4. Given real numbers $x, y$ satisfy
$$
2 x=\ln (x+y-1)+\ln (x-y-1)+4 \text {. }
$$

Then the value of $2015 x^{2}+2016 y^{3}$ is $\qquad$
4.8 060 .

Let $u=x+y-1, v=x-y-1$.
Then $u+v-2=\ln u+\ln v$
$$
\Rightarrow(\ln u-u+1)+(\ln v-v+1)=0 \text {. }
$$

Let $f(x)=\ln x-x+1$. Then
$$
f^{\prime}(x)=\frac{1}{x}-1=\frac{1-x}{x} \text {. }
$$

Thus, $f(x)$ is increasing on the interval $(0,1)$ and decreasing on the interval $(1,+\infty)$.
Hence $f(x) \leqslant f(1)=\ln 1-1+1=0$, i.e., $f(x) \leqslant 0$,

with equality if and only if $x=1$.
From $f(u)+f(v)=0, f(u) \leqslant 0, f(v) \leqslant 0$, it follows that $u=v=1$ must hold.
Thus, $x=2, y=0$.
Therefore, $2015 x^{2}+2016 y^{3}=8060$.