f'(×)= (3×+1) (×-1) dan f(-1)=6

Intergral

f(x) = ∫ F'(x) dx
f(x)= ∫(3x +1)(x-1) dx = ∫ 3x² - 2x - 1 dx
f (x)= x³ - x² - x + c
f(-1)= 6
(-1)³ -(-1) -(-1) + c = 6
-1 + 1 + 1 + c = 6
c = 5
f(x)= x³ - x²- x + 5