integral subtitusi dari∫ x (x+4)5 dx = .... a. 1/21 (3x+26)(x+4)^6 +C b. 1/21 (3x- 14)(x+4)^6 +C c. 1/21 (3x-10)(x+4)^6 +C d. 1/21 (3x+2)(x+4)^6 +C e. 1/21 (3x-
Matematika
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Pertanyaan
integral subtitusi dari∫ x (x+4)5 dx = ....
a. 1/21 (3x+26)(x+4)^6 +C
b. 1/21 (3x- 14)(x+4)^6 +C
c. 1/21 (3x-10)(x+4)^6 +C
d. 1/21 (3x+2)(x+4)^6 +C
e. 1/21 (3x-2)(x+4)^6 +C
a. 1/21 (3x+26)(x+4)^6 +C
b. 1/21 (3x- 14)(x+4)^6 +C
c. 1/21 (3x-10)(x+4)^6 +C
d. 1/21 (3x+2)(x+4)^6 +C
e. 1/21 (3x-2)(x+4)^6 +C
1 Jawaban
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1. Jawaban Anonyme
jawab
∫ x (x + 4)⁵ dx
substitusi
u = x+ 4 → x = u - 4
du = dx
∫ x (x + 4)⁵ dx = ∫ (u-4)(u)⁵ du
= ∫ u⁶ - 4u⁵ du
= 1/7 u⁷ - 4/6 u⁶ + c
= 1/7 (x+4)⁷ - 2/3 (x+4)⁶ + c
= 1/21 (x+4)⁶ { 3(x+4) - 14} + c
= 1/21 (x+4)⁶ (3x +12 - 14) + c
= 1/21 (3x -2 )(x+4)⁶ + c