\(\frac{4x}{x-1}\times\frac{x^2-1}{3x-3}=(*)\ \ \ \ |use:\ a^2-b^2=(a-b)(a+b)\\\\(*)=\frac{4x}{x-1}\times\frac{x^2-1^2}{3(x-1)}=\frac{4x}{x-1}\times\frac{(x-1)(x+1)}{3(x-1)}=\frac{4x}{x-1}\times\frac{x+1}{3}=\frac{4x(x+1)}{3(x-1)}\\\\=\boxed{\frac{4x^2+4x}{3x-3}}\leftarrow answer\)