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.
Sx-3(x+1)dx
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