Evaluate: ∫(x-3)√x²+3x-18dx

I=∫(x-3)√x²+3x-18dx
Put x²+3x-18=t ⇒(2x+3)dx=dt
Let x-3=A(2x+3)+B
Comparing coefficients:
We get A=1/2, B=-9/2
I=∫1/2(2x+3)√x²+3x-18dx -9/2∫√x²+3x-18dx
=1/2∫√tdt -9/2∫√x²+3x+(3/2)²-18-(9/4)dx
=1/2∫√tdt -9/2∫√(x+3/2)²-(81/4)dx
=1/3(x²+3x-18)^(3/2) -9/2[ (x+3/2)√(x+3/2)²-(81/4)/2 - (81/8)ln|(x+3/2)+√(x+3/2)²-(81/4)| ] +C
=1/3(x²+3x-18)^(3/2) -(9/4)(x+3/2)√x²+3x-18 - (729/16)ln|(x+3/2)+√x²+3x-18| +C