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Национальный авиационный университет

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DDR 2 p.64-131.pdf

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Solution. Let u sinn 1 x

and dv

sin x

dx . We evaluate

cosm x

du (n 1)sinn 2 x cos xdx ,

sin x

d(cos x)

( m 1 ),

(m 1) cos

m 1

cos

J n, m

sinn 1 x

n 1

sinn 2 x

dx .

(m 1) cosm 1 x

m 1

cosm 2 x

Thereby

J n, m

sinn 1 x

n 1

J n 2, 2 m .

1) cosm 1 x

m 1

Exercises for class and homework

T1.

Find the indefinite integrals.

(x 1)(2x 3)dx .

x3 8

dx .

x dx .

x 2

x 2 2x 3

dx .

x )dx .

x 1

dx .

sin 2

dx .

tg2 xdx .

9. (2 ctg x 1)(2 ctg x 1)dx .

10.

1 5 cos2 x dx .

11.

x 2 1 1

dx .

12.

1 cos 2x

16 x 2

13.

14.

e x (3 e x )dx .

15.

x 4

dx .

x 2 4

16.

6 x 1 8x

dx .

17.

18.

16x 2

19.

3 x

dx .

20. (3

x 1)(2 4

x 3)dx .

9 x2

Use appropriate substitutions to find the integrals.

21. а) sin 3 xd(sin x) ;

б) sin 3 x cos xdx ;

в) cos3 2x sin 2xdx .


22.	а) ln3 xd(ln x) ;								б)				ln3 x			dx .				23. а) 2tg x d (tg x) ;									б)	2tg x dx					.
																														2
														x																cos		x
24.	(3x 4)6 dx . 25.											3		2x 5dx .						26. 8		3 xd (3x) .						27.			xdx				.
																															2
																															5 x
28.			x 2 dx		.			29.				(2x 4)dx						.		30.			xdx			.		31.			xdx		.
			3									2											4
			4 x									5 4x x											4 x							1 x 4
32.			x3dx			.		33.		(3x 2				1)dx			.			34.		x 2 dx			.		35.		sin xdx					.

			1 x 4									x3 x 1										1 x6							4		cos x
36.		sin		xdx			.	37.			tg			xdx	.				38.		ln xdx			.			39. x 2 e x3 dx .
				x										x							x

40.		esin x				cos
							2
44.				dx		.
			e	x
				1
48.			(2x 7)dx
			(3x			2)	3

	(x 1)dx
					.
	x		x 2

xdx						x 1			dx				dx						xdx
xdx			41.			x			dx		42.		dx			43.			xdx
	.									.					.							.
x	.					x	2			.		1			.			x(x 1)				.
x						x						1		x				x(x 1)
45.			xdx	.	46. x tg(x 2 1)dx .											47.		x3dx			.
		1 x																	x 1
.		49. (1 2x)99 xdx .											50.				dx		.		51.
.		49. (1 2x)99 xdx .											50.			x	x	1	.		51.
																x	x	1
52.			dx		.			53.			dx		.	54.			x 2 dx			.
52.	1		3		.			53.			1 e x		.	54.						.
	1		x 1								1 e x						1 x 2

55.		dx	.		56.		1 x	2 dx		.
	x	x 2 4				4
							x
58.		dx		.	59.		e2x dx		.

	x	2 x 2 1					4 1 e x

Use integration by parts to find the integrals.

61.	x cos 4xdx .	62.	(2x 5) sin 2xdx .
64.	(x3 2x 2 4)e2x dx .		65. (x
66. x 2 2 x dx .		67.	ln(2x 1)dx .
69. ln(x 2 4)dx .		70.	x ln(x 1)dx .

57. x 2 9 x 2 dx .

60.			dx	.
	(1	x	2
			) arctg x

63. (x 3)e x dx .

3x 2 3x 1) cos xdx .

68. x ln xdx .

71.x ln(x 2 1)dx .

79.

72. x 3 ln 2 xdx .

75. x arcsin xdx .

78. arccos xdx .

81. e x cos 2xdx .

84. cosx sin3 xx dx .

73.ln 2 xdx .

76.x tg2 xdx .

x arctg x dx . 1 x2

82. sin ln xdx .

85. x 2 e x dx . (x 2)2

Answers

74.x arctg xdx .

77.x cos2 xdx .

80.lnx23 x dx .

83.e px sin qxdx .

86. arcsin 2 xdx .

						1.	2	x3		x2		3x C . 2. 1 x3										x2 4x C . 3.																	4	x7 / 4						C . 4.				3			x8 / 3						6	x5 / 3
						1.		x3		2		3x C . 2. 1 x3										x2 4x C . 3.																	7	x7 / 4						C . 4.				8			x8 / 3					5		x5 / 3
						3				2								3																					7											8								5
-		9		x2 / 3 C . 5.							4	x7 / 4					4		x9 / 4		C .					6. x 2									x C . 7. (x sin x) / 2 C .
	2									7						9
8.		tg x -x +C . 9. -4ctg x -5x +C .																								10.			tg x +5x +C . 11. ln																				x + 1+ x2											+
8.		tg x -x +C . 9. -4ctg x -5x +C .																								10.			tg x +5x +C . 11. ln																				x + 1+ x2											+
																																			2
																x															x2 4															14. 3ex x C .
+arctg x +C .										12. arcsin						x		C .				13. ln						x											C .
+arctg x +C .										12. arcsin						4		C .				13. ln						x											C .
15.			1 x3 - x +arctg x +C . 16.																	3x				4x				C . 17.									1	arctg				3x				+C . 18.										1	arcsin			4x	C .
15.			1 x3 - x +arctg x +C . 16.																	6ln 3			ln 4					C . 17.										arctg								+C . 18.										4	arcsin			3	C .
			3																	6ln 3			ln 4												6				2																	4				3
19. 2						3 x 2				3 x C .						20. x(-3 +								9		3 x -				8			4 x +						24 12						x7 ) +C . 21. а)													sin4 x C; б)
																							4							5									19																			4
sin4 x C; в)										1 cos4 2x C.									22.а) ln4 x								C ; б)								ln4 x				C .							23. а) 2tg x										C ; б)				2tg x C .
	4									8												4													4														ln 2											ln 2
24.			(3x 4)7						C . 25.					3	3	(2x 5)4						C .26.									9				3 (3 x)4									C . 27.								1			ln(x2			5) C .
						21							8																	4																				2
28.			1 ln(x3 4) C .										29. ln					5 4x x2								C . 30.									1	arctg					x2		C . 31.								1			arcsin x2 C .
																																			1						x2										1

			3																																4				2											2
			3																																4				2											2
32. 1 1 x4 C .33. 2 x3																		x 1 C .34.											1 ln			x3							1 x3							C .	35. 8ln(4													cos x )
32. 1 1 x4 C .33. 2 x3																		x 1 C .34.											1 ln			x3							1 x3							C .	35. 8ln(4													cos x )
					2																								3
																																																2			(ln x)3 / 2 C . 39.
2					cos x C . 36.								2cos						x C .						37. 2 ln									cos					x	C .						38.		2
2					cos x C . 36.								2cos						x C .						37. 2 ln									cos					x	C .						38.	3
1	ex				3	C . 40.				esin			x C			.		41.			2		x				1		3	C . 42. 2																	3									x ) C . 43.
1					3																2		x				1		3															x 2ln(1

3																					3						x
3																					3						x

2arctg x C .44.																										x ln(ex 1) C																					45. 2							x -2 arctg							x +C .								46.								1		ln			cos(x2				1)			C .
																																																																													1
																																																																													2
47.							2				(x					1)7								6		(x 1)5 2																(x 1)3 2												x 1					C	.	48.								2								2						25					C .
							2																	6																																													2														25
							7																	5																																										9(3x 2)															18(3x 2)2
							7																	5																																										9(3x 2)															18(3x 2)2
											(1 2x)100																													x 1 1																																						x 2
49.											(1 2x)100														C .				50. ln											x 1 1												C .			51. 2					x 2							2 arctg											x 2					C .					52.
49.													200												C .				50. ln											x 1 1												C .			51. 2					x 2							2 arctg										2						C .					52.
													200																											x 1 1																																					2
3(				t2					2t ln										t		) C , де t 1 3 x																										1 .					53. ln			ex				1 1				C .				54.					1			arcsin x


					2																																																		ex 1 1																	2
					2																																																		ex 1 1																	2
			x							1 x2 C .																55.				1		arccos									2				C .								56.						(1 x2 )3						C .							57.						81 arcsin									x
				2																										2											x																		3x3																		8									3
	x							9 x2								(9 2x2 ) C . 58.																										x2					1					C . 59.				4			4 (1 ex )7								4	4 (1 ex )3 C .																			60.
8								9 x2								(9 2x2 ) C . 58.																													x							C . 59.				7			4 (1 ex )7								3	4 (1 ex )3 C .																			60.
8																																													x											7											3
ln						arctg x									C .										61.			x		sin 4x													1					cos 4x C .								62.				5 2x					cos 2x									1		sin 2x C .											63.
ln						arctg x									C .										61.			x		sin 4x													1					cos 4x C .								62.				5 2x					cos 2x									1		sin 2x C .											63.
																										4															16																					2										2
																										4															16																					2										2
(x 2)ex C .																						64. (					x3							7		x2								7		x						9	)e2x C . 65. (3x2												2x 9)cos x
(x 2)ex C .																						64. (														x2							4			x						8	)e2x C . 65. (3x2												2x 9)cos x
																										2				4													4									8
+ (x3 x2																9x 3)sin x C .																								66.									x2 ln2 2 2x ln 2 2													2x C .							67.								xln(2x 1) x –
+ (x3 x2																9x 3)sin x C .																								66.																						2x C .							67.								xln(2x 1) x –
																																																						ln3 2
	1	ln(2x 1) C .																								68.			2		x x ln x																	4		x		x C .						69.		xln(x2 4) 2x 4arctg																									x		C . 70.
2																													3																		9																																				2
	x2						ln(x -1) -													1		(x +1)2							-				1		ln				x -1									+C . 71.							1		[(x2 1) ln(x2										1) x2										1] C . 72.
	x2																			1													1																						1

2																4														2																									2															1
2																4														2																									2
	3						3 x4 (8ln2 x 12 ln x 9) C .73. x(ln2 x 2ln x 2) C . 74.																																																		(x2					+1)arctg x -
32																																																																							2
-				x				+C .								75.							2x2 1						ar c sin x															1				x 1 x2 C . 76.														x tg x -							x2				+ln							cos x				+C .					77.
				x																			2x2 1																					1																									x2
				2																						4																	4																										2
				2																						4																	4																										2
	x2									x		sin 2x 1 cos 2x C . 78.																																	xarccos x										1 x2					C .				79.						1+ x2 arctg x -
4										4						8
-ln(x + 1+ x2 ) +C .																																	80.									2 ln2 x 2 ln x 1																	C .			81.					1 ex						(2sin 2x cos 2x) C .
-ln(x + 1+ x2 ) +C .																																	80.																		4x2								C .			81.					1 ex						(2sin 2x cos 2x) C .
																																																			4x2																5
							82.							x			(sin ln x									cos ln x) C . 83.																									e px ( psin qx q cos qx)																C.					84.
							82.						2				(sin ln x									cos ln x) C . 83.																													p2 q2												C.					84.
													2																																										p2 q2
								x									1	tg x С .													85.						x 2								ex C .86. x ar c sin2 x 2arcsin x																														1 x2 2x C .
	2cos2 x															2																						x 2

T1. Individual test problems

1.1. Find the integrals.

1.1.1.

3x 2

dx .

1.1.2. 4 x7 (2 x )dx .

4 x

1.1.3. 5 x3

(4 3

x )dx .

1.1.4. 3 x2

(1 4 x )dx .

1.1.5. 5 x4

(1 3 x )dx .

1.1.6. 6

x5 (3 3 x )dx .

1.1.7.

dx .

1.1.8.

x x 1

dx .

3 x4

1.1.9.

x 2

dx .

1.1.10. 3 x7

(5 5 x )dx .

5 x4

1.1.11. (5

1) 3 x4 )dx .

1.1.12. 4 x5

(2 6 x )dx .

1.1.13. 3

x8 (1 3 x )dx .

1.1.14. (4 x9

1) 3 x2 )dx .

1.1.15. 7

x5 (1 7 x )dx .

1.1.16.

x 1

dx .

3 x2

1.1.17.

3 x

x2 2

dx .

1.1.18.

dx .

1.1.19. 3 x7

(3 4 x )dx .

1.1.20. 3 x4

(1 5 x )dx .

1.1.21. 5

x3 (5

x )dx .

1.1.22.

x 4

dx .

3 x4

1.1.23.

2x3 1

dx .

1.1.24.

dx .

1.1.25. 4

x9 (1

x )dx .

1.1.26. (5 x 2) 3 x2 )dx .

1.1.27. 4

x3 (3 3 x )dx .

1.1.28.

x 1

dx .

3 x

1.1.29.

x 3x2 1

dx .

1.1.30.

x 2

dx .

1.2. Find the integrals.

1.2.1.2 3x dx .

x2 2

1.2.4.	6x 1							dx .
1.2.4.	2x		2					dx .
	2x				1
1.2.7.		5 3x										dx .
1.2.7.		2x2 1										dx .
		2x2 1
1.2.10.			1		5x								dx .
1.2.10.		25x				2							dx .
		25x								1
1.2.13.				x 3								dx .
1.2.13.		9x			2							dx .
		9x				7
1.2.16.		5 x							dx .
1.2.16.		x		2			2		dx .
		x					2
1.2.19.			5x 1										dx .
1.2.19.													dx .
				x2 3
1.2.22.				x 4									dx .
1.2.22.													dx .
				9 x2
1.2.25.		1 3x										dx .
1.2.25.												dx .
				x2 1
1.2.28.		8 2x									dx .
1.2.28.		3x			2	1					dx .
		3x				1

1.3. Find the integrals.

1.3.1.sin 2 (1 x)dx .

1.3.3.1 2 sin x 2 dx .

1.3.5.cos3 (1 3x)dx .

1.3.7.sin 2 32x dx .

1.3.9.cos3 (2x 3)dx .

1.3.11. 1 cos 2x 2 dx .

1.2.2.	3 5x									dx .			1.2.3.		8 13x								dx .
1.2.2.	1 x2									dx .			1.2.3.				x2 1						dx .
	1 x2																x2 1
1.2.5.		x 2									dx .		1.2.6.			3 7x									dx .
1.2.5.	2 x2										dx .		1.2.6.			1 4x2									dx .
	2 x2															1 4x2
1.2.8.	1 x								dx .				1.2.9.		3x					2		dx .
1.2.8.	2 x2								dx .				1.2.9.	2x				2				dx .
	2 x2													2x					1
1.2.11.	4x 3										dx .		1.2.12.				5x			1					dx .
1.2.11.	3x			2		4					dx .		1.2.12.					x2 6							dx .
	3x					4												x2 6
1.2.14.			5 3x									dx .	1.2.15.					4 2x									dx .
1.2.14.												dx .	1.2.15.					1 4x2									dx .
			4 3x2															1 4x2
1.2.17.			1 3x									dx .	1.2.18.				5 4x								dx .
1.2.17.												dx .	1.2.18.					1 x2							dx .
			4x2 1															1 x2
1.2.20.	1 3x									dx .			1.2.21.					x		5						dx .
1.2.20.	4x			2		1				dx .			1.2.21.				3			2x			2			dx .
	4x					1											3			2x
1.2.23.	2x				7				dx .				1.2.24.					7x 2									dx .
1.2.23.	x		2				5		dx .				1.2.24.														dx .
	x						5												x	2 1
1.2.26.	x 5							dx .					1.2.27.				3		7x					dx .
1.2.26.	x	2			7			dx .					1.2.27.				x		2		1			dx .
	x				7												x				1
1.2.29.	3x 7										dx .		1.2.30.					2x 1									dx .
1.2.29.		x2 4									dx .		1.2.30.														dx .
		x2 4															3x2 4

1.3.2.sin3 (1 2x)dx .

1.3.4.cos3 (5x 1)dx .

1.3.6.3 sin 2x 2 dx .

1.3.8.cos x 3 2 dx .

1.3.10.sin3 45x dx .

1.3.12. sin 2 (2x 1)dx .

1.3.13.sin3 6xdx .

1.3.15.sin 2 (4x 3)dx .

1.3.17. 1 2 cos x 2 dx . 1.3.19. sin2 (2x 1)dx .

1.3.21. 1 3cos x 2 dx .

1.3.23. sin3 (5x 1)dx .

1.3.25. cos2 (2x 1)dx .

1.3.27. cos2 7xdx .

1.3.29. sin3 4xdx .

1.4. Find the integrals 1.4.1. (4 3x)cos 2xdx .

1.4.3. (x2 4)e x dx .

1.4.5. x ln(4x 1)dx .

1.4.7. ln(x2 2x 2)dx . 1.4.9. (2 x2 )cos xdx . 1.4.11. (x 2)3 2x 1 dx . 1.4.13. x ln(x 3)dx .

1.4.15.3x 1 ctg2 3x 1 dx .

1.4.17.(2x 1)cos2 xdx .

1.3.25. cos2 (2x 1)dx .

1.3.27. cos2 7xdx .

1.3.29. sin3 4xdx .

1.3.14.sin 2 2x dx .

1.3.16.cos2 (1 2x)dx .

1.3.18.cos2 3xdx .

1.3.20.sin2 (1 x)dx .

1.3.22.cos2 25x dx .

1.3.24.cos2 (3 x)dx .

1.3.26.cos3 4xdx .

1.3.28.(sin x 5)2 dx .

1.3.30.sin 2 34x dx .

1.4.2.(x2 1)sin 3xdx .

1.4.4.(x2 3)e2x dx .

1.4.6.		2x 1 ln 2x 1 dx .
1.4.8.	ln(x2 2x 5)dx .

1.4.10.(3x 7)2 x dx .

1.4.12.arctg(3x)dx .

1.4.14. 4 x ln xdx . 1.4.16. x tg2 2x dx .

1.4.18.(3 x)sin2 2xdx .

1.3.26.cos3 4xdx .

1.3.28.(sin x 5)2 dx .

1.3.30. sin 2 34x dx .

1.4. Find the integrals 1.4.1. (4 3x)cos 2xdx .

1.4.3. (x2 4)e x dx .

1.4.5. x ln(4x 1)dx .

1.4.7. ln(x2 2x 2)dx . 1.4.9. (2 x2 )cos xdx . 1.4.11. (x 2)3 2x 1 dx . 1.4.13. x ln(x 3)dx .

1.4.15.3x 1 ctg2 3x 1 dx .

1.4.17.(2x 1)cos2 xdx .

1.4.19.	x arcsin x2dx .
1.4.21.	x ln(4 x)dx .
1.4.23.	ln(x2 4x 5)dx .
1.4.25.	(1 2x2 )cos 4xdx .
1.4.27.	x arccos3xdx .
1.4.29.	x ln(x2 3)dx .

1.4.2.(x2 1)sin 3xdx .

1.4.4.(x2 3)e2x dx .

1.4.6.		2x 1 ln 2x 1 dx .
1.4.8.	ln(x2 2x 5)dx .

1.4.10.(3x 7)2 x dx .

1.4.12.arctg(3x)dx .

1.4.14. 4 x ln xdx . 1.4.16. x tg2 2x dx .

1.4.18.(3 x)sin2 2xdx .

1.4.20.x arcsin x4dx .

1.4.22.3 4 3x ln 4 3x dx .

1.4.24.ln(x2 6x 13)dx .

1.4.26.(x2 6)3x dx .

1.4.28.x arctg(2x2 )dx .

1.4.30.x arccos x2dx .

Topic 2. Polynomials. The rational functions

Values of polynomial functions. Fundamental theorem of algebra. Factor theorem. The factored form of a function. Proper and improper rational fractions. Long division to transform to mixed-number form – polynomial plus proper function. Linear factors of a polynomial function.

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

T 2. Main concepts

2.1. Polynomial functions. Linear factors of a polynomial function

A polynomial function is a function that can be expressed in the form

Pn (x) = a0 xn +a1 xn-1 +...+an-1 x +an ,

Where the degree n is a nonnegative integer, the coefficients a0, a1, …, an are real numbers, and an ≠ 0.

If Pn (x0 ) =0 , the number x0 is called a root of the polynomial function

Pn(x).

Theorem 1. (Factor theorem). x -x0 is a factor of polynomial Pn (x) if and only if the polynomial equals 0 when x0 is substituted for x. On the other hand,

Pn (x) =(x -x0 )Qn-1 (x) ,

where Qn-1 (x) is a polynomial of n–1 degree.

Theorem 2. (Bezoo’s remainder theorem). If polynomial Pn(x) is divided by (x - ), then the remainder R is Pn( ). That is, if

Pn(x) = (x - ) Pn 1(x) + R

then

Pn( ) = R.

This is not difficult to show to be true, for if

Pn(x) = (x - ) Pn 1(x) + R

then

Pn( ) = ( - ) Pn 1( ) + R = R.

Theorem 3. (Fundamental theorem of algebra). If Pn(x) is a polynomial function of degree n 0, then Pn(x) has at least one complex zero.

Of course, a direct result of this is that the polynomial equation Pn(x) = 0 has at least one complex root.

Now, suppose that we have a polynomial function Pn(x) and that x1 is zero

of P. By the theorem 1, this implies that (x x1) is a factor of P and that we can write P as

Pn(x) = (x x1) Qn 1(x)

where Qn 1(x) is a polynomial function of degree n 1. Now, by the theorem 3, Qn 1(x) in tern has a zero x2, and Qn 1(x) can be written as

Qn 1(x) = (x x2) Qn 2(x)

where Qn 2(x) is a polynomial function of degree n 2. This implies that

Pn(x) = (x x1) (x x2) Qn 2(x). If we continue this process, eventually arrive at

Pn(x) = (x x1) (x x2) … (x xn)Q0(x)

where Q0(x) is a polynomial function of degree 0, namely, a constant function. We have the rational function it to you as an exercise to show that Q0(x) = a0, the leading coefficient of P.

Theorem 4. If Pn(x) is polynomial function of degree n 0 and leading coefficient a0, then

Pn(x) = a0 (x x1) (x x2) … (x xn)

where x1, x2, x3, … xn are the complex zeros of Pn(x). Particularly,

ax2 +bx +c = a(x -x )(x -x ) ,
1	2

where x1, x2, x3, … xn are the complex zeros of ax2 +bx +c =0 .

If P (x) is divided a without remainder by (x -x )k , but is not divided
n							0
by (x -x )k+1	, then		x is called the repeated k times roots of					P (x) .
0			0					n
In this case		P (x) =(x -x )k
		P (x) =(x -x )k			Q	(x) , Q	(x ) ¹0 .
		n		0	n-k	n-k	0
Whether a polynomial Pn(x) has the roots x1 , x2 , , xm (m £n) , with
repeated accordingly k1 , k2 ,…, km , then
		P (x) = a (x -x )k1 (x -x )k2 ...(x -x )km						.	(*)
		n	0	1		2	m

Theorem 5. (Identity equations). An identity is an equation that is true for all values of the variable.

Theorem 6. (Conditional equations). A conditional equation is one that is true for some value(s) of the variable and not true for other values of the variable

a0 =b0 , a1 =b1 , , an =bn .

Theorem 7. (Conjugate root theorem). If Pn(x) is a polynomial function of degree n 0 with real coefficients, and if a +bi is a zero of P, then a -bi is also a zero of P. Moreover, the number of the conjugate roots a +bi and a -bi

is the same.

Let us consider the product

(x -(a +bi))(x -(a -bi)) =((x -a) -bi)((x -a) +bi) =(x -a)2 +b2 = = x2 -2ax +a2 +b2 = x2 + px +q ,

where p =-2a, q = a2 +b2 .

Therefore, if a polynomial Pn(x) has a pair of conjugate complex roots a bi , we can substitute (see (*)) product (x -(a +bi))(x -(a -bi)) for a

quadratic trinomial x2 + px +q with real coefficients and a negative discriminant.

Theorem 8. Any polynomial with real coefficients can be given as a product of linear and quadratic factors with real coefficients, namely

Pn (x) a0 (x x1 )k1 ... (x xr )kr (x2 p1 x q1 )s1 ... (x2 pm x qm )sm .

____

Moreover, k1 k2 ... kr 2(s1 s2 ... sm ) n , Di pi2 4qi 0, i 1, m.

2.2. Rational functions

A rational function is a function that can be expressed as the ratio of two polynomial functions. That is, f(x) is a rational function if there exist polynomial functions Pn(x) and Qm(x) such that

f (x) Pn (x) . Qm (x)

The domain of f(x) is the set of all x such that the denominator Qm(x)is not zero (otherwise the ratio would be undefined). The zeros of f(x) are those zeros of the numerator Pn(x) that are in the domain of f(x) since a ratio is zero if and only if its numerator is zero.

If the degree of the numerator Pn(x) is less than the degree of the denominator Qm(x), then f(x) is a proper rational function. If, on the other hand, the degree of the numerator Pn(x) is larger than or equal to the degree of denominator Qm(x), then f(x) is an improper rational function.

Just as an improper fraction can be rewritten as the sum of an integer and a proper fraction (for example, 215 4 15 ), we can rewrite an improper rational

function as the sum of a polynomial function and a proper rational function. Suppose that Pn(x) and Qm(x) are nonconstant polynomial functions such

that the degree n of Pn(x) is greater than or equal to the degree m of Qm(x). There exist two unique polynomials Pn–m(x) and Pk(x) such that the degree k of Pk(x) is less than the degree of Qm(x) and

Pn (x)	P	(x)	Pk (x)	,	k m

Qm (x)	n m		Qm (x)

Or, equivalently,

Pn(x) = Qm(x) Pn–m(x) + Pk(x).

Qm(x) is the divisor, Pn–m(x) is the quotient, and Pk(x) is the remainder.

Here Pn m (x) is the integer part of the polynomial of ( n m )th degree),

Pk (x)	is a proper rational ratio.
Qm (x)

Given a rational algebraic expression in which a higher-degree polynomial is divided by a lower-degree polynomial, use long division to transform to mixed-number form – polynomial plus proper fraction.

Observe that the denominator consists only of distinct linear factors, each factor occurring exactly once. It can be shown that to each such factor x - a there corresponds a partial fraction of the form

A (A is a constant)

such that the composite fraction is the sum of the partial fractions. If there are n such distinct linear factors, there will be n such partial fractions. For instance,

x2 6x 4	A	B	C
				.
x x 1 x 2	x	x 1	x 2

To determine the constants A, B, and C, we first combine the terms on the right side:

x2 6x 4	A x 1 x 2 Bx x 2 C x x 1
		.
x x 1 x 2	x x 1 x 2

Since the denominators of both sides are equal, we may equate their numerators:

x2 + 6x – 4 = A x 1 x 2 Bx x 2 C x x 1 .

Although condition is not defined for x = 0, x = 1, and x = –2, we want to find values for A, B, and C that will make the last equation true for all values of x. That is, it will be an identity. By successively setting x equal to any three different numbers, we can obtain a system of equations which can be solved for A, B, and C. In particular, the work can be simplified by letting x be the roots of Qm(x) = 0, in our case x = 0, x = 1, and x = – 2. If x = 0 we have

–4 = A ( – 1) 2 + B 0 + C 0 = – 2A, so A = 2.

If x = 1,

3 = A 0 + B 3 + C 0, so B = 1.

If x = – 2,

–12 = A 0 + B 0 + C ( – 2) ( –3), so C = – 2.

Thus

x2 6x 4	2		1		2
							.
x x 1 x 2		x		x 1		x 2

If the denominator of Pn (x) contains only linear factors, some of which

Qm (x)

are repeated, then for each factor ( x - a)n, where n is the maximum number of times x - a occurs as a factor, there will correspond the sum of n partial fractions:

Mx N

Pn x	A	B	C	...	M	.
x a m	x a	x a 2	x a 3		x a m

Suppose a quadratic factor x2 + px + q occurs in Qm(x) and it 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. To each distinct irreducible quadratic factor that occurs exactly once in Qm(x), there will correspond a partial fraction of the form

x 2 px q .

If the denominator of

Pn (x)

contains linear and quadratic factors, some

Qm (x)

of which are repeated

Qm (x) a0 (x a) (x b) (x 2 px q) (x 2 lx s) ,

there will correspond the following sum of m (m = +…+ + 2( +…+ )

fractions:

P (x)

Qm (x)

(x a)

x b

x N

M x N

L x S

2 px q

(x 2 px q)

x2 lx s

(x 2 lx s)

Typical problems

T 2.

1. Rewrite the functions in the factored form:

а)

P (x) 3x2 x 14 ;

b) P (x) x3

2x2 x 2 .

c) P (x) x4

x3 6x2 14x 12 .

Solution. а) as is generally known, ax2 + bx + c = a(x – x1)(x – x2). The

roots of the quadratic equation 3x2

x 14 0 are

2, x

. So,

P (x)

3(x 2)(x 7 ) (x 2)(3x 7) ;

b) P3 (x) x2 (x 2) (x 2) (x 2)(x2 1) (x 2)(x 1)(x 1) .

c) use the following rule.

If Pn (x) with integer coefficients has the integer roots, they are among divisors of an .

There are divisors of an : 1; 2; 3; 6; 12 . It is true that

P4 (2) 16 8 24 28 12 0 , So, P4 (x) (x 2)Q3 (x) .

A rational algebraic expression such as

x4 x3 6x2 14x 12

x 2

can be thought of as an improper fraction too, since its numerator is of a higher degree than the denominator is. The same long-division process can be used to transform this expression to a new polynomial. Here, step by step, is how it works.

x 4 x3 6x 2 14x 12

x 2

x 4 2x3

x3 x 2 4x 6

x3 6x

x3 2x 2

14x

4x 2

6x 12

reminder

Hence,

P (x) (x 2)(x3

x 2 4x 6) .

A number

x 3

the root of the polynomial Q3 (x) x3 x 2

4x 6 ,

because of

Q3 ( 3) 27 9 12 6 0 . Using the long-division of

Q3 (x) to

х+3, we will have

x 4 x3

6x 2 14x 12

(x 2)(x 3)(x 2 2x 2) .

2. Express

2x2 3x 2

in the mixed-number form – polynomial plus

x2 4

proper fraction.

Solution. Transform the nominator in the following form. x4 2x2 3x 2 x2 (x2 4) 2x2 3x 2

x2 (x2 4) 2(x2 4) 3x 10 .

Then

x4 2x2 3x 2	x2	(x2	4) 2(x2 4) (3x 10)	x	2	2	3x	10	.
x2 4			x2 4				x2	4

3. Transform the given fraction into linear factors.

3x 2 21x

x3 3x 2 6x 8

Solution. There are roots –2; 1 and 4 of Q3 (x) x3

3x 2 6x 8 . Then

Q3 (x) (x 1)(x 2)(x 4) . So,

3x 2 21x

3x 2 6x 8

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

x 1

hence

3x2 21x

A(x 1)(x 4) B(x 2)(x 4) C(x 2)(x 1)

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

Unknown constants А, В and С are determined from expression: 3x2 21x A(x 1)(x 4) B(x 2)(x 4) C(x 2)(x 1) .

By theorem 5, an identity is an equation that is true for all values of the variable, namely

A+B +C =3 ; х: -5A-2B +C =-21 ; x0 : 4A-8B -2C =0,

we have A 3 ,

B 2 , C 2 .

Thus,

3x 2 21x

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

x 2

x 1

x 4

4. Express

x 4 3x 2 3x 2

in the mixed-number form.

x3 x 2 2x

Solution. Step 1. Since the degree of the numerator is not less than the degree of the denominator, carry out the integer part of a fraction:

x 4 3x 2 3x 2

x(x3

2x) x3 x2

3x 2

x3 x 2 2x

x3 x2 2x

x(x3

x2 2x) (x3

x2 2x) x 2

x 1

x 2

x3 x2

x 2

Step 2. If

there

are

such distinct linear factors, there

will

n such

partial fractions. In

our

case

x 2

and

have 3

x3 x2 2x

x x 2 x 1

distinct linear factors:

A(x 2)(x 1) Bx(x 1) Cx(x 2)

x 2

x(x 2)(x 1)

x 1

x(x 2)(x 1)

Therefore,

x 2 A(x 2)(x 1) Bx(x 1) Cx(x 2) .

Although condition is not defined for x = 0, x = – 1, and x = 2, we want to find values for A, B, and C that will make the last equation true for all values of x. That is, it will be an identity. By successively setting x equal to any three

different numbers, we can obtain they by letting x be the roots of Qm(x) = 0, in

our case x = 0, x = –1, and x = 2. If x = 0 we have

2 2A , then

A 1 ; if

x 2 , then

4 6B and B 2 / 3 . The

last,

if,

x 1 ,

then 1 3C and

C 1/ 3 . Thereby

x 2

x(x 2)(x 1)

3(x 2)

3(x 1)

Namely,

x 4 3x 2 3x 2

= x +1+

x3 x 2 2x

3(x -2)

3(x +1)

5. Indicate the form of the partial-fraction representation of	x2 2
		.
	(x 1)(x 1)2

Solution. The degree of the numerator is less than the degree of the denominator (which is 3). Since the denominator is already factored (a common

occurrence in practice), the denominator

(x 1)(x 1)2

has no second-degree

irreducible factors.

Since x + 1 appears twice in the factorization and x –

1 once,

x 2 2

(x 1)(x

1)2

x 1

(x 1)

To find the constants A, B and C remove the denominators by multiplying both sides by (x 1)(x 1)2 :

x 2 2 A(x 1)2 B(x 1)(x 1) C(x 1) .

Three equations are needed to find the three unknowns A, B and C. To obtain them, replace x by three different numbers in turn. Since x + 1 when x =

– 1, and x – 1 when x = 1, replace x by – 1 and then by 1. To obtain the third

equation, use x = 0. Thus

3 = 4A

setting x = 1;

3 = – 2C

setting x = – 1;

2 = A – B – C

setting x = 0.

Namely A

, C

and B

1 .

Thus,

x 2 2

1)(x

1)2

4(x 1)

2(x 1)2

6. Indicate the form of the partial-fraction representation of

(x 1)(x3 1)

Solution. As is generally known (x 1)(x3

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

(x 1)2 (x 2

x 1) . Since x

appears

twice

in the factorization and

x 2 x 1 is irreducible, we seek constants A, B, C, and D such that

x 2

Cx D

(x 1)2 (x 2

x 1)

(x 1)2

x 1

x 2 x 1

Then

x 2

A(x 2

x 1) B(x 2

x 1)(x 1) (Cx D)(x 1)2 or

x2.= (B + C)x3 + (A + 2C +D)x2 + ( – A + C + 2D)x + (A + B + D)

That is, coefficients of x3, x2 and x0 in both parts will be an identity.

Therefore

x3 : B + C = 0

x2 : A + 2C + D = 1

x : – A + C + 2D = 0 x0 : A + B + D = 0.

Thereby A 13 , B 13 , C 13 , D 0 .

Hence,

		x 2			1	1		1			x
												.
	2		2				2			2		.
(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

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