What is #int (-2x^3-x^2+6x+9 ) / (2x^2- x +3 )#?

Answer 1

Emily Ammerman

#int ((-2x^3-x^2+6x+9)*dx)/(2x^2-x+3)#

=#2Ln(2x^2-x+3)+(20sqrt23)/23*arctan[(4x-1)/sqrt23]-x^2/2-x+C#

#int ((-2x^3-x^2+6x+9)*dx)/(2x^2-x+3)#

#=-int ((2x^3+x^2-6x-9)*dx)/(2x^2-x+3)#

=-#int (x+1)*dx-int ((-8x-12)*dx)/(2x^2-x+3)#

=#-(x^2/2+x)+C+int ((8x+12)*dx)/(2x^2-x+3)#

=#-x^2/2-x+C+2*int ((4x+1)*dx)/(2x^2-x+3)+int (10*dx)/(2x^2-x+3)#

=#-x^2/2-x+C+2Ln(2x^2-x+3)+int (80*dx)/(16x^2-8x+24)#

=#2Ln(2x^2-x+3)-x^2/2-x+C+20*int (4*dx)/[(4x-1)^2+23]#

=#2Ln(2x^2-x+3)-x^2/2-x+C+(20sqrt23)/23*arctan[(4x-1)/sqrt23]#

=#2Ln(2x^2-x+3)+(20sqrt23)/23*arctan[(4x-1)/sqrt23]-x^2/2-x+C#

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Answer 2

Scarlett Allison

The integral of (\frac{-2x^3 - x^2 + 6x + 9}{2x^2 - x + 3}) with respect to (x) is (\frac{1}{4}x - \frac{3}{8}\ln(|2x^2 - x + 3|) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.