Take
[tex]u=\ln x\implies\mathrm du=\dfrac{\mathrm dx}x[/tex]
[tex]\mathrm dv=4x^2\,\mathrm dx\implies v=\dfrac43x^3[/tex]
Then
[tex]\displaystyle\int4x^2\ln x\,\mathrm dx=\frac43x^3\ln x-\frac43\int x^2\,\mathrm dx=\frac43x^3\ln x-\frac49x^3+C[/tex]
[tex]=\boxed{\dfrac49x^3(3\ln x-1)+C}[/tex]
