Final answer:
The integral ∫(37x⁵)(6x+1)(x-1) dx can be evaluated by first expanding the integrand into a polynomial and then integrating term by term, finally adding a constant of integration 'c'.
Step-by-step explanation:
To evaluate the integral ∫(37x⁵)(6x+1)(x-1) dx, we will follow a step-by-step approach. First, we need to expand the integrand. We can distribute 37x⁵ across (6x+1) and then across (x-1). Once the expression is fully expanded, we can integrate term by term.
The expanded form of the integrand is 37x⁵ multiplied by each term in the binomials, giving us a polynomial. After integrating this polynomial, we add the constant of integration, which we denote as 'c'.
In this case, the use of absolute values does not apply directly to the integration process, because the function we are integrating is a polynomial, which does not involve functions that result in different behavior over separate domains such as logarithmic or square root functions.