#I=int(-2x^3-2x^2+6x+9)/(2x^2-x+3)dx#
#=-int(x^3+x^2-3x-9/2)/(x^2-1/2x+3/2)dx#
#=-int(x(x^2-1/2x+3/2)+3/2x^2-9/2x-9/2)/(x^2-1/2x+3/2)dx#
#=-int(x(x^2-1/2x+3/2)+3/2(x^2-1/2x+3/2)-15/4x-27/4)/(x^2-1/2x+3/2)dx#
#=-int(x+3/2-3/4*(5x+9)/(x^2-1/2x+3/2))dx#
#=-intxdx-3/2int1dx+15/4intx/((x-1/4)^2+23/16)dx+27/4int1/((x-1/4)^2+23/16)dx#
Let #x-1/4=sqrt(23/16)tan(theta)#
#dx=sqrt(23/16)sec(theta)^2d theta#
So :
#I=-x^2/2-3/2x+60/23int(sqrt(23/16)tan(theta)+1/4)/((tan(theta)^2+1))*sqrt(23/16)sec(theta)^2d theta+101/23int1/(tan(theta)^2+1)*sqrt(23/16)sec(theta)^2d theta#
Because #tan(theta)^2+1=sec(theta)^2#,
#=-x^2/2-3/2x+60/23sqrt(23/16)int(23/16tan(theta)+1/4)d theta+101/23sqrt(23/16)int1d theta#
#=-x^2/2-3/2x+15/4sqrt(23/16)inttan(theta)d theta+15/23sqrt(23/16)int1d theta+101/23sqrt(23/16)theta#
We have #inttan(theta)d theta=-ln(|cos(theta)|)#, you can see the proof here.
#=-x^2/2-3/2x-(25sqrt23)/16ln(|cos(theta)|)+(15sqrt23)/92theta+(101sqrt23)/92theta#
Because #theta=tan^(-1)(4/sqrt23(x-1/4))#, and #cos(Arctan(x))=1/sqrt(x^2+1)# (you can see the proof here),
#I=-x^2/2-3/2x+(25sqrt23)/32ln(|16/23(x-1/4)^2+1|)+29/24sqrt23/92tan^(-1)(4/sqrt23(x-1/4))+C#, #C in RR#
\0/ Here's our answer !
