= Solution
Put $x=h\nu/(k_BT)=hc/(\lambda k_BT)$. In the long-<wavelength> limit $x\ll1$, the <Taylor expansion> $e^x-1=x+O(x^2)$ in <Planck's law> gives the <Rayleigh-Jeans law>,
$$
\boxed{U_\nu\simeq\frac{8\pi k_BT}{c^3}\nu^2.}
$$
This is the classical <equipartition theorem> result: each electromagnetic mode carries mean <energy> $k_BT$. Extending it to arbitrarily high <frequency> would make the total <energy density> diverge, the <ultraviolet catastrophe>.
In the short-<wavelength> limit $x\gg1$, $e^x-1\simeq e^x$, giving the <Wien approximation>,
$$
\boxed{U_\nu\simeq\frac{8\pi h}{c^3}\nu^3e^{-h\nu/(k_BT)}.}
$$
Large <photon energies> are thermally suppressed, removing the classical high-<frequency> divergence.
If the requested “form” is expressed per unit <wavelength>, the change of variables includes its <Jacobian determinant>: $U_\lambda=U_\nu|d\nu/d\lambda|=U_\nu c/\lambda^2$. Thus
$$
U_\lambda=\frac{8\pi hc}{\lambda^5}\frac1{e^{hc/(\lambda k_BT)}-1},\qquad
\boxed{U_\lambda\simeq\begin{cases}8\pi k_BT/\lambda^4,&x\ll1,\\8\pi hc\,e^{-x}/\lambda^5,&x\gg1.\end{cases}}
$$
Substituting $\nu=c/\lambda$ without the <Jacobian determinant> would give an incorrect spectral density.
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