From the given integrand, divide the numerator by the denominator. So that
#(x^3−2x^2+6x−3)/(−x^2+9x+2)=-x-7-(71x+11)/(x^2-9x-2)#
We have to simplify the third term, so that
#(71x+11)/(x^2-9x-2)=((71/2)(2x-9+9+(2*11)/71))/(x^2-9x-2)#
that is to force into the numerator a term #(2x-9)# which is the derivative of #(x^2-9x-2)#, enabling us to use #int((du)/u)=ln u+C#.
We can also use #int (du)/(u^2-a^2)=1/(2a)*ln((u-a)/(u+a))+C# afterwards
the continuation is
#(71x+11)/(x^2-9x-2)=((71/2)(2x-9+661/71))/(x^2-9x-2)#
#(71x+11)/(x^2-9x-2)=((71/2)(2x-9)+(71/2)(661/71))/(x^2-9x-2)#
Now, we are ready to integrate
#int(x^3−2x^2+6x−3)/(−x^2+9x+2)dx=int(-x-7-(71x+11)/(x^2-9x-2))dx#
#int(x^3−2x^2+6x−3)/(−x^2+9x+2)dx=#
#int(-x-7-((71/2)(2x-9)+(71/2)(661/71))/(x^2-9x-2))dx#
#int(-x-7-((71/2)(2x-9))/(x^2-9x-2)-((71/2)(661/71))/(x^2-9x-2))dx#
also by "completing the square"
#int(-x-7-((71/2)(2x-9))/(x^2-9x-2)-((661/2))/((x-9/2)^2-(sqrt89/2)^2))dx#
and
#-x^2/2-7x#
#-71/2*ln(x^2-9x-2)-(661sqrt(89))/178*ln((2x-9-sqrt89)/(2x-9+sqrt89))+C#
God bless....I hope the explanation is useful.
