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Verified answer
Jwabintegralsubstitusi
1. ∫ x ( 3x² -5)³ dx
= (x / 6x) . (1/4) (3x² -5)⁴ + c
= 1/6 . 1/4. (3x² - 5)⁴ + c
= 1/24 (3x² - 5)⁴ + c
2. ∫ (6x² + 8x + 3) ( 2x³ + 4x² + 3x +1)⁷ dx
= (6x² +8x +3)/(6x² +8x +3) .(1/8) (2x³ +4x² + 3x+1)⁸+c
= (1)(1/8)(2x³ + 4x² + 3x+ 1)⁸ + c
= 1/8 (2x³ + 4x² + 3x+ 1)⁸ + c