Integral x per akar x +4 dx?
Matematika
AdrianJr6
Pertanyaan
Integral x per akar x +4 dx?
1 Jawaban
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1. Jawaban Anonyme
klo dgn substitusi
u = (x+4)^1/2
u^2 = x+ 4
x = u^2 - 4
dx/du = 2u
dx = 2u du
int x/ (x+4)^1/2 dx = int {(u^2 -4)/u } (2u. du )
= int 2 (u^2 - 4) du
= 2 (1/3 u^3 - 4u) + c
= 2/3 u^3 - 8u
= 2/3 u( u^2 - 12)
= 2/3 (u^2 -12) (u)
= 2/3 (x+4-12) (x+4)^1/2
= 2/3 (x - 8)(x+4)^1/2
klo parsial
int x (x+4)^-1/2 dx
= 2x (x+4)^(1/2) - 4/3 (x+4)^(3/2)
= 2/3 (x+4)^(1/2) [ 3x - 2(x+4)]
= 2/3 (3x-2x-8)(x +4)^1/2
= 2/3 (x-8)(x + 4)^1/2