Pertanyaan
integral (8x-20)(x∧2-5x+3)∧4
Ditanyakan oleh: USER8914
34 Dilihat
34 Jawaban
Jawaban (34)
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= ∫ (8x - 20).(x² - 5x + 3)⁴ dx
misal :
u = x² - 5x + 3
du / dx = 2x - 5 → kalikan 4 → 4 du/dx = 8x - 20
maka :
= ∫ (8x - 20).(x² - 5x + 3)⁴ dx
= ∫ (4.du/dx).(u)⁴ dx
= 4.∫ u⁴ du
= 4. 1/5. u⁵ + c
= 4/5. u⁵ + c
= 4/5.(x² - 5x + 3)⁵ + c
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
)|(
FZA
= ∫ (8x - 20).(x² - 5x + 3)⁴ dx
misal :
u = x² - 5x + 3
du / dx = 2x - 5 → kalikan 4 → 4 du/dx = 8x - 20
maka :
= ∫ (8x - 20).(x² - 5x + 3)⁴ dx
= ∫ (4.du/dx).(u)⁴ dx
= 4.∫ u⁴ du
= 4. 1/5. u⁵ + c
= 4/5. u⁵ + c
= 4/5.(x² - 5x + 3)⁵ + c
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
)|(
FZA