Find: the integral (x^2/4 - 4) dx
I know that the answer is x^3/12 -4 +c but I do not know how to get this
o I think I know. Is it because you add 1 to the x^2 which is x^3 and the inverse 1/3 you multiply by 4 to get 12
(1/4) integral(x^2-16) dx
= (1/4) [ integral x^2 dx -integral dx]
= (1/4 [ x^3/3 - 16 x + C]
= x^3/12 - 4 x + c