We need to simplify the expression
The denominator is #x^2+x-6=(x+3)(x-2)#
Let's do a long division
#color(white)(aaaa)##2x^2+9x-6##color(white)(aaaa)##∣##x^2+x-6#
#color(white)(aaaa)##2x^2+2x-12##color(white)(aaaa)##∣##2#
#color(white)(aaaaaa)##0+7x+6#
#(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)#
We can now do the decomposition into partial fractions
#(2x+6)/(x^2+x-6)=(7x+6)/((x+3)(x-2))#
#=A/(x+3)+B/(x-2)#
#=(A(x-2)+B(x+3))/((x+3)(x-2))#
Therefore,
#7x+6=A(x-2)+B(x+3)#
Let #x=2#, #=>#. #20=5B#, #=>#, #B=4#
Let #x=-3#, #=>#, #-15=-5A#, #=>#, #A=3#
#(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)=2+3/(x+3)+4/(x-2)#
#int((2x^2+9x-6)dx)/(x^2+x-6)=int2dx+3intdx/(x+3)+4intdx/(x-2)#
#=2x+3ln(∣x+3∣)+4ln(∣x-2∣)+C#