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Ax + B

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 = ∫ ax2 + bx + c dx .

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)

+ B −

J 2 =

dx =

∫

dx =

∫ ax 2

+ bx + c

ax 2 + bx + c

∫

(2ax + b)

∫

dx +

−

+ bx + c

∫

d(ax2 + bx + c)

+ B −

J1 =

+ bx + c

+ B

−

+ bx + c

where an integral

is as seen above.

Mx + N

Namely, the integral of the third form

∫

( p

− 4q < 0 ) is

+ px + q

equal to

∫

Mx + N

x + p 2

dx =

ln(x

+ px + q) + N

−

arctg

+ C.

+ px + q

−

q −

ІV.

Integral

Mx + N

dx ,

where

n > 1

and

p 2 − 4q < 0 .

∫ (x2 + px + q)n

Introduce t = x +

. Then dt = dx and:

In = M ∫

∫

p 2

dt +

−

= q −

+ a

)

+ a

)

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

101

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2.3. 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 of
	Qm (x)

simpler (partial) fractions follow these steps:

•іf the degree of Pn (x) is not less than the degree of Qm (x) divide Pn (x)

by Qm (x) to obtain a quotient and remainder:

Pn (x) = Pn−m (x) Qm (x) +Pk(x).

Where the degree of Pk(x) is less than the degree of Pn (x)

or else Pk(x) = 0.

Then

Pn (x)

= Pn−m (x) +

Pk (x)

k < m .

Qm (x)

Pk (x)

Apply the remainder steps to

Qm (x)

• іf 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;

• іf 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

Т.3

∫

+ 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

(x + 1)

+ 1

∫

+ 4x + 3

102

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Solution. Since 4x 2 + 4x + 3 = (2x + 1)2 + 2 і

dx =

d (2x + 1) , therefore

∫

d(2x + 1)

2 =

arctg

2x +1

+C .

+ 4x + 3

(2x + 1)

+ ( 2)

2 2

∫

− 6x + 1

D = 36 − 4 9 = 0 , therefore

Solution.

Given

the

discriminant

9x 2 − 6x + 1 = (3x − 1)2

and

∫

= ∫

∫

d(3x − 1)

= −

+ C .

− 6x + 1

3(3x − 1)

(3x − 1)

∫

− 2x − 8

Solution.

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

∫

= ∫

d(x − 1)

x − 1− 3

+ C =

x − 4

+ C .

− 2x − 8

− 1)

− 3

x − 1+ 3

x + 2

The second method. Since

x 2 − 2x − 8 = (x + 2)(x − 4) ,

1 (x + 2) − (x − 4)

1 1

−

(x + 2)(x − 4)

6 (x + 2)(x − 4)

Then

6 x − 4

x + 2

∫

−

∫

x − 4

−

x + 2

+ C =

x − 4

x + 2

− 2x − 8

x − 4

+ C .

x + 2

5. ∫

+ x − 1

Solution. Complete the square:

2x 2 + x

− 1 = 2(x 2 +

−

) = 2((x + 1/ 4)2 − 1/16 − 1/ 2) =

= 2(x + 1/ 4)2 − (3 / 4)2 ) .

Then

103

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							∫	dx
								2x 2 + x − 1
=	1		1		4		x + 1/ 4 − 3 / 4
						ln
	2						x + 1/ 4 + 3 / 4
		2		3

∫

d(x + 1/ 4)

(x + 1/ 4)

− (3 / 4)

+ C =

x −1/ 2

+ C =

2x −1

+C .

x + 1

x +1

∫2x − 3

6.12x − 9x 2 − 2 dx .

Solution. Given the derivative

(12x − 9x 2 − 2)′ = 12 − 18x , the numerator is

transformed to the form:

2x − 3 = −

(12 − 18x) +

− 3 = −

(12 − 18x) −

Then

∫

2x − 3

dx =

−

∫

(12 − 9x 2 − 2)′

−

∫

12x − 9x

− 2

12x − 9x

− 2

= −

12x − 9x 2

− 2

J1 ,

d(x −

)

3x − 2 − 2

where J1

+ C .

∫

2 2 3x − 2

−

3 x +

x −

−

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

7. Find ∫

x3dx

x − 5

Solution:

(x − 5)(x2 + 5x + 25)+ 125

∫

x3dx

= ∫

x3 − 125 + 125

dx =

∫

dx =

x − 5

125

= ∫ x

+ 5x

+ 25

dx =

25x + 125ln

x − 5

+ C.

x − 5

8. I = ∫

3x 2 − 21x

dx .

−

− 6x + 8

Solution. We have the integrand

P (x)

3x 2 − 21x

Q3 (x)

x3 − 3x 2 − 6x + 8

104

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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 +

x − 1

x −

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

3x 2 − 21x

−

(x + 2)(x − 1)(x − 4)

+ 2

x − 1

x − 4

Hence,

I = 3∫

2∫

− 2∫

= 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.

+ 7x − 2

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 − 8x

x2 + 3x

– 3x3+21x2 − 6x

−2x − 7

is the remainder.

Thus,

10x3 + 19x2 − 8x − 7

= x

+ 3x −

2x + 7

x2 + 7x − 2

+ 7x − 2

Then

2x + 7

3x2

2x + 7

t = x2 + 7x − 2

∫ x

+ 3x −

dx =

− ∫

dx =

dt = (2x + 7)dx

+ 7x − 2

−∫

3x2

− ln

+ C =

3x2

− ln

x2 + 7x − 2

+ C.

10.

I = ∫

x 4 − 3x 2 − 3x

− 2

dx .

− x

− 2x

Solution. The degree of Pn (x) is not less than the degree of Qm (x)

Pn (x)

by Qm (x) to obtain a quotient and remainder:

105

32x2 −

. Divide

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By Example 4 (see topic 2):

x 4 − 3x 2 − 3x − 2

= x +1+

−

x3 − x 2 − 2x

3(x −2)

3(x +1)

Then

I =

(x + 1)2 + ∫

−

∫

−

∫

(x + 1)2

+ ln

−

x − 2 3

x + 1

− 32 ln x − 2 − 13 ln x + 1 + C .

∫x 2 + 2

11.(x − 1)(x + 1)2 dx .

Solution. By Example 5 (see topic 2)

x 2 + 2

−

(x − 1)(x + 1)2

4(x − 1)

4(x + 1)

2(x + 1)2

Thus

∫

x 2 + 2

∫

−

∫

− 1)(x +

x − 1

(x + 1)

x + 1

x − 1

x + 1

+ C .

x + 1

I = ∫

+ 9x3 − 36x2 − 16x − 20

12.

x2 (x − 5)(x2 + 2x + 2)

dx .

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

x4 − 11x3 − 3x2 − 16x − 20

A B

Dx + E

x2 (x − 5)(x2 + 2x + 2)

x − 5

x2 + 2x + 2

A(x3

− 3x2

− 8x −10)

B(x4 − 3x3 − 8x2 −10x)

C(x4 + 2x3 + 2x2 )

x − 5

(B + C + D)x4 = 1;

(Dx + E)(x

)

(A − 3B + 2C − 5D + E)x3 = −11;

− 5x

(− 3A − 8B

+ 2C − 5E)x

= −3;

+ 2x + 2

(− 8A − 10B)x = −16;

− 10A = −20.

106

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= 2;

C + D = 1;

= 0;

2C − 5D + E = −13;

2C − 5E = 3;

C = −1;

B = 0;

D = 2;

= −1.

A = 2.

Then I = ∫

2x − 1

−

dx = −

− ln

x − 5

+C1 +

−

+ 2x

d (x2 + 2x + 2)

+ ∫

2x + 2 − 3

+ C1 + ∫

dx = −

− ln

x − 5

−

x2 + 2x + 2

−3∫

= −

− ln

x − 5

+ ln (x

+ 2x + 2)− 3arctg (x + 1) + C.

(

)

x + 1

+ 1

13. I = ∫

x 2

dx .

(x + 1)(x

+ 1)

Solution. By Example 6 (see topic 2)

x 2

−

(x +

− x + 1)

(x + 1)

x + 1

− x

+ 1

Thus

xdx

∫

−

∫

x + 1

(x + 1)

− x + 1

= −

−

x + 1

I1 ,

where

3(x + 1)

xdx

x −

= t,

(t +

)dt

∫

∫ x 2 − x + 1

−

)

dx = dt

= ∫

tdt

∫

ln(t2 +

) +

arctg

+C =

ln(x2 −x +1) +

arctg

2x −1

+C .

107

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Thus

I = −

−

x + 1

ln(x2 −x +1) +

arctg

2x −1

+C .

3(x + 1)

3 3

14. ∫

x2 − 2x + 2

dx .

− 3x

+ 6x − 7

Solution. In this case

x2 − 2x + 2 =

(x3 − 3x2 + 6x − 7)′ . Then

∫

x2 − 2x + 2

dx =

∫

d (x3 − 3x2 + 6x − 7)

− 3x

6x − 7

− 3x

+ 6x − 7

ln | x3 − 3x2 + 6x − 7 | +C .

Т.3 Exercises for class and homework

In Exercises 1 to 9 complete the square to find the integral.


1.	∫				dx			.			2.	∫				dx			.			3.		∫				dx	.
1.	∫			x	2			.			2.	∫	x	2				4	.			3.		∫	x	2	+ 8x + 20		.
				x		+	6x						x	− 4x +				4							x		+ 8x + 20
4.	∫			(x − 2)dx						.	5.	∫	(2		x + 3)dx					.		6.		∫				dx	.
4.	∫			2						.	5.	∫	2							.		6.		∫	2x		2	+ 5x − 7	.
				x		− 4x + 3							x		+ 2x + 10										2x			+ 5x − 7
7. ∫			(3x + 4)dx						.		8. ∫					e x dx					.	9. ∫				cos xdx				.
7. ∫		3x			2	− x − 4			.		8. ∫		e	2x		− 2e	x	+ 6			.	9. ∫	sin		2					.
		3x				− x − 4							e			− 2e		+ 6					sin			x − 2 sin x + 2

Express the integrant as the sum of partial fractions and determine the resulting integral.

10.

∫

11. ∫

x3dx

(x − 2)(x − 3)

(x + 2)(x − 4)

12.

∫

x 2 − 4x

dx .

13. ∫

(x2 − 10x + 7)dx

+ 5x −

(x + 1)(x − 2)(x − 5)

14.

∫

(2x − 3)dx

15. ∫

−

+ 2x

+ x

(x − 3)

16.

∫

17. ∫

1− 2x + 5x 2 − 2x

dx .

1)(x

(x − 1)

+ 1)

108

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18. ∫

1+ x3

dx .

19. ∫

1+ x 4

dx .

− 4x + 5)

− x

+ x − 1

20. ∫

21. ∫

+ x

1+ x

22. ∫

(x3 − 6)dx

23. ∫

xdx

+ 6x

+ 8

(1+ x

)

+ x

)

Answers

+ C .

2. −

+ C .

arctg

x + 4

+ C .

x2 − 4x + 3

+ C .

+ 6

−

(3x − 4)8

x2 + 2x + 10

x + 1

+ C

x − 1

+ C .

arctg

2x + 7

x + 1

arctg ex − 1 + C . 9. arctg(sin x − 1) + C . 10. ln

x − 3

+ C . 11.

(x + 2)2 +

x + 2

x − 2

x − 4

+ C .

12. x −

x − 1

−

x + 6

+ C .

13. ln

(x + 1)(x − 2)/(x − 5)

+ C .

x − 3

x + 1

x +1

14.

−

+ C .

15. 2ln

−

+ C

16.

+ 1

x − 1

2(x − 1)

4 + x2

arctg

+C .

17. −

+ arctg x + C .

18.

ln | x2 − 4x + 3 | +

arctg(x − 2) −

3(x − 1)3

x −

−

3x − 5

+ C.

19.

x2 + x + ln

x − 1

− arctg x + C .

20. −

+ C .

2(x2 − 4x +

1+ x2

4x4

2x2

1+ x2

21.

(x + 1)2

arctg

2x + 1

+ C . 22.

arctg

+ ln

x2 + 4

−

3 2

arctg

x 2

+ C .

x2 − x + 1

2 + x2

23.

+ 1

−

+ C.

+ 4

18(x

+ 1)

12(x

Т.3 Individual test problems

3.1. Complete the square to find the integral.

3.1.1. ∫

. 3.1.2. ∫

3.1.3. ∫

− 5x + 4

− 4x + 10

− 7x + 3

109

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3.1.4. ∫

3.1.5. ∫

+ x −

+ 6x +

3.1.7. ∫

3.1.8. ∫

− 11x +

+ x + 2

3.1.10. ∫

3.1.11. ∫

+ 3x

− 5x + 6

3.1.13. ∫

3.1.14. ∫

− 8x

−

2x − x

3.1.16. ∫

3.1.17. ∫

+ 4x + 25

− 4x + 15

3.1.19. ∫

3.1.20. ∫

− 2x + 5

−

3.1.22. ∫

3.1.23. ∫

− 8x +

+ 8x +

3.1.25. ∫

3.1.26. ∫

− x +

+ 6x

+ 3

3.1.28. ∫

3.1.29. ∫

− 2x − x

+ 3x +

3.2. Complete the square to find the integral.

3.2.1. ∫

(x + 1)dx

2.2.2. ∫

+ 6)dx

+ 3x − 4

+ 2x + 1

3.2.4. ∫

xdx

3.2.5. ∫

+ 5)dx

+ 2x + 5

+ x − 2

3.2.7. ∫

+ 4)dx

3.2.8. ∫

(5x

− 2)dx

− 6x −

− 5x + 2

3.2.10. ∫

(x + 1)dx

3.2.11. ∫

(x − 4)dx

+ 4x + 10

− 2x − 3

3.2.13. ∫

(5x + 1)dx

3.2.14. ∫

xdx

− 4x + 6

+ 2x − 4

3.2.16. ∫

(2x − 1)dx

3.2.17. ∫

− x)dx

+ 4x − 3

x − 4x − 5

3.2.19. ∫

(2x − 1)dx

3.2.20. ∫

(x − 4)dx

+ x − 2x

+ x

3.2.22. ∫

− 3)dx

3.2.23. ∫

(2x + 3)dx

+ 4x + 5

+ 2x + 7

110

3.1.6. ∫

− 4x +

3.1.9. ∫

− 12x + 3

3.1.12. ∫

− 3 − 4x

3.1.15. ∫

− 6 − x

3.1.18. ∫

− 9x +

3.1.21. ∫

− 6x +

3.1.24. ∫

− 8x +

3.1.27. ∫

− 6x +

3.1.30. ∫

+ 5x +

3.2.3. ∫

(2x − 1)dx

− 2x − 1

3.2.6. ∫

(3x − 2)dx

+ 3x − 2

3.2.9. ∫

(4x − 1)dx

− 4x + 5

3.2.12. ∫

x + 8)dx

x + 6x + 10

3.2.15. ∫

(x − 3)dx

− 5x + 4

3.2.18. ∫

(2x − 1)dx

− 6x − 9

3.2.21.∫ (3x + 1)dx .

x2 − 4x − 2

3.2.24. ∫	(x		− 5)dx	.
	2x	2	+ x − 3

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