Reading that as
[tex]F(x)=\displaystyle\int_1^{3x}(\ln t)^2\,\mathrm dt[/tex]
By the fundamental theorem of calculus, the first derivative is
[tex]F'(x)=3(\ln(3x))^2[/tex]
and so the second derivative would be
[tex]F''(x)=\dfrac{6\ln(3x)}x[/tex]
which means
[tex]F''(1)=6\ln3[/tex]
