Sx-3(x+1)dx

Answer 1

To solve the integral of s(x-3)(x+1)dx, we need to first expand the expression:

s(x-3)(x+1) = sx^2 - 3sx + sx - 3s

Now we can integrate each term separately:

∫sx^2 dx - ∫3sx dx + ∫sx dx - ∫3s dx

Integrating each term:

= (1/3)sx^3 - (3/2)sx^2 + (1/2)sx^2 - 3sx + C

= (1/3)sx^3 - (1/2)sx^2 - 3sx + C

Therefore, the integral of s(x-3)(x+1)dx is (1/3)sx^3 - (1/2)sx^2 - 3sx + C, where C is the constant of integration.