46: Applications of Partial Fractions

© Christine Crisp

“Teach A Level Maths”

Vol. 2: A2 Core Modules

There are 2 topics where partial fractions are useful.

• finding some binomial expansions

• differentiating some algebraic fractions

• integrating some algebraic fractions

Method:

We could write this as 11 )2()1)(8( xxxWe would then need to expand the 2 binomial expressions, finding 4 terms for each, and finally multiply all 3 sets of brackets !Using partial fractions means we only have to add or subtract 2 expansions.

e.g. 1 Find the expansion, in ascending powers of x, up to and including the term in of the following:

)2)(1(

8

xx

x

3x

x

B

x

A

xx

x

21)2)(1(

8Solution:

)1()2(8 xBxAx Multiply by :)2)(1( xx

:2x 2)3(6 BB

:1x 3)3(9 AA

So,

xxxx

x

2

2

1

3

)2)(1(

8

We expand the 2 fractions separately. We have:

1)1(31

3

xx

...12312

1)1(32

nx )1( nnnx x )2)(1( nnn x

Replace n by )1(

!! 3

)3)(2)(1(

2

)2)(1()1(13)1(3

321 xx

xx

3213 xxx

xxxx

x

2

2

1

3

)2)(1(

8

111

21)2(2)2(2

xx

x22

The 2nd fraction is 1)2(2 xFor the binomial we must have , so we take outside the brackets:

1)2( 1.)..1(

so we can save time by replacing x by in that.

2

x

The 1st fraction gave

321 1)1( xxxx

321

2221

21

xxxx

so,

8421

32 xxx

1

21

2

2

x

32 3333 xxx 842

132 xxx

842

132 xxx

1 1x

So,

xxxx

x

2

2

1

3

)2)(1(

8

1

21

x1)1(3 x

3213 xxx

1 1 2 22

xx

32

8

23

4

13

2

54 xxx

1 1x Both series must be valid so we need the most stringent condition which is

1 1for x

SUMMARYTo find a binomial expansion for an algebraic fraction with a denominator that factorises into linear factors:• Find the partial

fractions.• For each partial fraction, write the denominator with a negative power.

• Expand each part and write the values for which the series converges.

• Find the validity of the entire expansion by choosing the most stringent of the restrictions on x.

• Combine the expansions.

• For expressions of the form , remove from the brackets.

nbxa )( na

Exercises

1.x

B

x

A

xx 213)21)(3(

5

2.x

C

x

B

x

A

xx

xx

2)1(1)2()1(

411822

2

Express each of the following in the form given and hence obtain the expansions in ascending powers of x up to and including the term in .

2xGive the values of x for which each expansion is valid.

1.x

B

x

A

xx 213)21)(3(

5

Solutions:

The partial fractions are

xx 21

2

3

1

so we need 11 )21(2)3( xx

11

1 )21(23

1)3(

x

x

11

)21(23

13

1

x

x

2

9

1

3

11 xx

2421 xx

)421(29

1

3

11

3

1 22 xxxx

11

)21(23

13

1

x

xSo,

121 x 222

)2)(1(2)1(1 xx

!

1 1 3 33

xx

1 11 2 1

2 2x x

1

31

x2

32

)2)(1(

3)1(1

xx

!

3

1

1

31

x 2 121 x

22 84227

1

9

1

3

1xxxx

2

27

215

9

35

3

5xx

)421(29

1

3

11

3

1 22 xxxx

1 1

2 2x Valid for

2.x

C

x

B

x

A

xx

xx

2)1(1)2()1(

411822

2

The partial fractions are

xxx

2

2

)1(

1

1

22

121 )2(2)1()1(2 xxx1

121

21)2(2)1()1(2

xxx

121

21)1()1(2

xxx

!2

))(2)(1())(1(1)1(

21 x

xx

121

21)1()1(2

xxx

!2

))(3)(2())(2(1)1(

22 x

xx

2321 xx

By replacing x with in

2

x 21 1)1( xxx

21 xx

1

21

x2

221

xx 2

421

xx

1 1x

1 1x

x 1 1 2 x 2

2

121

21)1()1(2

xxx

2321 xx )1(2 2xx2

421

xx

421321222

222 xxxxxx

2

4

21

2

94 xx

1 x 1 For all 3 series to be valid we need

( 1 1 )x ( 2 2 )x ( 1 1 )2

x