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

To evaluate the definite integral \[ \int_{0}^{\sqrt{3 / 3}} \frac{1}{1+9 x^{2}} d x, \] we can start by simplifying the limits of integration. We have \[ \sqrt{\frac{3}{3}} = \sqrt{1} = 1. \] So, we can rewrite the integral as \[ \int_{0}^{1} \frac{1}{1 + 9x^2} \, dx. \] Next, we will utilize a substitution to simplify the integrand. Let's take \[ u = 3x \quad \Rightarrow \quad du = 3 \, dx \quad \Rightarrow \quad dx = \frac{du}{3}. \] Now, we also need to change the limits of integration: - When \( x = 0 \), \( u = 3(0) = 0 \). - When \( x = 1 \), \( u = 3(1) = 3 \). With this substitution, the integral transforms to \[ \int_{0}^{3} \frac{1}{1 + 9\left(\frac{u}{3}\right)^2} \cdot \frac{du}{3} = \int_{0}^{3} \frac{1}{1 + \frac{9u^2}{9}} \cdot \frac{du}{3} = \int_{0}^{3} \frac{1}{1 + u^2} \cdot \frac{du}{3}. \] Therefore, the integral becomes \[ \frac{1}{3} \int_{0}^{3} \frac{1}{1 + u^2} \, du. \] The integral \[ \int \frac{1}{1 + u^2} \, du = \tan^{-1}(u). \] Now, evaluating the definite integral, we have \[ \frac{1}{3} \left[ \tan^{-1}(u) \right]_{0}^{3}. \] Calculating the limits: \[ \tan^{-1}(3) - \tan^{-1}(0) = \tan^{-1}(3) - 0 = \tan^{-1}(3). \] Putting it all together, we have \[ \frac{1}{3} \tan^{-1}(3). \] Thus, the value of the definite integral is \[ \boxed{\frac{1}{3} \tan^{-1}(3)}. \]