For a bosonic degree of freedom, the supplied integral gives
$$
I_B=\int_0^\infty\frac{y^3}{e^y-1}\,dy=\frac{\pi^4}{15}.
$$
For a fermionic degree of freedom, use
$$
\frac1{e^y+1}=\frac1{e^y-1}-\frac2{e^{2y}-1}.
$$
Changing variable $u=2y$ in the second integral gives
$$
I_F=I_B-\frac18I_B=\frac78I_B.
$$
This is the energy-density case of the <fermion-to-boson thermal integral ratio>.

Summing the result of part (a) over all particle species therefore gives
$$
\epsilon
=\frac{4\pi g_*}{(hc)^3}(kT)^4\frac{\pi^4}{15}
=\boxed{\frac{4\pi^5}{15(hc)^3}g_*(kT)^4},
$$
where the <effective number of relativistic energy degrees of freedom> is
$$
\boxed{g_*=\sum_{\rm bosons}g_i+\frac78\sum_{\rm fermions}g_i.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 9B / b / Solution

{title2=$g_*$}

For relativistic species in thermal equilibrium at a common temperature $T$, the total energy density can be written
$$
\epsilon=\frac{4\pi^5}{15(hc)^3}g_*(k_BT)^4,
\qquad
g_*=\sum_{\rm bosons}g_i+\frac78\sum_{\rm fermions}g_i.
$$
The factor $7/8$ is the <fermion-to-boson thermal integral ratio> for energy density.

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 Incoming links: Effective number of relativistic energy degrees of freedom