Example 1 :
Resolve the following rational expressions into partial fractions.
1/(x2-a2)
Solution :
Let us decompose the denominator into linear factors.
1/(x2-a2) = [A(x + a) + B(x - a)]/(x2-a2)
1 = A(x + a) + B(x - a)
When x = -a 1 = B(-a –a) 1 = B(-2a) B = -1/2a |
When x = a 1 = A(a +a) 1 = A(2a) A = 1/2a |
Hence the solution is
Example 2 :
Resolve the following rational expressions into partial fractions.
(3x + 1)/(x - 2) (x + 1)
Solution :
3x + 1 = A(x + 1) + B(x - 2)
When x = -1 3(-1) + 1 = B(-1 - 2) -3 + 1 = B(-3) -2 = -3B B = 2/3 |
When x = 2 3(2) + 1 = A(2 + 1) 6 + 1 = A(3) 7 = 3A A = 7/3 |
Hence the solution is
Example 3 :
Resolve the following rational expressions into partial fractions.
x/(x2 + 1)(x - 1)(x + 2)
Solution :
x = A(x+2)(x2+1) + B(x2+1)(x-1) + (Cx + D)(x-1)(x+2)
When x = 1 1 = A(3)(2) 1 = 6A A = 1/6 |
When x = -2 -2 = B(5)(-3) -2 = -15B B = 2/15 |
When x = 0
x = A(x+2)(x2+1) + B(x2+1)(x-1) + (Cx + D)(x-1)(x+2)
0 = A(2)(1) + B(1)(-1) + D(-1)(2)
0 = 2A - B - 2D
By applying the values of A and B, we get
0 = (1/3) - (2/15) - 2D
2D = 3/15
D = 1/10
When x = -1
-1 = A(1)(2) + B(2)(-2) + (-C+D)(-2)(1)
-1 = 2A - 4B + 2C - 2D
By applying the values of A, B and D
-1 = (1/3) - (8/15) + 2C - (1/5)
-1 = ((5 - 8 - 3)/15) + 2C
-1 = -6/15 + 2C
-1 + (2/5) = 2 C ==> -3/5 = 2C ==> C = -3/10
Hence the solution is
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