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Национальный авиационный университет

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DDR 3 p.132-189

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Topic 5. Integration of irrational functions

A rational function of x and a x2 ,

a x2 ,

x2 a , n ax b and so

on can be integrated by using the algebraic and trigonometric substitutions. Integration of binomial differentials. Euler substitutions. Method by Ostrogradskiy.

Literature: [1, section 6, ch. ch. 6.6], [2, section 2, ch. 2.1], [3, гл. 7, §1], [4, section 7, §22], [6, section 8], [7, section 10, §§10–11], [9, 1 part, §33].

T5 Main concepts

5.1. Integration of the quadratic irrationality.

Consider the integrals

dx	,	Ax B	dx , ( a 0 , D 0 ).
ax 2 bx c		ax 2 bx c

These integrals can be found by the method illustrated in Examples 5, 6, 8 and 9 (see topic 3). First complete the square of the denominator and use

the substitution

x +

. The first integral in this case transforms to the

form:

C ,

if a 0 ,

z z 2 m2

z 2 m2

arcsin

C , if

a 0 ,

and the second integral is expressed as the sum of simpler integrals.

5.2. Integration of

ax b n

ax b s

R x, x n

,..., x s ;

R x, ax b n ,..., ax b s

,...,

;

cx d

Any rational function of x and x n ,..., x

can be transformed into rational

function of t by the substitution x tk , dx ktt 1dt.

Here k is the smallest

multiple of the fractional denominators

m ,...,

132

A rational function of x and ax b n ,..., ax b

can be transformed

into rational function of t by the substitution ax + b = tk, dx =

tk – 1dt. Here

k is the smallest multiple of the fractional denominators

m ,...,

ax b n

ax b s

The function R x,

,...,

.can be transformed into

cx d

rational function of t by the substitution

ax b

. Here k is the smallest

cx d

multiple of the fractional denominators mn ,..., rs .

5.3. Integration of binomial differentials xm(a + bxn)pdx .

Here m, n and p are rational numbers.

Case 1: p is an integer number. Let x = tk. Here k is a less general denominator of fractional powers m and n.

Case 2:	m 1	is an integer number, let a + bxn = tk. Here k is a
	n

denominator of fraction p.

Case 3:	m 1	p	is an integer number. Let	a bxn	t	k	. Here k is a
	n			xn

denominator of fraction p.

5.4. Integration of R x, ax2 bx c .

We shall consider three methods:

The first Euler substitution ax2 bx c t x a, a 0 .

The second Euler substitution

ax2

bx c tx

c, c 0 .

The third Euler substitution

ax2 bx c t

x x

(if x1 is the root of

equation).

5.5. Integrals in the form

(x m)

bx c

The substitution

x m 1

turns this integral into the easy integral.

133

5.6. Integrals in the form				f x dx				.
			F x ax			2	bx c

Here	f (x)	is a rational function of a single variable. This integral can

	F(x)
be computed by partial fractions of				f (x)	.

				F(x)
5.7. Three trigonometric substitutions.
Integrals in the form R(x,			ax 2 bx c )dx can be found by using a

trigonometric substitution, if the integrand is a rational function of x and:

1.	a x2	let	x	a sin t.
2.	a x2	let	x	a tg t.
3.	x2 a let		x	a sect.

The motivation behind this general procedure is quite simple.

1. If you replace x in		a x2 by	x a sin t you obtain:
a x2	a	a sin t 2	a a sin2 t	a 1 sin2 t

a cos2 t a cost.

The important thing is that the square root sign disappears.

2.	a x2	a a tg2 t			a	1 sin2 t
							cos2 t
			a	cos2 t sin2 t					a	a sec t.
					cos2 t			cos t

3.	x2 a	a	a		a	1 cos2 t			a tg t.
		cos2 t					cos2 t

	T5.	Typical problems
Find the integrals
1.		dx	.
		1 x 2x2

Solution. Complete the square:

134

	1 x 2x2 2(x2																x				1 ) 2[(x 1 )2														1					1		] 2[					9		(x 1 )2							] .
	Then														2						2								4					16						2							16						4
	Then				dx																		dx											1										d(x 1/ 4)
					dx																		dx											1										d(x 1/ 4)
		1 x 2x2															2[		9			(x 1 )2 ]												2							3					2						1 2
		1 x 2x2															2[					(x 1 )2 ]																			3											1 2
															16												4																			x
															16												4														4											4
																																									4											4
												1		arcsin							x 1/ 4								1			arcsin					4x 1								C .
												2									3 / 4								2											3
2.						(x 2)dx										.
2.																.
				4x 2 4x 17
	Solution. Given the derivative																										(4x 2 4x 17) 8x 4 ,																							the			numerator
transforms to the form:
	x 2					1 (8x 4)									4			2					1 (8x +4) -										5	=		1			(4x2 +4x +17)														¢-	5	.
	x 2					1 (8x 4)									8			2					1 (8x +4) -											=		8			(4x2 +4x +17)														¢-	2	.
	Then					8									8								8						2							8																		2
	Then																				1								5
																					1								5
				(x 2)dx											(						8 (8x						4)				)dx			1							d(4x 2 4x 17)
				(x 2)dx											(						8 (8x						4)			2	)dx			1							d(4x 2 4x 17)
																																		8
			4x 2 4x 17																			4x 2 4x 17												8									4x 2 4x 17
	5								dx											1					4x 2 4x 17													5								d(2x 1)
	5																			1					4x 2 4x 17											4
		2					4x	2	4x					17							4															4								(2x 1)						2	16
							4x		4x					17																														(2x 1)							16
					1				4x 2 4x 17																5																							C.
																									5	ln		2x 1 4x 2 4x 17
																										ln		2x 1 4x 2 4x 17
						4																	4
	We			used							the				given									table						integrals:															du						u		C			and
	We			used							the				given									table						integrals:													2								u		C			and
																																											2			u
ò	du																							+C .
	du						=ln			u + u2 a2
	u2 a2						=ln			u + u2 a2
	u2 a2
3.						3 x						dx .
3.				x		3		x	2			dx .
				x				x
	Solution. The smallest multiple of 2 and 3 is 6. This suggests the
substitution						x t 6 ,							dx 6t 5 dt . Then

135

t 2

dt 6

t 4

dt 6

(t 4 1) 1

t 1

x 2

t 1

(t 2

1)(t 1)(t 1) 1

6 (t 2 1)(t 1)dt

t 1

6 (t 3 t 2 t 1)dt

t 4

2t 3 3t 2 6t 6 ln

t 1

x 2

2 x 3 3

x 6 6 x 6 ln

6 x 1

C .

4. I

3 x

dx.

t18 t4 12t11dt

Solution.

x t12 ;

dx 12t11dt;

6t3

2 t

2x2 12 x3

2x12 x

2x2 4 x

2x12 x

5. I

3x 1 x

dx.

3x 1

3x 1 t6

Solution.

dx 2t5dt

3x 1 6

3x 1

3x 1 3 3x 1 2

3 3x 1 2 C.

x 2

1 x

dx .

Solution. Let1 x t 4 , then

x t 4 1 ,

dx 4t 3 dt .

136

Consequently,

x 2

1 x

(t 4

1)2

t 2

dt 4 ((t

4 (t10 2t 6 t 2 t 4 )dt

t11

8 t 7

t 3

t 5

(1 x)11

8 4

(1 x)7

(1 x)3

(1 x)5 C .

4 (x 1)3 (x 2)5

Solution. In

this

case

4 (x 1)3 (x 2)5 (x 1)(x 2) 4

x 2

and an

x 1

integrand is a rational function of

and 4

x 2

. The substitution

x 2

t 4

x 1

is appropriate:

t 4

x 1

x 2

3t 4

12t 3dt

Then

t 4

t4 1

4 1

(t 4 1)2

4 (x 1)3 (x 2)5

1)(x

2) 4

x 2

t 4 1

t 4

12t 3

x 1

C .

x 2

8.I x 2 x 1dx.

x x 2


	x 2		x 1
Solution. I	x 2		x 2	dx


	x	x 2
		x 2


1	x 1
	x 2	dx


	x

137

		x 1		t2 x 1 xt2 2t2
		x 2		t2 x 1 xt2 2t2																													1 t t t2 1 dt
		x 2
		xt		2	x							2t		2				1 x					2t2 1							6
		xt			x							2t						1 x					t2		1					6			(2t2			1)(t2						1)2
		dx								6tdt								;
		dx						t2 1 2										;
					6												t 1 tdt										6						tdt
					6					(2t2 1)(t 1)(t 1)																	6				(2t2 1)(t 1)
			2t 1								1												2
			2t 1								1												2
2															dt ln							2t		1					2 arctg(				2t) 2 ln								t 1		C
2			2t2	1					t		1				dt ln							2t		1					2 arctg(				2t) 2 ln								t 1		C
																										2
																																		2(x 1)
												3x									x		1
												3x									x		1																		C.
						ln																		1							2arctg
						ln																		1							2arctg
											x 2										x 2														x 2
											x 2										x 2														x 2
Integration of the binomial differentials
9.		x (3 3 x )2 dx .
Solution. Multiplying the integrand we get
																		1						1				2						1						5		7
	x (3 3 x )2 dx x 2 (9 6x 3 x																											3	)dx (9x 2 6x 6 x 6 )dx
																		3						11							13
															6x					36 x					6			6		x 6 C .
																			2									6
																			2									13
																				11								13
10. I						dx											.
10. I					x 3 x 1 2												.
Solution. This p is an integer number and is equal to																																					2. Thus:
			I x								1					1			2								x t			6	;		6			t	2	dt
											1					1			2											6							2
											2					3
													x					1				dx
													x					1				dx								5	dt;				1	t			2	2
																									dx 6t						dt;				1	t
							3t					3arctgt C															36 x					3arctg 6					x C.
					1		t2					3arctgt C																				3arctg 6					x C.
					1		t2																			1 3				x

138

	5					xdx
11. I												.
11. I					25 x3							.
	3				25 x3
Solution. Multiplying the integrand we get
																5	xdx								1					3		1
									I							5	xdx						x5 (3					2x5 )2 dx.
									I							3 25 x3							x5 (3					2x5 )2 dx.
																3 25 x3
Here m = 1					; n		3				and					m 1 2							– an integer number.
	5						5									n
																								3														5
																																						5
																																	3 t				2 3
					1								3			1			3 2x					5	t		2						3 t						;
					1								3	2					3 2x						t			x							2				;
Thus I x					5					x			5			dx																	2		2
Thus I x						3 2				x						dx																	2
																											5			t		2		3
																							dx				5		3	t					dt;
																							dx				3	t		2					dt;
																											3			2
										1																	2
						3 t2 3									1			5					3 t2									5
						3 t2 3									1			5					3 t2				3					5						2
													t					t										dt					3 t dt
						2							t					t						2				dt				6	3 t dt
						2												3						2								6
				5		t	5				t3		C				5			3			25 x3						3	5		3 25 x3 C.

				2				18									18													2
12.			dx						.
12.		x11 1 x 4							.
		x11 1 x 4
Solution. Transforming the integrand we get
						I										dx				x 11 (1 x 4 ) 12 dx .
						I				x11						1 x 4				x 11 (1 x 4 ) 12 dx .
										x11						1 x 4
Here		p 1				,	m 11, n 4 . Thereby																						m 1 p 3 is the integer
	2																												n
number, we have the third case. Transforming the integrand we get
					I											dx													dx							.
					I				x11 x 2																	x13										.
									x11 x 2									x 4 1								x13				x 4 1
This suggests the substitution x 4																					1 t 2 , 4x 5 dx 2tdt . Therefore,
x 4	t 2 1,					x 5 dx 1 tdt .																Multiplying									both						numerator and
																2

denominator by x 5 , we get

139

x 5 dx

2 tdt

x 4 1

(x 4 ) 2

x 4 1

(t 2 1) 2 t 2

(t 2 1)2 dt

1 (t 4

2t 2 1)dt

t 5

1 t 3

1 t C

4 5

x 4 3

1 1 x 4

C .

Euler substitutions

13.

x2 2x 2

Solution. Here a 1 0 . We can use the substitution

x 2 2x 2 t x .

Then

2 t

2tx x

, 2x 2

2tx ,

t 2 2

2(1 t)

t 2 2t 2

dt , 1 x 2 2x 2 1 t

t 2 2

t 2

2(1 t)2

2(1 t)

Thereafter we get an integral

( t 2

2t 2)

2(1 t)

t 2 2t 2

dt .

x2 2x 2

2(1 t)2

( t 2 )

(1 t)t 2

To express the integrand as the sum of partial fractions:

2 2t 2

(1 t)t 2

t 2

1 t

With equality

t 2 2t 2 A(1 t) Bt(1 t) Ct 2

computing the unknown coefficients:

t 2 : B C 1 ,

t 0 : A 2 ;

t 1: C 1;

B 0 .

Consequently,

t 2 2t 2

dt 2

t 1

C .

1 t

(1 t)t

140

Replacing t by

x2 2x 2 x . Thus

1 x x 2 2x 2

C .

1 x 2 2x 2

x 2 2x 2

14. I

9 x2

dx.

x 2

Solution. Here the integrand is polynomial in which a 1 0 and c =

9. We use the second Euler substitution:

9 x2

tx 3

9 x

6xt

9 x

6 t

6 t2 1

dt.

t2 1

t2 1 2

Hence

6t2

1 dt

3 t

I 6

dt.

1 t

Multiplying the integrand we get

I 9

9 t

3 t

9 t

2t 3

4arctgt C,

when t

9 x2

5 ln

2t 3 5

t2 1

15. I

(x 1) x

2 4x 2

Solution. This suggests the substitution

x 1

1 and

, hence

t 2

arcsin

t 1

C =

1 2t t2

2 (t 1)2

141

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