Let's try factoring out. We can call the denominator D

#D = x^6 - x^4 = (x^4)(x^2-1) = x^4(x-1)(x+1)#

The numerator N is #x^5 + 1#. It is zero for #x=-1# so it contains the factor #(x-(-1)) = x+1#

Namely #N=x^5 + 1 = (x+1)(x^4-x^3+x^2-x+1)#

Now we are ready to divide

#N/D = (x^4-x^3+x^2-x+1)1/(x^4)(x-1)#

Now if the last term #1# wasn't there on the numerator, we could divide in a straightforward way. So we rewrite #N# as

#N = N -1 + 1#

#N = x^3(x-1) + x(x-1) + 1#

and we obtain

#N/D = \frac{1}{x} + \frac{1}{x^3} + \frac{1}{x^4(x-1)}#

#N/D=\frac{1}{x} + \frac{1}{x^3} - \frac{x^3+x^2+x+1}{x^4}+\frac{1}{x-1}#

and in the end

#N/D = -\frac{1}{x^2} - \frac{1}{x^4} +\frac{1}{x-1}#

And

#I = \frac{1}{x} -\frac{1}{2x^2}+ log|x-1|#