Taking a Dip

The substitution $t=x+n$ gives us $\begin{aligned} I(2n)&=\int_{-n}^n\ln(n-x)\ln(n+x)\ dx \\&=\int_0^{2n}\ln(2n-t)\ln(t)\ dt. \end{aligned}$ Using the solution of the problem Dive , we see that the minimum value of the given integral is $I\left(e^\frac\pi{\sqrt6}\right)=2\left(1-\dfrac\pi{\sqrt6}\right)e^\frac\pi{\sqrt6},$ making the answer $6+1=\boxed7$ .