Question 637491
<pre>
We write the sum of all fractions with a denominator which is a
divisor of the denominator:

{{{(4x^2+2x-1)/(x^2(x+1))}}} = {{{A/x}}} + {{{B/x^2}}} + {{{C/(x+1)}}}

Clear of fractions:

4x² + 2x - 1 = Ax(x+1) + B(x+1) + Cx²

Substitute x=-1 to make the (x+1)'s 0

4(-1)² + 2(-1) - 1 = A(-1)(-1+1) + B(-1+1) + C(-1)²

4(1) - 2 - 1 = A(-1)(0) + B(0) + C(1)
   4 - 2 - 1 = 0 + 0 + C
           1 = C

Substitute x=0 to make the 1st and 3rd terms on the right 0:

4x² + 2x - 1 = Ax(x+1) + B(x+1) + Cx²

4(0)² + 2(0) - 1 = A(0)(0+1) + B(0+1) + C(0)

0 + 0 - 1 = A(0)(1) + B(1) + 0
       -1 = 0 + B
       -1 = B

Substitute B = -1 and C = 1

4x² + 2x - 1 = Ax(x+1) + (-1)(x+1) + (1)x²

4x² + 2x - 1 = Ax(x+1) - (x+1) + x²

Substitute x=1  (Note: we could use ANY number here that we haven't used):

4(1)² + 2(1) - 1 = A(1)(1+1) - (1+1) + 1²

 4 + 2 - 1 = 2A - 2 + 1
         5 = 2A - 1
         6 = 2A
         3 = A

So A = 3, B = -1, C = 1

Answer:

{{{(4x^2+2x-1)/(x^2(x+1))}}} = {{{3/x}}} - {{{1/x^2}}} + {{{1/(x+1)}}}


Edwin</pre>