, we can substitute \(a = 9\) and \(b = 1\):
\[ \int \frac{dx}{9x^2 + 1} = \frac{1}{1} \cdot \frac{1}{3} \tan^{-1} \left( \frac{3}{1} x \right) + C \]

3. **Simplify the expression**:
This simplifies to:
\[ \frac{1}{3} \tan^{-1}(3x) + C \]

4. **Final answer**:
Therefore, the integral \(\int \frac{dx}{9x^2 + 1}\) is:
\[ \frac{1}{3} \tan^{-1}(3x) + C \]