Gimana caranya menyelesaikannya?

∫ (9x^2 + 2/(x^3) - 4) dx
= ∫ (9x^2 + 2x^-3 - 4) dx
= 9/3 x^3 + 2/(-2) x^-2 - 4x | batas -1 sampai 2
= 3x^3 - 1/(x^2) - 4x | batas - 1 sampai 2
= (3(2)^3 - 1/(2^2) - 4(2)) - (3(-1)^3 - 1/(-1)^2 - 4(-1))
= (24 - 1/4 - 8) - (-3 - 1 + 4)
= 16 - 1/4 - (0)
= 15 + 1 - 1/4
= 15 + 3/4
= 15 3/4

9/3x^2+1 + 2/-2x^-3+1 -4x
= 3x^3 - x ^-2 -4x
subs nilai x=2 kurangi dengan x=-1
[3(2)^3 - (1/2)² - 4(2)]- [3(-1)^3+ 1+4]
= 24-1/4 -8 - (-3-1+4)
= 15 3/4 -0
= 15 3/4 <E>
Smoga membantu :)