Let $\mathcal{I}$ denote the value of the triple integral,
$$\mathcal{I}:=\int_{0}^{4}\mathrm{d}x\int_{x}^{4}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,\frac{6}{1+48z-z^{3}}.$$
Having a cubic function with non-nice roots in the denominator of the integrand usually guarantees you an algebraically induced headache if you attempt to solve the integral by brute force. Since this is a homework problem, and assuming your professor isn't an outright sadist, you can expect there to be some trick to solving the integral with a lot less effort.
Sure enough, by changing the order of integration, we can easily reduce the triple integral to a single-variable integral:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{4}\mathrm{d}x\int_{x}^{4}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,\frac{6}{1+48z-z^{3}}\\
&=\int_{0}^{4}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\int_{0}^{y}\mathrm{d}z\,\frac{6}{1+48z-z^{3}}\\
&=\int_{0}^{4}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\int_{0}^{y}\mathrm{d}x\,\frac{6}{1+48z-z^{3}}\\
&=\int_{0}^{4}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,\frac{6y}{1+48z-z^{3}}\\
&=\int_{0}^{4}\mathrm{d}z\int_{z}^{4}\mathrm{d}y\,\frac{6y}{1+48z-z^{3}}\\
&=\int_{0}^{4}\mathrm{d}z\,\frac{48-3z^{2}}{1+48z-z^{3}}\\
&=\int_{1}^{129}\mathrm{d}u\,\frac{1}{u};~~~\small{1+48z-z^{3}=u}\\
&=\ln{\left(129\right)}.\blacksquare\\
\end{align}$$
