Question 621023
<pre>
f(x) = {{{(x^2-1)/(x(x-2))}}} 

When the degree of the numerator is greater than or equal to the
degree of the denominator, the fraction must be divided out first
by long division:

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

           <u>           1</u>
x² - 2x + 0)x² - 0x - 1
            <u>x² - 2x + 0</u>
                 2x - 1

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

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

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

2x-1 = A(x-2) + Bx

This must be true for all values of x so we choose x=2
so the first term on the right will be 0

2(2)-1 = A(2-2) + B(2)
 4 - 1 = A(0) + 2B
     3 = 0 + 2B
     3 = 2B
     {{{3/2}}} = B

Now we choose x=0 so the second term on the right will be 0

2(0)-1 = A(0-2) + B(0)
   0-1 = A(-2) + 0
    -1 = -2A 
    {{{(-1)/(-2)}}} = A
    {{{1/2}}} = A

{{{A/x}}} + {{{B/(x-2)}}} = {{{expr((1/2)/x)}}} + {{{(3/2)/(x-2)}}} = {{{expr((1/2)/x)}}}·{{{2/2}}} + {{{(3/2)/(x-2)}}}·{{{2/2}}} = {{{1/(2x)}}} + {{{3/(2(x-2))}}}

Therefore f(x) = {{{(x^2-1)/(x(x-2))}}} = 1 + {{{1/(2x)}}} + {{{3/(2(x-2))}}}


Edwin</pre>