Answer
$4x^4(\ln x)^2-2x^4\ln x+\dfrac{x^4}{2}+C$
Work Step by Step
Let $I=\int16x^3 \space (\ln x)^2dx=16[\dfrac{x^4(\ln x)^2}{4}-\dfrac{2}{4}\int x^3\ln x \space dx]\\=4x^4(\ln x)^2-8\int x^3\ln x \space dx \\=4x^4(\ln x)^2-8 [\dfrac{x^4\ln x}{4}-(1/4) \int x^3dx \\=4x^4(\ln x)^2-2x^4\ln x+2\int x^3\space dx \\=4x^4(\ln x)^2-2x^4\ln x+\dfrac{x^4}{2}+C$
