\( \left. \begin{array} { l } { \frac { d } { d x } \int _ { 0 } ^ { x ^ { 2 } } \frac { d t } { t ^ { 2 } + 3 } } \\ { \frac { d } { d x } \int _ { 0 } ^ { x ^ { 2 } } \frac { d t } { t ^ { 2 } + 3 } = \square } \end{array} \right. \)

To solve the problem, we apply the Fundamental Theorem of Calculus along with the chain rule. First, let's denote the integral as \( F(x) = \int_{0}^{x^2} \frac{dt}{t^2 + 3} \). When we differentiate \( F(x) \) with respect to \( x \), we need to consider that the upper limit is a function of \( x \), specifically \( x^2 \). Therefore, by the chain rule: \[ \frac{d}{dx} F(x) = \frac{d}{dx} \int_{0}^{x^2} \frac{dt}{t^2 + 3} = \frac{d}{d(x^2)} \int_{0}^{u} \frac{dt}{t^2 + 3} \cdot \frac{d(x^2)}{dx} \] where \( u = x^2 \). The derivative of \( x^2 \) with respect to \( x \) is \( 2x \). Thus, we have: \[ \frac{d}{dx} F(x) = \frac{1}{(x^2)^2 + 3} \cdot 2x = \frac{2x}{x^4 + 3} \] So the final result is: \[ \frac{d}{dx} \int_{0}^{x^2} \frac{dt}{t^2 + 3} = \frac{2x}{x^4 + 3} \]