Evaluate Limit of Integral with Logarithmic Function

To find the value of the given limit:
$L = \lim_{x \to 0} \frac{1}{x^3} \int_0^x \frac{t \ln (1+t)}{t^4 + 4} \, \mathrm{d}t = \lim_{x \to 0} \frac{\int_0^x \frac{t \ln (1+t)}{t^4 + 4} \, \mathrm{d}t}{x^3}$

As $x \to 0$, the numerator $\int_0^0 \frac{t \ln (1+t)}{t^4 + 4} \, \mathrm{d}t = 0$ and the denominator $x^3 \to 0$. Since this is in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Differentiating the numerator with respect to $x$ using Leibniz's rule of differentiation under the integral sign:
$\frac{\mathrm{d}}{\mathrm{d}x} \left( \int_0^x \frac{t \ln (1+t)}{t^4 + 4} \, \mathrm{d}t \right) = \frac{x \ln (1+x)}{x^4 + 4} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(x) - 0 = \frac{x \ln (1+x)}{x^4 + 4}$

Differentiating the denominator with respect to $x$:
$\frac{\mathrm{d}}{\mathrm{d}x} (x^3) = 3x^2$

Applying L'Hôpital's Rule:
$L = \lim_{x \to 0} \frac{\frac{x \ln (1+x)}{x^4 + 4}}{3x^2}$

Simplifying the expression:
$L = \lim_{x \to 0} \frac{x \ln (1+x)}{3x^2 (x^4 + 4)} = \lim_{x \to 0} \frac{\ln (1+x)}{3x(x^4 + 4)}$

We can split the limit into known standard limits:
$L = \frac{1}{3} \cdot \left( \lim_{x \to 0} \frac{\ln (1+x)}{x} \right) \cdot \left( \lim_{x \to 0} \frac{1}{x^4 + 4} \right)$

Using the standard logarithmic limit $\lim_{x \to 0} \frac{\ln (1+x)}{x} = 1$:
$L = \frac{1}{3} \cdot 1 \cdot \frac{1}{0^4 + 4} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$

Thus, the correct option is B.