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- a

X

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To any

A fraction of the form ; to any factor of the first degree, as x b, occurring twice in the denominator there correspond two

B

с partial fractions, and

If x b occurs three times, - 6

D there is an additional fraction,

and so on.

(x-7) quadratic factor x2+px+q there corresponds a partial fraction

Px+Q of the form

i if the factor xo+px +q occurs twice, x2 + px +?

Rx+S there is a second partial fraction,

and so on.

(x2+px +9) Here the quantities A, B, C, D, P, Q, R, S are all independent of x.

We shall make use of these results in the examples that follow.

5x - 11 Example 1. Separate

into partial fractions.

2x2+x-6 Since the denominator 2x2+x-6=(x+2)(2x-3), we assume

5x - 11 A B

+ 23c2+x-6

3+2 2.-3 where A and B are quantities independent of a whose values have to be determined. Clearing of fractions,

5x—11=A(2x – 3) + B(x+2). Since this equation is identically true, we may equate coefficients of like powers of Q; thus

2A+B=5, -3A+2B=-11; whence

A=3, B=-1. 5x – 11 3

1 2x2+x--6 +2 2x - 3

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ma+n
Example 2. Resolve

into partial fractions.
(-a) (x+b)
mx+n

A B
Assume

+
(2-a) (x+b) X-a X+ b
.. mx+1=A(x+6) + B(xa)

.(1). We might now equate coefficients and find the values of A and B, but it is simpler to proceed in the following manner :

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X-a

.(1);

3+3

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Since A and B are independent of x, we may give to w any value we please. In (1) put -a=0, or x=a; then

A=ma+n;

atb

mb-n putting a+b=0, or x=-6, B=

a+b ma+n

1 ma+n mb-n

+ (x-a)(x+b) a+b

2+

23x – 11x2 Example 3. Resolve

into partial fractions.
(2x - 1)(9—32)
23x – 11x2

A B C
Assume

+ + (2x— 1)(3+2)(3 — ) 2x-1

3 ... 23x – 11x2=A(3+x)(3-2) +B(2x − 1)(3-X)

+ C(2:— 1)(3+2). By equating the coefficients of like powers of x, or putting in succession 2x-1=0, 3+3=0, 3—3=0, we find that

A=1, B=4, C=-1. 23x – 11x2

1 4 1

+
(2x - 1)(9—32) 22 - 1 3+x 3-

3x2+x-2 Example 4. Resolve

into partial fractions.
(x - 2)2 (1–2x)
3x2 + 3—2 A B

С
Assume

+ + (2-2)2 (1-2x) 1-23

(2-2) :: 3x2+x—2=A(2—2)2+B(1–2x)(x-2) + C(1–2x). Let 1-2x=0, then

;

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let 3—2=0, then

C=-4.
To find B, equate the coefficients of x2; thus

5
3=A-2B; whence B=-

3
3x2+x-2

1
5

4
(oc — 2)2 (1–2x) 3(1-23) 3(3-2)(-2)2

42-19 Example 5. Resolve

into partial fractions. (2c2+1)(x-4)

42 – 19x
Assume

Ax+ B с
+

i
(22+1)(C-4) 2c2+1
:. 42-193=(Ax+B)(x-4) + C(2+1).

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Let x=4, then

C= -2; equating coefficients of x2, 0=A+C, and A=2; equating the absolute terms, 42=-4B+C, and B=-11,

42 - 19x 2x – 11 2

(2c2+1)(2—4) 22+1 2-4 464. In all the preceding examples the numerator has been of lower dimensions than the denominator; if this is not the case, we divide the numerator by the denominator until a remainder is obtained which is of lower dimensions than the denominator.

6x3 +522—7 Example. Resolve

into partial fractions.

3x2 – 2x – 1
By division,
6x3 + 5x2 — 7

8x -4
=2x+3+
3x2–2x1

3x2–2x-1
8x -4

5

1 and

+ ; 3x2 - 2x - 1 3x+1 -1 6x3 +5X2—7

5 1 2x+3+

+ 3x2-25-1

3x+1

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465. We shall now explain how resolution into partial fractions may be used to facilitate the expansion of a rational fraction in ascending powers of x.

3x2+x-2 Example. Find the general term of

when ex

(x - 2)2 (1–2x)
panded in a series of ascending powers of a.
By Ex. 4, Art. 463, we have
3x2+x-2

1
5

4
(x-2)(1-22) 3(1-2x) 3(x-2) (2-2)2

1
5

4
+
3(1-2x). 3(2-x) (2—2)2

--1-2) +(1-1)-(-1)

Hence the (r+1)th or general term of the expansion is

2 5 1

+

466. The following example sufficiently illustrates the method to be pursued when the denominator contains a quadratic factor.

;

7 + 3 Example. Expand

ascending powers of a.

(1 + 2) (1 + x2) and find the general term.

7 + 20

A BX + c
Assume

+
(1 + 2) (1 + x2)

1+30

1+ 22
.:7 + = A(1 + 2x2)+(Bx + C)(1+x).
Let 1+2= 0,

then A= 3;
equating the absolute terms, 7 = A + C, whence C = 4;
equating the coefficients of x2, 0 A + B, whence B = - 3.
7 + 3

3 4 - 3x

+
(1 + 2) (1 + x2) 1 + x 1 + x2

= 3(1 + 3)-1 + (4 - 3x)(1+ 22)-1
= 3{1 – X + x2 +(-1)PWP +...}

+(4 – 3x) {1 – 2 + + - ......+(-1)px2p+...}.
To find the coefficient of **:
(1) If is even, the coefficient of 2 in the second series

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is 4(-1); therefore in the expansion the coefficient of 2* is 3 +4(-1).

(2) If r is odd, the coefficient of an in the second series is – 3(- 1)?, and the required coefficient is 3(-1) ? – 3.

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r+1

EXAMPLES XLIV. d.

Resolve into partial fractions :

1.

7 2 - 1

46 + 13x

2.
1 – 5x + 6x2 12 2 – 112 – 15

22 – 10x + 13
4.

5. (3C — 1) (22 – 5x + 6)

9 6.

7. (x - 1) (3 + 2)2

1+ 3x + 2x2 3.

(1 – 2 x)(1 – 22) 2x3 + x2 – 3 – 3 X(3 – 1) (2x + 3) 24 – 3 x3 – 3 22 + 10

( + 1)(3-3)

8.

26 x2 + 208 x
(x2 + 1)(a + 5)
3 x3 – 8 x2 + 10

(2 — 114

2 22 – 11x + 5 (x – 3)(x2 + 2 x 5) 5 23 + 6x2 + 5x (22 – 1)(x + 1)3

10.

11.

Find the general term of the following expressions when expanded in ascending powers of x. 1+ 3 x

5x + 6

X2 + 7x + 3 12. 13.

14. 1 + 11x + 282 (2 + x) (1 - 3) 22.+ 7 x + 10 2x - 4

4 + 3x + 2 x2 15.

16. (1 - 2) (1 - 2x)

(1 – 2) (1 + x 2 2c2) 3 + 2 x – 22

4 + 7 x 17.

18. (1 + x)(1 – 4 x)2

(2 + 3x)(1 + x)2 2 x + 1 19.

20.

1- x + 2x2 (x - 1)(2c2 + 1)

(1 – 2 )3 1

3 – 2 x2 21.

22. (1-ax)(1- br)(1-cx)

(2 – 3 x + x2)

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