Question 1043286
<pre>
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:

        
        <u>        1x³+0x²+1x+0</u>
1x²+0x-1)1x&#8309;+0x&#8308;+0x³+0x²+0x+2
         <u>1x&#8309;+0x&#8308;-1x³</u>
             0x&#8308;+1x³+0x²
             <u>0x&#8308;+0x³+0x²</u>
                 1x³+0x²+0x
                 <u>1x³+0x²-1x</u>
                     0x²+1x+2
                     <u>0x²+0x+0</u>
                         1x+2


So 

{{{(x^5+2)/(x^2-1)}}}{{{""=""}}}{{{x^3+x+(x+2)/(x^2-1)}}}

Now we break the fraction part into partial fractions:

First factor the denominator:

{{{(x+2)/(x^2-1)}}}{{{""=""}}}{{{(x+2)/((x-1)(x+1))}}}

Then

{{{(x+2)/((x-1)(x+1))}}}{{{""=""}}}{{{A/(x-1)+B/(x+1)}}}

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

{{{x+2}}}{{{""=""}}}{{{A(x+1)+B(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: 

{{{1+2}}}{{{""=""}}}{{{A(1+1)+B(1-1)}}}

{{{3}}}{{{""=""}}}{{{2A}}}

{{{3/2}}}{{{""=""}}}{{{A}}}

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

{{{-1+2}}}{{{""=""}}}{{{A(-1+1)+B(-1-1)}}}

{{{1}}}{{{""=""}}}{{{-2B}}}

{{{-1/2}}}{{{""=""}}}{{{B}}}

So 

{{{(x+2)/((x-1)(x+1))}}}{{{""=""}}}{{{A/(x-1)+B/(x+1)}}}

becomes:

{{{(x+2)/((x-1)(x+1))}}}{{{""=""}}}{{{"3/2"/(x-1)+"-1/2"/(x+1)}}}

Simplify those by multiplying tops and bottoms by 2

{{{(x+2)/((x-1)(x+1))}}}{{{""=""}}}{{{3/(2(x-1))-1/(2(x+1))}}}

Substituting in

{{{(x^5+2)/(x^2-1)}}}{{{""=""}}}{{{x^3+x+(x+2)/(x^2-1)}}} 

{{{(x^5+2)/(x^2-1)}}}{{{""=""}}}{{{x^3+x+3/(2(x-1))-1/(2(x+1))}}}    

Edwin</pre>