Question

Tentukan integral 8x (1-2x)³ dx dengan menggunakan rumus integral parsial
Yang tau tolong dijawab kak/bang :) thanks

Answer 1

Verified answer

INTEGRAL PARSIAL

∫ u dv = u v - ∫ v du

∫ (ax + b)^n dx = 1/(a(n + 1)) . (ax + b)^(n + 1) + C

∫ 8x (1 - 2x)^3 dx
Misal :
u = 8x
du = 8 dx

dv = (1 - 2x)^3 dx
v = 1/(-2(4)) . (1 - 2x)^4 = -1/8 (1 - 2x)^4

∫ 8x (1 - 2x)^3 dx = u . v - ∫ v du
= 8x . -1/8 (1 - 2x)^4 - ∫ -1/8 (1 - 2x)^4 . 8 dx
= -x (1 - 2x)^4 + ∫ (1 - 2x)^4 dx
= -x (1 - 2x)^4 + 1/(-2(5)) . (1 - 2x)^5 + C
= -x (1 - 2x)^4 - 1/10 (1 - 2x)^5 + C


bisa disederhanakan lagi menjadi
= -x (1 - 2x)^4 - 1/10 (1 - 2x)^5 + C
= (1 - 2x)^4 [-x - 1/10 (1 - 2x)] + C
= (1 - 2x)^4 [(-10x - 1(1 - 2x)]/10 + C
= (1 - 2x)^4 (-10x - 1 + 2x)/10 + C
= (1 - 2x)^4 (-8x - 1)/10 + C
= -1/10 . (8x + 1) (1 - 2x)^4 + C