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T 2. Exercises for class and homework

Rewrite the functions in the factored form

1.	P (x) x2	2x 8 .	2. P (x) 2x2 x 10 .
	2		2
3.	P (x) x3	x2 x 1.	4. P (x) x5	16x .
	3		5
5.	P (x) x4	3x3 x2 3x 2 .	6. P (x) x4	3x3 6x2 5x 3 .
	4		4

Express the given fractions in the mixed-number form.

7.		x3 1		.		8.	x3		x 1	.			9.		x3	.
	x2 5x		6				x2		x 1					x 2

Express the given expressions as the sum of partial fractions.
10.		x2	6x 7		.	11.		x4 4x3 8x2 4				.	12.		8x3	2x 2		.
		(x2 4x 5)(x 3)							(x 1)(x2		2x)2					x4	1

Answers

1.(x +2)(x -4) . 2. (x 2)(2x 5) . 3. (x +1)(x2 +1) .

4.x(x +2)(x -2)(x2+4) . 5. (x 1)(x 1)2 (x 2) . 6. (x2 x 1)(x2 2x 3) .


7. x 5			26			7	.		8. x 1		x 2		.	9. x2 2x 4				8		.
		x 3			x 2						x2 x 1								x 2
10.	1		1			1		.	11.		1	1			1	.	12.		2		3	3x 1	.
	x 1		x	3		x	5			(x 2)2		x2			x 1				x 1		x 1	x2 1

Summary of this section

This flow chart describes the procedure for representing a rational function as a sum of partial fractions.

How to represent Pn x by partial fractions: Qm x

100

Is degree of

Divide Qm(x) into Pn(x):

Pn(x) less

Pn x

= Pn - m(x) +

Pk x

than degree

, degree

Qm x

of Qm(x) ?

of Pk(x) is less than degree of Qm(x).

yes

for

Pk x

For each (ax + b)n write

Qm x

Express Qm(x) as product

of powers of first-degree

(l)

and irreducible second-

i 1

ax b

degree polynomials.

For each (ax + bx + c)m write

c j x d j

(ll)

Find constants ki, cj, dj such that the

ax2 bx c

j 1

Pn x

Pk x

sum of (l) and (ll) is

Q x

Useful facts. The number of unknown constants equals the degree of Qm(x). (Check that you have the right number.)

If b2 – 4ac 0, ax2 + bx + c is irreducible. If b2 – 4ac 0, ax2 + bx + c is reducible.

T 2. Individual test problems

2.1. Rewrite the functions in the factored form

2.1.1. P (x) = x3	+x2 -2x -8 .			2.1.2.	P (x) = x3	-x2 -2x +8 .
3					3
2.1.3. P (x) = x3	-3x2 -4x +12 .			2.1.4.	P (x) = x3	+4x2 +5x +2 .
3					3
2.1.5. P (x) = x3	-7x2 +16x -12 .			2.1.6. P (x) = x3		+5x2 +8x +4 .
3					3
2.1.7. P (x) = x3	+3x2 +6x -10 .			2.1.8. P (x) = 2x3			-7x2 +8x -3 .
3					3
2.1.9. P (x) =3x3		+4x2	-2x -5 .	2.1.10. P (x) = 2x3			+3x2 -6x -7 .
3					3
2.1.11. P (x) = x4		+3x2	-4 .	2.1.12. P (x) = x4			+2x2 -8 .
4					4
2.1.13. P (x) = x4		-2x3	-x2 +2x .	2.1.14. P (x) = 2x4				+7x2 -9 .
4					4
2.1.15. P (x) = x3		+6x2	+13x +10 .	2.1.16. P (x) =5x4				-3x2 -8 .
3					4
2.1.17. P (x) = x4		-6x3 +11x2 -6x .		2.1.18. P (x) = 4x4				-4x2 +1 .
4					4
2.1.19. P (x) = 4x3 +x2			-3x -2 .	2.1.20. P (x) =9x4				+6x2 +1 .
3					3

101

2.1.21. P (x) x3	3x2 2x 8 .			2.1.22. P (x) = x3	-x2 -4x -6 .
3				3
2.1.23. P (x) = x3	+2x2		+5x +24 .	2.1.24. P (x) = 2x3		+x2 -3x +6 .
3				3
2.1.25. P (x) =-x3		-7x2 +6x +2 .		2.1.26. P (x) = 2x4		+3x2 -5 .
3				4
2.1.27. P (x) =3x3		+x2	-9x -10 .	2.1.28. P (x) = 4x4		-20x2 +25 .
3				4
2.1.29. P (x) =-x3		-x2	+11x +3 .	2.1.30. P (x) 9x4		12x2 4 .
3				4

Тopic 3. Integration of rational functions by partial fractions

Integrating of partial fractions. Integrating of rational functions.

Literature: [1, section 6, ch. ch. 6.4], [2, section 2, ch. 2.1], [3, section 7, §1], [4, section 7, §22], [6, section 8], [7, section 10, §§7–9], [9, §31].

T3. Main concepts

Occur the following partial fractions:

І.

;

ІІ.

;

ІІІ.

Mx N

;

ІV.

Mx N

(x a)n

x 2 px q

(x 2 px q)n

where a, p, q, A, M , N are

real

numbers, n = 2,

, D p 2

4q 0 ,

namely

x 2 px q is irreducible.

Let’s consider the integrals of the partial fractions:

І.

d (x a)

A ln

x a

C .

x a

ІІ.

dx A (x a) n d(x a)

C ( n 1 ).

(x a)

n)(x a)

n 1

ІІІ.

( a 0 ).

ax2 bx c

Suppose a quadratic factor ax2 + bx + c cannot be expressed as a product of two linear factors with real coefficients. Such a factor is said to be an irreducible quadratic factor over the real numbers.

First perform algebraic manipulations, as is shown:

b 2

bx c

a x

4a2

102

		b 2		2	,
a	x		m
a	x		m
		2a

where

b 2

b 2 4ac

. There is a sign of

being

4a 2

opposite of a sign of a discriminate D . Then

d(x

)

ax2 bx

)

where z x 2ba .

There are three cases:

1)if D 0 , then J1 am1 arctg mz C ;

2)if D 0 , then J1 az1 C ;

3)if D 0 , then J1 2am1 ln zz mm C .

Now let’s consider a more general form of this integral

J2 Ax B dx ax2 bx c

If the numerator were Ax + B, it would be the derivative of the denominator. The problem would then be covered by the formula

ff dx = ln f + C.

Ax B

(2ax

B Ab

J 2

ax 2 bx c

(2ax b)

bx c

d(ax2 bx c)

dx B

Ab J

ax2 bx c

Ab J

ax2 bx c

where an integral J1 is as seen above.

103

Namely, the integral of the third form

Mx N

( p

4q 0 ) is

px q

equal to

Mx N

ln(x2 px q)

x p 2

arctg

px q

ІV.

Integral

Mx N

dx , where

n 1 and

p 2 4q 0 .

(x2

px q)n

Introduce t = x 2p . Then dt = dx and:

p 2

dt N

, a

)

The first integral is a table integral and for the second integral we can use a reduction formula ( 2.3 ).

3.2. Integration of rational fractions

Now that we have had some practice in determining indefinite integrals, suppose we consider some problems of a greater degree of difficulty.

When you are integrating fractions sometimes a preliminary division is needed to get familiar with an integration form.

To express	Pn (x)	where Pn (x)	and Qm (x) are polynomials as the sum
	Qm (x)

of simpler (partial) fractions follow these steps:

If the

degree

of Pn (x) is not

less

than

the

degree of Qm (x) divide

Pn (x)

Qm (x) to

obtain

quotient

and

remainder:

Pn (x) = Pn m (x) Qm (x) +Pk(x). Where

the

degree

Pk(x) is less

than the

degree of P (x)

or else Pk(x) = 0. Then

Pn (x)

(x)

Pk (x)

, k m .

Qm (x)

n m

Qm (x)

Pk (x)

Apply the remainder steps to

Qm (x)

If ax+b appears exactly n times in the factorization of Qm(x), form the

sum:

Pk x

Qm x

ax b n

ax b n 1

ax b

where the constant c1,c2, …cn are to be determined later.

104

ax2 + bx + c appears exactly n times in the factorization of Qm(x),

then write the sum as follows:

Pk x

c1 x d1

c2 x d2

cn x dn

Qm x

ax2 bx c n

ax2 bx c

n 1

ax2 bx c

where the constants c1, c2,…cn

and d1, d2…dn

are known.

Typical problems

T3.

2x 2

Solution. Given the formula

x2 2ax (x a)2

a 2 , the complete square

is shown as (x a)2 . In our case we have one:

d(x 1)

arctg(x 1) C .

2x 2

Solution. Since

4x 2 4x 3 (2x 1)2 2 і

dx 1 d(2x 1) , therefore

d(2x 1)

2x +1

arctg

+C .

(2x 1)

( 2)

2 2

4x 3

6x 1

Solution.

Given

the

discriminant

D 36 4 9 0 , therefore

9x 2 6x 1 (3x 1)2

and

d(3x 1)

C .

3(3x 1)

6x 1

(3x 1)

2x 8

Solution.

The first method. Complete the square: x 2 2x 8 (x 1)2 9 , then

		dx	d (x 1)				1	ln	x 1	3	C	1	ln	x 4	C .

x	2	2x 8	(x 1)	2	3	2	6		x 1	3		6		x 2

105

The second method. Since																			x 2 2x 8 (x 2)(x 4) ,
									1							1 (x 2) (x 4)														1 1						1
																																							.
					(x 2)(x								4)			6			(x 2)(x							4)													.
					(x 2)(x								4)			6			(x 2)(x							4)				6 x 4						x 2
Then				dx							1				dx					1					dx			1						1
				dx							1				dx					1					dx			1			x 4			1			x 2			C
																												6 ln						6 ln
		x 2 2x 8									6				x 4					6			x 2
		x 2 2x 8									6				x 4					6			x 2
																				1 ln				x 4			C .
																								x 4
																								x 2
																				6				x 2
5.						dx				.
5.			2x	2						.
			2x			x 1
Solution. Complete the square:
			2x 2 x 1 2(x 2																x			1 ) 2((x 1/ 4)2 1/16 1/ 2)
			2x 2 x 1 2(x 2														2					1 ) 2((x 1/ 4)2 1/16 1/ 2)
																	2					2
															2(x 1/ 4)2 (3 / 4)2 ) .
Then														dx							1					d(x 1/ 4)
														dx							1					d(x 1/ 4)
										2x			2	x							2				(x 1/ 4)				2	(3 / 4)			2
										2x				x		1					2				(x 1/ 4)					(3 / 4)
1 1						4			x 1/ 4 3 / 4									C 1 ln									x 1/ 2					C =			1 ln			2x -1			+C
						4	ln		x 1/ 4 3 / 4																		x 1/ 2											2x -1
							ln
	2 2 3							x 1/ 4 3 / 4																3			x 1							3				x +1				1
6.						2x 3						dx .
6.			12x 9x						2		2	dx .
			12x 9x								2

Solution. Given the derivative (12x 9x 2 2) 12 18x , the numerator is transformed to the form:

2x 3 19 (12 18x) 129 3 19 (12 18x) 53 .

Then

2x 3			dx		1		(12 9x 2	2)		dx			5	dx
12x 9x	2		dx		9		12x 9x	2	2	dx			3	12x 9x	2
12x 9x		2			9		12x 9x		2				3	12x 9x		2
				1 ln		12x 9x 2		2			5	J1 ,
											5

				9							27
				9							27

106


			dx				d(x	2	)			3		3x 2 2
where	J1		dx				d(x	3	)			3	ln	3x 2 2		C .
where	J1	x 2	4	x	2	x	2 2			2	2	2 2	ln	3x 2	2	C .
			3		9

										3
							3			3

In Exercises 7 to 14 express the integrand as the sum of simpler (partial) fractions to find the integral.

7. Find

Solution:

x 5 x2

5x 25 125

x3dx

x3 125 125

x 5

125

5x2

5x 25

x 5

25x 125ln

x 5

3x 2 21x

dx .

Solution. We have the integrand

P (x)

3x 2 21x

Q3 (x)

x3 3x 2

6x 8

The roots of Q3 (x) x3

3x 2

6x 8 are the real and differs numbers

–2; 1 and 4.

Then Q3 (x) (x 1)(x 2)(x 4) . In our case we have one

3x 2 21x

3x 2 6x 8

(x 2)(x 1)(x 4)

x 1

This expression can be found by the method illustrated in Example 3 (see topic 2)

3x 2 21x

(x 2)(x 1)(x 4)

x 1

x 4

Hence,

3 ln

x 2

2 ln

x 1

2 ln

x 4

C .

x 2

x 1

x 4

9. Find

10x3 19x2 8x 7

dx.

107

Solution. Since the degree of the numerator is not less than the degree of the denominator, carry out a long division:

x4 + 10x3 + 19x2 8x 7

x2 + 7x 2

x4 +7x3

2x2

3x3+21x2

x2 + 3x

– 3x3+21x2

2x 7

is the remainder.

Thus,

10x3

19x2 8x 7

2x 7

7x 2

Then

2x 7

t x2 7x 2

dt 2x 7 dx

7x 2

3x2

x2 7x

10.

x 4

3x 2 3x 2

dx .

Solution. The degree of Pn (x) is not less than the degree of Qm (x) . Divide (x) by Qm (x) to obtain a quotient and remainder:

By Example 4 (see topic 2):

			x 4 3x 2 3x 2									= x +1+					1	-				2			-		1		.
Then			x3 x 2 2x									= x +1+					x	-		3(x -2)					-	3(x +1)			.
			x3 x 2 2x														x			3(x -2)						3(x +1)
	1	(x 1)2 dx							2				dx		1				dx				1
I																								(x 1)2 ln						x


	2					x			3 x 2						3 x 1								2
	2					x			3 x 2						3 x 1								2
							2	ln		x 2				1 ln			x 1				C .
							2

11.		x 2 2			3									3
					3									3
						dx .
	(x	1)(x		1)	2	dx .
	(x	1)(x		1)
Solution. By Example 5 (see topic 2)
			x 2 2									3						1							3			.
			(x 1)(x 1)2								4(x 1)					4(x 1)							2(x 1)2					.
			(x 1)(x 1)2								4(x 1)					4(x 1)							2(x 1)2

108

Thus

x 2 2				dx			3				dx	3		dx			1	dx
(x 1)(x	1)		2	dx			4			x 1		2	(x 1)		2		4	x 1
(x 1)(x	1)						4			x 1		2	(x 1)				4	x 1
		3	ln		x	1			3		1	1 ln		x 1		C .
		3	ln		x	1			3		1	1 ln		x 1		C .
	4							2			x 1	4
	4							2			x 1	4

12.I x4 9x3 36x2 16x 20 dx .

x2 x 5 x2 2x 2

Solution. The denominator has degree 5, while the numerator has the same degree 5. Hence:

x4 11x3 3x2 16x 20

Dx E

x2 x 5 x2 2x 2

A x3

x 5

x2 2x 2

3x2

8x 10

B x4 3x3 8x2

10x

C x4

2x3 2x2

x 5

C D x

3B 2C

5D E x3 11;

Dx E x

3A 8B 2C 5E x2 3;

x2 2x

8A 10B x 16;

10A 20.

C D 1;

A 2;

5D E 13;

5E 3;

B 0;

D 2;

2x 1

Then I

x 5

x2 2x 2

2x 2 3

d x2 2x 2

x 5

x2 2x 2

x 5

ln x

2x 2 3arctg x 1

x 1

109

13. I											x 2										dx .
13. I											3							1)			dx .
						(x 1)(x												1)
Solution. By Example 6 (see topic 2)
												x 2															1							1										1							x
																																																									.
				(x 1)							2	(x		2			x 1)										3			(x 1)								2				x 1						x	2	x 1
				(x 1)								(x					x 1)										3			(x 1)												x 1						x		x 1
Thus												1						dx									1				dx								1						xdx
												1						dx									1				dx								1						xdx
												3			(x 1)									2			3				x 1								3					x	2
												3			(x 1)												3				x 1								3					x			x 1
																								1							1 ln					x 1						1				I1 ,
																								1
																						3(x 1)
where																						3(x 1)								3														3												1 )dt
where
										xdx																xdx														x			1		t,							(t
																																											1									(t
I1																																											2													2
I1																										1			2				3										2											2				3
					x 2 x 1																		(x			1			2				3																				t	2				3
																											2)												dx dt
																																	4						dx dt																			4
				tdt															dt														4																					2t				4
				tdt								1							dt								1					2						3					1				2							2t
																											2 ln(t							+						) +								arctg									+C =
		t	2				3					2					t	2					3				2 ln(t							+				4		) +			2				3	arctg							3		+C =
		t					4										t						4
										=			1 ln(x2 -x +1) +																				1			arctg							2x -1					+C .
										=			1 ln(x2 -x +1) +																					3		arctg												+C .
Thus													2																					3												3
Thus							1									1 ln												1 ln(x2 -x +1) +																				1									2x -1
I																					x 1																														arctg								+C .

						3(x
											1)					3												6																			3 3											3
											1)					3												6																			3 3											3
14.					x2 2x 2														dx .
14.	x	3		3x					2		6x							7	dx .
	x			3x							6x							7
																				2										1							3							2
Solution. In this case x																							2x 2 3 (x																3x						6x					7)			. Then
Solution. In this case x																							2x 2 3 (x																3x						6x					7)			. Then
											x2 2x 2															dx				1					d (x3 3x2 6x 7)
								x		3	3x					2		6x							7	dx				3								x	3		3x				2		6x			7
								x			3x							6x							7					3								x			3x						6x			7
																			1 ln \| x3 3x2 6x 7 \| C .
																			3

110

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