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 !