Using properties of the logarithm, you can rearrange the integrand as
[tex]\dfrac12\ln(2x^2)=\dfrac{\ln2+\ln x^2}2=\dfrac{\ln2+2\ln x}2=\ln\sqrt2+\ln x[/tex]
Then recall that the integral of [tex]\ln x[/tex] is
[tex]\displaystyle\int\ln x\,\mathrm dx=x(\ln x-1)+C[/tex]
(or you can find that out by integrating by parts) and so
[tex]\displaystyle\int\frac12\ln(2x^2)\,\mathrm dx=(\ln\sqrt2)x+x(\ln x-1)+C[/tex]
