Diberikan integral ((1-x)/(1-x(^1/4)) dx = x + 2/K x^(3/2)

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Pembahasan

Untuk menyelesaikan integral ((1-x)/(1-x^(1/4))) dx, kita bisa menggunakan substitusi. Misalkan u = x^(1/4), maka u^4 = x dan 4u^3 du = dx.
Integral menjadi: integral ((1-u^4)/(1-u)) * 4u^3 du
Kita bisa memfaktorkan (1-u^4) sebagai (1-u^2)(1+u^2) = (1-u)(1+u)(1+u^2).
Maka, integral menjadi: integral ((1-u)(1+u)(1+u^2) / (1-u)) * 4u^3 du
= integral (1+u)(1+u^2) * 4u^3 du
= integral (u+u^3+u^2+u^5) * 4u^3 du
= integral (4u^7 + 4u^6 + 4u^5 + 4u^4) du
= (4/8)u^8 + (4/7)u^7 + (4/6)u^6 + (4/5)u^5 + C
= (1/2)u^8 + (4/7)u^7 + (2/3)u^6 + (4/5)u^5 + C

Substitusikan kembali u = x^(1/4):
= (1/2)(x^(1/4))^8 + (4/7)(x^(1/4))^7 + (2/3)(x^(1/4))^6 + (4/5)(x^(1/4))^5 + C
= (1/2)x^2 + (4/7)x^(7/4) + (2/3)x^(3/2) + (4/5)x^(5/4) + C

Bandingkan dengan bentuk yang diberikan: x + 2/K x^(3/2) + 4/L x^(5/4) + 4/M x^(7/4) + C
Terjadi ketidaksesuaian dalam koefisien dan pangkat x. Mari kita periksa kembali soalnya, atau coba metode lain.

Metode lain: Pembagian Polinomial.
Misalkan y = x^(1/4), maka x = y^4.
Integral menjadi: integral ((1-y^4)/(1-y)) * 4y^3 dy
(1-y^4)/(1-y) = (1-y)(1+y)(1+y^2) / (1-y) = (1+y)(1+y^2) = 1 + y + y^2 + y^3

Integral menjadi: integral (1 + y + y^2 + y^3) * 4y^3 dy
= integral (4y^3 + 4y^4 + 4y^5 + 4y^6) dy
= (4/4)y^4 + (4/5)y^5 + (4/6)y^6 + (4/7)y^7 + C
= y^4 + (4/5)y^5 + (2/3)y^6 + (4/7)y^7 + C

Substitusikan kembali y = x^(1/4):
= x + (4/5)x^(5/4) + (2/3)x^(6/4) + (4/7)x^(7/4) + C
= x + (4/5)x^(5/4) + (2/3)x^(3/2) + (4/7)x^(7/4) + C

Bandingkan dengan bentuk yang diberikan: x + 2/K x^(3/2) + 4/L x^(5/4) + 4/M x^(7/4) + C

Dari perbandingan:
Koefisien x^(3/2): 2/K = 2/3 => K = 3
Koefisien x^(5/4): 4/L = 4/5 => L = 5
Koefisien x^(7/4): 4/M = 4/7 => M = 7

Maka, K + L + M = 3 + 5 + 7 = 15.