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19.	Find ∫				dx			.
19.	Find ∫				2			.
		1+ 9x																	dx
Solution. ∫					dx				=	1	∫		dx			=	1	∫	dx				=	1	arctg 3x + C .
Solution. ∫			1		+ 9x		2		=	9	∫	1/ 9		+ x	2	=	9	∫	(1/ 3)	2	+ x	2	=	3	arctg 3x + C .
			1		+ 9x					9		1/ 9		+ x			9		(1/ 3)		+ x			3
20.	Find ∫			x 2 dx		.
20.	Find ∫			2		.
			x		+ 1

Solution. As the power of the numerator is not less than that of the

denominator, we use long division:

x 2 dx

(x 2

+ 1) − 1

1−

dx =

∫ x 2

∫ x 2 + 1

∫

+ 1

x2 + 1

= ∫ dx − ∫

= x − arctg x + C .

21. Find ∫

5x + 3

Solution. ∫

ln | 5x + 3 | +C.

5x + 3

22. Find ∫cos3xdx .

Solution. ∫cos 3xdx =

sin 3x + C .

23. Find ∫(2x − 1)5 dx .

Solution. ∫(2x − 1)5 dx =

1 (2x − 1)6

+ C =

(2x − 1)6

+ C .

24. Find ∫

xdx

x + 2

(x + 2) − 2

Solution. ∫

xdx

∫

dx =

∫ dx − 2∫

x + 2

= x −2 ln

x +2

+C .

25. Find ∫

5dx

5 − (3x

+ 8)2

Solution. By formula 13 and (2.1): ∫

5dx

arcsin

3x + 8

+ C.

5 − (3x + 8)2

26. Find ∫

4dx

− 6x

+ 7

Solution. First perform algebraic manipulations on an integrand, as is shown:

						4x	2	– 6x + 7				2x −	3	2	19
											=			+		.
											=		2	+	4	.
Now it is easy to find the integral:													2		4
Now it is easy to find the integral:
		4dx				1 2						2x −	3			4		4x − 3
∫		4dx				1 2						2x −	2			4		4x − 3
						= 4				arctg				+ C	=		arctg		+ C.
	2x −	3	2	+	19	= 4	2		19	arctg		19		+ C	=	19	arctg	19	+ C.

												2
		2			4
		2			4

ІІ. Using substitution

27. Find ∫

2xdx

x2 + 7

Solution. Note that 2x is the derivative of x2 + 7. Introduce u = x2+7. Then

2xdx

du = 2xdx, and ∫

= ∫ u .

x2 + 7

Now it’s easy to find

∫ duu .

yields ln|x2 + 7|. Since x2 + 7 is always

Replacing u by x2 + 7 in ln|u|

positive we can omit the absolute-value bars:

∫

2xdx

= ln (x

+ 7)+ C.

x2 + 7

2xdx

x2 + 7

)

We can write this as

∫

= ∫

(

= ln (x2 + 7)+ C.

x2 + 7

28. Find ∫ x 2 − xdx .

Solution. Using the substitution

2 − x = t , we get

2 − x = t 2 ,

x = 2 − t 2

and

dx = −2tdt .

Then evaluate the new indefinite integral:

∫ x 2 − xdx = ∫(2 − t 2 ) t (−2t)dt = 2∫(t

4 − 2t 2 )dt =

t 5

−

t 3

+ C .

Replacing t by

2 − x we have

∫ x

2 − xdx = 2

(2 − x)5

− 4

(2 − x)3 + C .

29. Find ∫

− e

Solution. Using the substitution

1− e x

= t . In this case it’s more useful to

find dt : dt = −e x dx

and just as − e x

= t − 1 so dt = (t − 1)dx , or

dx =

− 1

Since

∫

= ∫

t − (t − 1)

d (t − 1)

dt = ∫

−

dt = ∫

− ∫

− e

t(t − 1)

t − 1

= ln | t − 1| − ln | t | +C = ln ex − ln 1− ex + C = x − ln1− e x + C .

30. Find ∫ 4 − x 2 dx .

Solution. Using the substitution x = 2 sin t , dx = 2 cos tdt . Thereby

4 − x 2 = 4 − 4 sin 2 t = 2 1− sin 2 t = 2 cos t .

Therefore

∫ 4 − x 2 dx = ∫ 2 cos t 2 cos tdt = 4∫cos2 tdt =
= 2∫(1+ cos 2t)dt = 2t + sin 2t + C .
Replacing t, we have:
sin t =	x	, t = arcsin		x	,		sin 2t = 2 sin t cos t =
sin t =		, t = arcsin			,		sin 2t = 2 sin t cos t =
2			2
= 2 sin t 1− sin 2 t			= 2		x			1− x 2	= x	4 − x 2 ,
					2			4	2
∫ 4 − x 2 dx = 2 arcsin							x	+ x	4 − x 2	+ C .
∫ 4 − x 2 dx = 2 arcsin						2		+ x	4 − x 2	+ C .
						2		2

Frequently the basic integration form (2.2) is useful. It is provided to perform the following algebraic manipulations of the differential.

dx = d(x + b) , b is a constant,


dx =	1		d (ax + b) ,		а, b are the constants,

	a
xdx =		1		d (x2 ) ,		dx	= d(ln x) ,
	2					x
cos xdx = d (sin x) ,					sin xdx = −d (cos x) ,
dx			= d (tg x) ,			dx			= −d (ctg x)
cos2						sin2
	x							x

31. Find ∫ x 1 + x2 dx .

Solution. As in our case xdx = 12 d (1 + x2 ) , for the original integral we can

state

∫ x 1 + x2 dx = 12 ∫ 1 + x2 d (1 + x2 ) .

				1						2				3
										2
The table integral is provided ∫u 2 dx =													u 2 + C , we get
The table integral is provided ∫u 2 dx =												3	u 2 + C , we get
		∫ x 1 + x2 dx =							1					1
											∫(1 + x2 )						d (1 + x2 ) =
																2
									2							2
				3										3
		=	1	2	(1 + x2 )					+ C =				1	(1 + x2 )								+ C .
			1	2				2						1						2
				3				2						3						2
		2		3										3
32. Find ∫ x3e− x4 dx .																			1
Solution. Finding d(− x4 ) = −4x3dx , then														x3dx = −					1			d (− x4 ) . Hence
Solution. Finding d(− x4 ) = −4x3dx , then														x3dx = −								d (− x4 ) . Hence
						1								4
	∫ x3e− x4 dx = −					1	∫e− x4 d(− x 4 ) = −											1			e− x4			+ C .
	∫ x3e− x4 dx = −						∫e− x4 d(− x 4 ) = −														e− x4			+ C .
		dx		4										4
33. Find ∫		dx		.
33. Find ∫	x 1	− ln 2 x		.
	x 1	− ln 2 x

Solution. As in our case d(ln x) =

dx , for the original integral we can state

d ln x .

∫

= ∫

x 1 − ln 2 x

1 − ln 2 x

We get the table integral ∫

= arcsin u + C . Therefore,

1 − u 2

= arcsin(ln x) + C .

∫ x

arctg x

1− ln2 x

34. Find ∫

dx .

1+ x

arctg2 x

Solution.

∫

arctg x

dx = ∫ arctg xd (arctg x) =

+ C .

1+ x

35. Find

∫ x2 x3 − 7dx .

Solution. We can write this as ∫ x2 (x3 −7)2 dx . Notice that the integrand

contains a power of the function x3 – 7. If u = x3 – 7 then du = 3x2dx.

Since the constant factor is 3 in du, the constant 7 does not appear in the integrand, this integral does not have the form of ∫un du . However we can put

the given integral in this form by first multiplying and dividing it by 3. This does not change the value. Thus:

													1			1				u = x3 − 7
		∫ x2 x3 − 7dx =											1		∫(x3 − 7)2 3x2dx =											=
		∫ x2 x3 − 7dx =											3		∫(x3 − 7)2 3x2dx =					du = 3x2dx						=
				1								3						3							3
		1		1			1			u 2						2		3		2			3		3
=		1	∫u 2 du			=	1			u 2				+ C =		2	u 2		+ C =	2		(x	3	− 7)2		+ C.
=		3	∫u 2 du			=		3		3				+ C =		3 3	u 2		+ C =	9		(x		− 7)2		+ C.
											2
																				3
Or ∫ x2 x3 − 7dx =					1	∫			x3 − 7d (x3 − 7)=									2	(x3 − 7)			+ C.
					1													2			2
					3													9			2
					3													9
36. Find ∫		4x3 − 3x							dx .
36. Find ∫	2x		4	− 3x	2	+	7		dx .

Solution.

If u = 2x4– 3x2 + 7 then du = (8x3 – 6x)dx which is two times greater then the numerator.

We multiply the integrand by 2 and compensate it by the factor 12 putting it before the integral sign.

1 2(4x3 − 3x)dx 1

(8x3 − 6x)dx

u = 2x4 − 3x2 + 7

1 du

∫

du = (8x3 − 6x)dx

∫

2x4 − 3x2 + 7

+ C =

2x4

− 3x2 + 7

+ C.

37.

Find

∫

sin xdx

2 + cos2 x

Solution.

∫

sin xdx

t = cos x

= −∫

(− sin x)dx

= −∫

2 + cos x

dt = − sin xdx

2 + cos2 x

= − ln

t + 2 + t2

+ C = − ln

cos x + 2 + cos2 x

+ C.

2 + t2

38.

Find

∫

2dx

x ln

Solution.

2dx

t = ln x

∫

dt =

2∫

= 2∫t−3dt = −

t−2 + C = C

−

x ln3 x

ln2 x

ІІІ. Using integration by parts

39. Find	∫ x cos xdx .
Solution.	To use the formula (2.1), it is necessary to write ∫ x cos xdx in the

form ∫udu . The integrand is so simple that there is not much choice. Try u = x and dv = cos x dx; that is break up the integrand this way:

∫x cosx dx u dv

Then find du and v. Since u = x, it follows that du = dx. Since dv = cos x dx, we choose v = sin x. Of course v could be sin x + C for any constant C, but choose the simplest v whose derivative is cos x. Applying integration by parts yields:

∫ x cos x dx = x sin x − ∫ sinx dx.

u dv

u v

v du

∫v du is simpler than the original integral ∫u dv.

∫ sin x dx = −cos x − C.

Hence ∫ x cos x dx = x sinx + cosx + C. Save the constant of integration till the end. The reader may check this by differentiation.

We denote: u = cos x; dv = x dx. Let us see what this method gives:

∫ x cos xdx =	u = cos x du = − sin xdx		= −	x2 cos x	+ ∫	x2	sin xdx.

		x2
	dv = xdx v =
			2		2
		2

The new integrand is harder than the one we started with. Though sinx is not

harder than cosx; x2 sin x is definitely harder than xcosx for the exponent has 2

increased from 1 to 2. 40. Find ∫ x ln xdx.

Solution.

Setting dv = lnxdx is not a wise move since v = ∫ln xdx is not immediately

apparent. But setting u = lnx is promising since du = dxx is much easier than lnx. This second approach goes through smoothly:

u = ln x du =

∫ x ln xdx =

− ∫

dv = xdx v =

x2 ln x

−

∫ xdx =

ln x −

+ C.

The result may be checked by differentiation.

Remark 1. The key to applying integration by parts is the correct choice of u and dv. Usually three conditions should be met:

First: v can be found by integrating and should not be too messy.

Second: du should not be messier than u.

Third: ∫vdu should be easier than the original ∫udv .

For instance ∫ x2e2x dx , with u = x2 and dv = e2xdx, meets these criteria. In this case v = e2x can be found and is not too messy; du = dx2 = 2xdx is easier than u as example 25 shows. ∫vdu is indeed simpler than ∫udv .

41. Find ∫ln xdx .

Solution.

∫ln xdx =

u = ln x du =

= x ln x − ∫ x

= x ln x − x + C.

dv = dx v = x

42. Find ∫ln2 xdx.

Solution. ∫ln2 xdx =

u = ln

x du

2ln xdx

= x ln2 x −

dv = dx v = x

− 2∫

x ln xdx

= x ln x2 − 2∫ln xdx.

By example 41 ∫ln xdx = x ln x − x + C. Thus: ∫ln2 xdx = x ln2 x − 2x ln x + x − C.

43. Find. ∫arctg 2xdx.

Solution. ∫arctg 2xdx =

u = arctg 2xdx du =

2dx

1+ 4x2

dv = dx v = x

= x arctg 2x

− ∫

2xdx

1+ 4x2 = t

=x arctg 2x −

∫

8xdx

1+ 4x

dt = 8xdx

1+ 4x

= x arctg 2x −

∫

= x arctg 2x −

+ C =

= x arctg 2x −

ln (1+ 4x2 )+ C .

44. Find ∫ x2e2 xdx.

Solution. ∫ x

2 x

u = x2

du = 2xdx

dx =

dv = e2 xdx v =

e2x

−

− ∫ 2 e

2xdx =

2 x

− ∫ xe

dx =

u = x du = dx

= 2 x

−

2 x

dv = e

2xdx v = 1 e

2 x

1 2

−

xe2x + ∫

e2x dx =

x2e2x −

xe2x +

e2x

+ C.

Integration by parts with u = x3 could be used to express

∫ x3e2xdx

in terms

of ∫ x2e2 xdx. Another integration by parts with u = x2, then expresses

∫ x2exdx

in terms of

∫ xexdx, as was done in Example 25. Each time integration by parts

lowers the exponent by 1.

45. Find ∫(x2 − 2x + 5)e− x dx .

dv = e− x dx , therefore

v = −e− x

Solution. In this case

u = x2 − 2x + 5 and

and du = (2x − 2)dx .Hence

∫(x2 − 2x + 5)e− x dx = −(x2 − 2x + 5)e− x + 2∫(x − 1)e− x dx .

Using another integration by parts to the last integral:

u = x − 1 , dv = e− x dx , du = dx , v = −e− x ,

∫(x − 1)e− x dx = − (x − 1)e− x + ∫e− x dx = −(x − 1)e− x − e− x + C1 = − xe− x + C1 . Thus

∫(x2 − 2x + 5)e− x dx = −(x2 − 2x + 5)e− x − 2xe− x + C = −(x2 + 5)e− x + C .

Remark 2. If P(x) is a polynomial of the form

∫ P(x)eαx dx = Q(x)eαx + C ,

where Q(x) is a polynomial which has the same degree of P(x) , then the

indeterminate coefficient method is used. The following example shows how to integrate in this case.

46. Find ∫(x3 + 18)e2x dx .
Solution.		∫(x3 + 18)e2x dx = (Ax3 + Bx 2 + Cx + D)e2x + C1 ,									where
A, B, C, D are unknown coefficients.
Differentiating both parts of the expression with respect to х:
(x3 + 18)e2x = (3Ax2 + 2Bx + C)e2x + 2(Ax3 + Bx2 + Cx + D)e2x ,
then
x3 + 18 = (3Ax 2 + 2Bx + C) + 2(Ax3 + Bx2 + Cx + D) ,
x3 + 18 = 2Ax3 + (3A + 2B)x 2 + (2B + 2C)x + C + 2D .
Since, the coefficients of the same degrees of х are equal and we have
		1 = 2A, 0 = 3A + 2B, 0 = 2B + 2C, 18 = C + 2D,
therefore A =	1	, B = −	3	, C =	3	,	D =		69	.
2		4			4		8
Hence							1	(4x3 − 6x2 + 6x + 69)e2x + C1 .
	∫(x3 + 18)e2x dx =						1
	∫(x3 + 18)e2x dx =						8

Remark 3. Reduction or recursion formulas. Many formulas in a table of integrals express the integral of a function that involves the n-th power of some expression in terms of the integral of a function that involves the (n – 1)-th or lower power of the same expression. These are reduction formulas. Usually they are obtainned by an integration by parts as examples 47 and 48 shows.

47. Find ∫cosln xdx .

Solution. Letting u = cosln x and dv = dx we have

du = − sin ln x	dx	, v = x ,	∫cosln xdx = x cosln x + ∫sin ln xdx .
	x

Using other integration by parts to the last integral:

u = sin ln x , dv = dx , du = cosln x dxx , v = x ,

∫sin ln xdx = x sin ln x − ∫cosln xdx .

Since

∫cosln xdx = x cosln x + x sin ln x − ∫cosln xdx , 79

therefore

∫cos ln xdx =

(x cos ln x + x sin ln x) + C .

48. Find ∫

(x2 + a)n

Solution. ∫

∫

a + x2 − x

dx =

(x2 + a)n

a + x2

x2dx

x xdx

∫

dx − ∫

∫

−

∫

(x2 + a)n

(x2 + a)n−1

(x2 + a)n

u = x du = dx

∫

xdx

v =

dv =

−

2(1− n)(x2 + a)n−1

(x2 + a)n−1

(x2 + a)n

∫

−

2(1− n)(x2 + a)n−1

2(1− n)

(x2 + a)n−1

−

∫

2a(1− n)(x2 + a)n−1

2a(1− n)

(x2 + a)n−1

2(1− n)+ 1

∫

−

2a(1− n)

(x2 + a)n−1

2a(1− n)(x2 + a)n−1

Hence

2n − 3

∫

(2.3)

(x2 + a)n

2a(n − 1)(x2 + a)n−1

2a(n − 1)

(x2 + a)n−1

This is a reduction formula.

For example, this formula is spelled in the following case.

J1 =

arctg

+ C is the table integral;

∫ x2 + a2

J 2 =

∫

J1 =

(x2 + a 2 )2

2a 2

x 2 + a 2

a 2

arctg

+ C etc.

2a 2

2a3

x 2 + a 2

49. Find J n, −m =

sin n x

dx .

∫ cosm x

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