PARTIAL FRACTIONS DECOMPOSITION EXAMPLES

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)

 =  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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