Tentukan integral (8 x3-3) dx

[tex] \int\limits {8 x^{3}-3 } \, dx \\ = \frac{8}{3+1} x^{3+1}- \frac{3}{1} x^{1}+C\\ = \frac{8}{4} x^{4}-3x+C\\ \bold{=2 x^{4}-3x+C} [/tex]

∫ (8x³ - 3) dx

∫ 8x³ dx - ∫ 3 dx

8x⁴/4 - 3x

= 2x⁴ - 3x + C

semoga jelas dan membantu ya