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

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Answer 1 · 2017-11-18T17:54:22

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

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

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

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

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

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

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

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

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

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

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

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