PARTIAL FRACTIONS CUBIC DENOMINATOR

Example 1 :

Resolve the following rational expressions into partial fractions.

x/(x-1)3

Solution :

By decomposing the denominator, we get

x/(x-1)3  =  (A(x - 1)2 + B(x - 1) + C)/(x-1)3

 =  A(x - 1)2 + B(x - 1) + C

If x=1

1  =  C

C  =  1

If x = 0

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

0 = A - B + 1

A - B  =  -1 ----(1)

If x = 2

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

2 = A + B + 1

A + B  =  1 ----(2)

(1) + (2)

A - B + A + B  =  -1 + 1

2A  =  0

A  =  0

Applying the values of A in the first equation, we get

0 - B  =  -1

B  =  1

Hence the solution is

Example 2 :

Resolve the following rational expressions into partial fractions.

1/(x4 - 1)

Solution :

By decomposing the denominator, we get

1 = A(x - 1)(x2 + 1) + B(x+1) (x2+1) + (Cx+D)(x+1)(x-1)

If x  =  1

1  =  B(2)(2)

4B  =  1

B  =  1/4

If x  =  -1

1 = A(-2)(2)

-4A  =  1

A  =  -1/4

If x  =  0

1 = A + B - D

1 = (1/4) + (1/4) - D

1  =  (2/4) - D

D  =  (1/2) - 1

D  =  -1/2

Hence the solution is

Example 3 :

Resolve the following rational expressions into partial fractions.

(x - 1)2/(x3 + x)

Solution :

By decomposing the denominator, we get

(x - 1)2  =  A(x2 + 1) + (Bx + C) x 

If x  =  0

1  =  A

If x  =  1

0  =  2A + B + C

0  =  2 + B + C

B + C  =  -2 ---(1)

If x  =  -1

4  =  2A + (-B + C)(-1)

4  =  2A + B - C

4  =  2 + B - C

2  =  B - C  ---(2)

(1) + (2)

-2 + 2   =  B + C + B - C

0  =  2B

B  =  0

By applying the value of B in the first equation, we get

C  =  -2

Hence the solution is

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