1. We need to evaluate

$$L=\lim_{x\to 0}\frac{48}{x^4}\int_0^x \frac{t^3}{t^6+1}\,dt.$$

2. As $t\to 0$,

$$\frac{t^3}{t^6+1}\to t^3$$

because $t^6+1\to 1$. So near $t=0$, the integrand behaves like $t^3$.

Hence,

$$\int_0^x \frac{t^3}{t^6+1}\,dt \sim \int_0^x t^3\,dt = \frac{x^4}{4}.$$

Therefore,

$$L=\lim_{x\to 0}\frac{48}{x^4}\left(\frac{x^4}{4}\right)=48\cdot \frac14=12.$$

3. For a fully rigorous evaluation, we can also compute the integral exactly by substitution. Let

$$u=t^6+1 \implies d\nu=6t^5dt,$$

which is not directly convenient. Instead, observe the local expansion:

$$\frac{1}{1+t^6}=1-t^6+t^{12}-\cdots$$

so

$$\frac{t^3}{1+t^6}=t^3-t^9+t^{15}-\cdots$$

Integrating from $0$ to $x$,

$$\int_0^x \frac{t^3}{1+t^6}\,dt n=\frac{x^4}{4}-\frac{x^{10}}{10}+\frac{x^{16}}{16}-\cdots$$

Thus,

$$\frac{48}{x^4}\int_0^x \frac{t^3}{1+t^6}\,dt =48\left(\frac14-\frac{x^6}{10}+\frac{x^{12}}{16}-\cdots\right)\to 12.$$

Therefore, the required integer is

$$\boxed{12}.$$