Calculus: Early Transcendentals (2nd Edition)

It can be rewritten as $$\int_1^6(2x^3-4x)dx=\int_1^62x^3dx-\int_1^64xdx$$.
In the problem, it is asked only to rewrite the integral as the difference of two integrals, not to evaluate it. By the difference rule given in table 5.4 we have $$\int_1^6(2x^3-4x)dx=\int_1^62x^3dx-\int_1^64xdx.$$