# PARTIAL FRACTIONS EXAMPLES

Partial Fractions Examples :

Here we are going to see some example problems on partial fractions.

## Partial Fractions Examples

Example 1 :

Resolve into partial fractions. Solution :

1/(x - 1) (x + 1)  =  A/(x - 1) + B/(x + 1)

1/(x - 1) (x + 1)  =  A(x + 1) + B(x - 1)/(x - 1)(x + 1)

Since we have same denominators on both side, we can equate the numerators.

1  =  A(x + 1) + B(x - 1)  ------(1)

The constants A and B can also be found by successively giving suitable values for x.

To find A, put x = 1 in (1)

To find B, put x = -1 in (1)

 1  =  A(1 + 1) + B(1 - 1)1 = 2A + B (0)1 = 2A + 02A  =  1  ===> A = 1/2 1  =  A(-1 + 1) + B(-1 - 1)1 = A(0) + B (-2)1 = 0 - 2B-2B  =  1  ===> B = -1/2 Example 2 :

Resolve into partial fractions. Solution :

7x - 1/(x2-5x + 6)  =  7x - 1/(x - 3) (x - 2)

7x - 1/(x - 3) (x - 2)  = A/(x - 3)  + B/(x - 2)

7x - 1/(x - 3) (x - 2)  = A(x - 2) + B(x - 3)/(x - 3)(x - 2)

Since we have same denominators on both side, we can equate the numerators.

7x - 1  =  A(x - 2) + B(x - 3)  ------(1)

The constants A and B can also be found by successively giving suitable values for x.

To find A, put x = 3 in (1)

To find B, put x = 2 in (1)

 7(3) - 1 = A(3-2)+B(3-3)21 - 1 = 1A + B (0)20 = A + 0 A = 20 7(2) - 1 = A(2-2)+B(2-3)14 - 1 = A(0) + B (-1)13 = 0 - B B = -13 Example 3 :

Resolve into partial fractions. Solution :

(x2+ x + 1)/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)

Taking L.C.M

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

Since we have same denominators on both side, we can equate the numerators.

x2+ x + 1  =  A(x-2)(x-3) + B(x-1)(x-3) + C(x-2)(x-1) ----(1)

The constants A, B and C can also be found by successively giving suitable values for x.

To find A, put x = 1 in (1)

To find B, put x = 2 in (1)

To find C, put x = 3 in (1)

 If we put x = 1, both B and C will become zero. 12+ 1 + 1  =  A(1-2)(1-3)3  =  A(-1)(-2)2A = 3  ==> A = 3/2 If we put x = 2, both A and C will become zero. 22+ 2 + 1  =  B(2-1)(2-3)4 + 2 +1  =  B(1)(-1)7= -B  ==> B = -7 If we put x = 3, both C and B will become zero. 32+ 3 + 1  =  C(3-2)(3-1) 9 + 3 + 1  =  C(1)(2)13 = 2C  ==> C = 13/2 Example 4 :

Resolve into partial fractions. Solution :

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

Taking L.C.M

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

Since we have same denominators on both side, we can equate the numerators.

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

By equating the coefficients of x2, x and constant terms respectively, we get

A + B = 0  -------(2)

4A + B + C  = 0   -------(3)

4A - 2B - C  =  1  -------(4)

By adding the equations (3) and (4) we get,

00000000000004A + B + C  = 000000000000000000000

00000000000004A - 2B - C  =  10000000000000000000

000000000000-------------------0000000000000000000

0000000000008A - B  =  1------(5)00000000000000000

(5) + (2)

A + B + 8A - B = 0 + 1

9A  =  1 ==> A = 1/9

 Substituting the value of A in (2), we get1/9 + B  = 0B  =  -1/9 Substituting the values of A and B in (3), we get4(1/9) + (-1/9) + C  = 04/9 - 1/9 + C = 0(4 - 1 + 9C)/9 = 03 + 9C = 09C = -3C  =  -3/9 ==> -1/3 Example 5 :

Resolve into partial fractions. Solution :

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

Taking L.C.M

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

Since we have same denominators on both side, we can equate the numerators.

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

By equating the coefficients of x2, x and constant terms respectively, we get

A + B = 0  -------(2)

-2A + B + C  = 1   -------(3)

A - 2B + 2C  =  -2  -------(4)

2 x (3) - (4)

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

-4A - A + 2B + 2B + 2C - 2C  = 2 + 2

-5A + 4B  =  4  -------(5)

5 x (2) + (5)

5A + 5B + (-5A + 4B)  = 0 + 4

5A + 5B -5A + 4B  = 4

9B  =  4  ==> B = 4/9

 Substituting the value of B in (2), we getA + 4/9  =  0A  =  -4/9 Substituting the values of A and B in (3), we get-2(-4/9) + 4/9 + C  = 18/9 + 4/9 + C  =  1(8 + 4 + 9C)/9 =  112 + 9C  =  99C  =  -3C  =  -1/3  After having gone through the stuff given above, we hope that the students would have understood, how to decompose a rational expression into partial fractions.

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