Write #(2x-5)# as #(2x+2-7)#
Then the integral is
#I=int((2x-5)dx)/(x^2+2x+2)=int((2x+2-7)dx)/(x^2+2x+2)#
#=int((2x+2)dx)/(x^2+2x+2)-int(7dx)/(x^2+2+2)#
#=I_1+I_2#
Calculate the integral #I_1# by substitution
Let #u=x^2+2x+2#, #=>#, #du=(2x+2)dx#
#I_1=int((2x+2)dx)/(x^2+2x+2)=int(du)/u#
#=ln(u)#
#=ln(x^2+2x+2)#
For the integral #I_2#, complete the square of the denominator
#x^2+2x+2=x^2+2x+1+1=(x+1)^2+1#
Let #u=x+1#, #=>#, #du=dx#
#I_2=int(7dx)/(x^2+2+2)=7int(dx)/((x+1)^2+1)#
#=7int(du)/(u^2+1)#
#=7arctan(u)#
#=7arctan(x+1)#
And finally,
#I=ln(x^2+2x+2)-7arctan(x+1)+C#
