To do the division we insert all the necessary
zero place holders for all the missing powers
of x.  It isn't necessary to put in the 1
coefficients but I thought I'd do it anyway:

        
                1x³+0x²+1x+0
1x²+0x-1)1x⁵+0x⁴+0x³+0x²+0x+2
         1x⁵+0x⁴-1x³
             0x⁴+1x³+0x²
             0x⁴+0x³+0x²
                 1x³+0x²+0x
                 1x³+0x²-1x
                     0x²+1x+2
                     0x²+0x+0
                         1x+2


So 



Now we break the fraction part into partial fractions:

First factor the denominator:



Then



Multiply through by the LCD = (x-1)(x+1)



This must be true for all values of x.

So substitute x=1 to make the last term on the right
become 0: 







Substitute x=-1 to make the first term on the right
become 0:







So 



becomes:



Simplify those by multiplying tops and bottoms by 2



Substituting in

 

    

Edwin