Product rule with a composite factor

Let $u(x) = (x^2+1)^2$ and $v(x) = x-1$. By the chain rule, $u'(x) = 2(x^2+1)\cdot 2x = 4x(x^2+1)$, and clearly $v'(x) = 1$.

By the product rule, $f'(x) = u'(x)v(x) + u(x)v'(x) = 4x(x^2+1)(x-1) + (x^2+1)^2$. Factoring out $(x^2+1)$ gives $f'(x) = (x^2+1)\left[4x(x-1) + (x^2+1)\right] = (x^2+1)(5x^2 - 4x + 1)$.