Let #I=int(-3x^2-5x+4)/(x^3-2x+1)dx=-int(3x^2+5x-4)/(x^3-2x+1)dx#.
Observe that, #d/dx(x^3-2x+1)=3x^2-2#, so,
#I=-int(3x^2-2)/(x^3-2x+1)dx-int(5x-2)/(x^3-2x+1)dx#
Using the Result #int(f'(x))/f(x)dx=ln|f(x)|+c#, we get the First
Integral, #-ln|x^3-2x+1|+c_1.#
Let #J=int(5x-2)/(x^3-2x+1)dx#
The sum of the co-effs. of #Dr.# poly. is #0#
#:. Dr.=(x-1)(x^2+x-1)#.
For #J#, we will have to use the Method of Partial Fraction . So, let,
#(5x-2)/{(x-1)(x^2+x-1)}=A/(x-1)+(Bx+C)/(x^2+x-1); A,B,C in RR.#
The const. #A# can be easily found by Heavyside's Method as
#A=[(5x-2)/(x^2+x-1)]_(x=1)=3/(1+1-1)=3,# &, so, by the last eqn.,
#(5x-2)/{(x-1)(x^2+x-1)}-3/(x-1)=(Bx+C)/(x^2+x-1),# i.e.,
#(5x-2-3x^2-3x+3)/{(x-1)(x^2+x-1)}=(Bx+C)/(x^2+x-1),# or,
#(-3x^2+2x+1)/{(x-1)(x^2+x-1)}={(x-1)(-3x-1)}/{(x-1)(x^2+x-1)}=(Bx+C)/(x^2+x-1).#
#Clearly, B=-3, C=-1," &, with, "A=3,# we have,
#J=int3/(x-1)dx-int(3x+1)/(x^2+x-1)dx#
#=3ln|x-1|-int{3/2(2x+1)-1/2}/(x^2+x-1)dx#
#=3ln|x-1|+3/2int{d/dx(x^2+x-1)}/(x^2+x-1)dx+1/2int1/{(x+1/2)^2-(sqrt5/2)^2}dx#
#=3ln|x-1|+3/2ln|x^2+x-1|+(1/2)(1/(2sqrt5/2))ln|(x+1/2-sqrt5/2)/(x+1/2+sqrt5/2)|#
Finally, #I=-ln|x^3-2x+1|-3ln|x-1|-3/2ln|x^2+x-1|
-1/(2sqrt5)ln|(2x+1-sqrt5)/(2x+1+sqrt5)|+K.#
