Question

∫(-6x^3+9x^2+4x-3)dx

Answer 1

Answer:

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

❤

Answer 2

$∫(-6x^3+9x^2+4x-3)dx$

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

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

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

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

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