Since #x^2-8x+17 = x^2 -2 xx 4 x+4^2+1 = (x-4)^2+1 # has no real zeros, it can not be factorized without using complex numbers. This is not a problems as far as evaluating the integral is concerned, since we can use the substitution #t=x-4# makes its evaluation rather simple:
#int 8/{x^2 - 8 x+17} dx = int 8/{(x-4)^2+1} dx = int 8/{1+t^2} dt = 8 tan^{-1} t+C = 8 tan^{-1} (x-4)+C#
If we insist on using partial fractions to reduce the integral further ( though I wouldn't particularly recommend it ), we can make use of
#x^2-8x+17 = (x-4+i)(x-4-i)#
so that
#1/{x-4-i}-1/{x-4+i} = {2i}/{x^2-8x+17}#
and thus
#int 8/{x^2 - 8 x+17} dx = 4/i int (1/{x-4-i}-1/{x-4+i} ) dx = 4/i (ln(x-4-i) -ln(x-4+i))+C' = 4/i ln({x-4-i }/{x-4+i})+C' = 4/i ln({-i}/i {1+i(x-4)}/{1-i(x-4)})+C'= 4/i ln( {1+i(x-4)}/{1-i(x-4)})+C'+4/i ln(-1)#
To simplify this further we have to resort to the polar form :
# 1 pm i(x-4)= sqrt{(x-4)^2+1} exp(pm i tan^{-1}(x-4))#
so that
# {1+i(x-4)}/{1-i(x-4)} = exp(2 i tan^{-1}(x-4)) #
and the integral becomes
#4/i 2i tan^{-1}(x-4) +C#
where #C = C'+4 pi #
