no 5 nya dong ka bantu

^ tanda pangkat
(x^2/3 + ⅓x^6 + 7x) dx
= {1/(2/3+1)} x^(2/3+1) + (1/3)(1/7)x^7 + (7/2)x²
= (3/5)x^5/3 + (1/21)x^7 + (7/2)x²
= (3/5).³√x^5 + (1/21)x^7 + (7/2)x²
masukan 2 dan 1
= (3/5).³√2^5 + (1/21)2^7 + (7/2)2² - {(3/5).3√1^5 + (1/21)1^7 + (7/2)1²}
= (3/5)2.³√2² + (1/21)(128) + 14 - {(3/5)3 + (1/21) + (7/2)}
= (6/5).³√4 + (128/21) + 14 - {(9/5) + (1/21) + 7/2}
= (6/5).³√4 + (127/21) + 14 - (9/5) - (7/2)
itung ajah lanjutannya