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∫ (3x²+4-2)dx Can anyone walk me through this?
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the answer is 6x
if u take the derivative of 4dx - 2dx u'll get nothing
∫ (3x²+4-2)dx = ∫ (3x²+2)dx = 3(1/3)x^3 +2(x) +c = x^3 +2x +c
i doubt if your question is ∫ (3x²+4x-2)dx ...
gahh damnit i mistyped the problem. supposed to be +4x not +4. Sorry
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oh so its 6x+4
\[\int (3x²+4-2)dx =\frac{3x^{2+1}}{2+1}+\frac{4x^{1+1}}{1+1}-2x^{0+1}\] \ \[=x^3+2x^2-2x\]
∫ (3x²+4x-2)dx thats what the problem should look like
satellite73 helps you solve your problem :)
Thanks for the help everyone
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