Question

∫(-6x^3+9x^2+4x-3)dx​

Answer 1

$∫(-6x^3+9x^2+4x-3)dx$

$= > ∫(-6x^3)dx+ ∫ (9x^2)dx+ ∫ (4x)dx- ∫ (3)dx$

$= > - 6 ∫(x^3)dx+9 ∫ (x^2)dx+ 4∫ (x)dx-3 ∫ (1)dx$

$= > - 6( \frac{ {x}^{4} }{4} ) + 9( \frac{ {x}^{3} }{3} ) + 4( \frac{ {x}^{2} }{2}) - 3x + c$

$= > \frac{ - 3 {x}^{4} }{2} + 3 {x}^{3} + 2 {x}^{2} - 3x + c$

Answer 2

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Step-by-step explanation:

hope it helps you ✌️✌️