Integral (10x3-1)(5x4-2x+9)^3 dx

pakai integral substitusi
misalkan
5x^4 - 2x + 9 = u
(20x^3 - 2) dx = du
2(10x^3 - 1) dx = du
dx = du / 2(10x^3 - 1)

(10x^3 - 1)(5x^4 - 2x + 9)^3 dx
ganti dengan u dan du
(10x^3 - 1) u^3 du / 2(10x^3 - 1)
= {(10x^3-1)/2(10x^3-1)} u^3 du
= (1/2) u^3 du
baru integral kan
= (1/2)(1/(3+1)) u^(3+1) + C
= (1/2)(1/4) u^4 + C
= (1/8) u^4 + C
kembalikan u ke x
= (1/8)(5x^4 - 2x + 9)^4 + C

Integral substitusi
∫ (10x³ - 1) (5x⁴ - 2x + 9)³ dx
u = 5x⁴ - 2x + 9
du / dx = 20x³ - 2 → dx = du / (20x³ - 2)
= ∫ (10x³ - 1) u³ du / (20x³ - 2)
= ∫ (10x³ - 1) u³ du / [2 (10x³ - 1)]
= 1/2 ∫ u³ du
= 1/2 (1/4 u⁴) + C
= 1/8 (5x⁴ - 2x + 9)⁴ + C