Evaluate the definite integral. \[ \int_{0}^{\sqrt{3 / 3}} \frac{1}{1+9 x^{2}} d x \]

To evaluate the definite integral \[ \int_{0}^{\sqrt{3 / 3}} \frac{1}{1+9 x^{2}} \, dx, \] we can use a trigonometric substitution. The integrand can be rewritten to fit a standard form that resembles the derivative of the arctangent function. Specifically, we know that \[ \int \frac{1}{1 + a^2 x^2} \, dx = \frac{1}{a} \tan^{-1}(ax) + C, \] for \(a = 3\). Here, \(a = 3\), and thus we have: \[ \int \frac{1}{1 + 9x^2} \, dx = \frac{1}{3} \tan^{-1}(3x) + C. \] Now we can proceed to evaluate the definite integral. We must evaluate \[ \left[ \frac{1}{3} \tan^{-1}(3x) \right]_{0}^{\sqrt{3/3}}. \] Calculating at the upper limit: 1. **Evaluate at \(x = \sqrt{3/3} = 1\)**: \[ \frac{1}{3} \tan^{-1}(3 \cdot 1) = \frac{1}{3} \tan^{-1}(3). \] 2. **Evaluate at the lower limit \(x = 0\)**: \[ \frac{1}{3} \tan^{-1}(3 \cdot 0) = \frac{1}{3} \tan^{-1}(0) = 0. \] Thus, we can substitute back into our evaluation: \[ \int_{0}^{\sqrt{3 / 3}} \frac{1}{1+9 x^{2}} \, dx = \frac{1}{3} \tan^{-1}(3) - 0 = \frac{1}{3} \tan^{-1}(3). \] In conclusion, the value of the definite integral is \[ \frac{1}{3} \tan^{-1}(3). \]