Integration techniques

INTEGRATION TECHNIQUES

1. Review of formulas and techniques
Summary of known integral formulas.
ˆ

xr+1
x dx =
+ c, for r = −1 (power rule)
r+1

ˆ

r

ˆ

ˆ
sin x dx = − cos x + c

cos x dx = sin x + c

ˆ

ˆ
2

sec x dx = tan x + c

sec x tan x dx = sec x + c

ˆ

ˆ
2

csc x dx = − cot x + c

csc x cot x dx = − csc x + c

ˆ

ˆ
x

e−x dx = −e−x + c

x

e dx = e + c
ˆ

ˆ
√

tan x dx = − ln | cos x| + c
ˆ

1
dx = ln |x| + c for x = 0
x

ˆ

1
dx = tan−1 x + c
2
1+x

1
dx = sin−1 x + c
2
1−x

1
√
dx = sec−1 x + c
2−1
|x| x

Example 1.1. (A Simple Substitution)
ˆ
Evaluate sin(ax)dx, for a = 0.
Example 1.2. (Generalizing a Basic Integration Rule)
ˆ
1
Evaluate
dx, for a = 0.
2 + x2
a
Example 1.3. (An Integrand That Must Be Expanded)
ˆ
Evaluate (x2 − 5)2 dx.
Example 1.4. (An Integral Where We Must Complete the Square)
ˆ
1
√
dx.
Evaluate
−5 + 6x − x2
Example 1.5. (An Integral Requiring Some Imagination)
ˆ
4x + 1
Evaluate
dx.
2 + 4x + 10
2x
1

2


2. Integration by Parts
Remember that if u and v are differentiable functions of x, then u · v is also differentiable and
d
d
d
(u · v) = u · v + v · u.
dx
dx
dx
Using differential forms this may be written as
d(uv) = udv + vdu.
Taking the integral of the both sides of (1), we obtain
ˆ
ˆ
ˆ
d(uv) = udv + vdu.
Note that

´

(1)

(2)

d(uv) = uv. Therefore (2) can be written as
ˆ
ˆ
uv = udv + vdu,

or, equivalently,

ˆ

ˆ
udv = uv −

vdu.

Remark. This technique requires the separation of the integral into two parts as u and dv
(which includes the differential of integral, say dx) such that
• u gets simpler when differentiated.
• dv is easy to integrate.
Example 2.1. Evaluate the following integrals.
ˆ
(1)
x sin x dx
ˆ
(2)
x2 ex dx
ˆ
(3)
ex cos x dx
ˆ
(4)
x ln x dx
ˆ
(5)
ln x dx
ˆ
(6)
sin−1 x dx
ˆ
2
(7)
x3 ex dx
ˆ
(8)
x sec2 x dx
ˆ
(9)
x3x dx
ˆ
(10)
cos(ln x) dx
ˆ
(11)
x4 ex dx


ˆ
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)

ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ

3

√
ln x dx
√

e

x

dx

x2 sin x dx
xe2x dx
xe−x dx
e2x cos 3x dx
tan−1 x dx
2

ˆ

x5 e−x dx
−1

esin x dx
ˆ
√
(21)
x 1 − x dx
ˆ 2
(22)
x3 ln x dx
(20)

1

3. Trigonometric Integrals
3.1. Products of Powers of Sines and Cosines. If we are given an integral of the form
ˆ
sinm x cosn x dx, where m, n ≥ 0,
There are three possible cases:
Case 1. If m is odd, we write m as 2k + 1 and use the identity sin2 x = 1 − cos2 x to obtain
sinm x = sin2k+1 x = (sin2 x)k sin x = (1 − cos2 x)k sin x,
and, then

ˆ

ˆ
m

n

sin x cos x dx =

(1 − cos2 x)k cosn x sin x dx.

Using the substitution u = cos x, we get
ˆ
ˆ
ˆ
m
n
2
k
n
sin x cos x dx = (1 − cos x) cos x sin x dx = − (1 − u2 )k un du.
Now, the integral is easy to evaluate.
Case 2. If m is even and n is odd, we write n as 2k + 1 and use the identity cos2 x = 1 − sin2 x
and repeat what we have done in Case 1 changing the places of sine and cosine.
Case 3. If both m and n are even, we substitute
1 − cos 2x
1 + cos 2x
, cos2 x =
2
2
to reduce the integrand to one in lower powers of cos 2x.
sin2 x =

4


3.2. Integrals of Powers of tan x and sec x. We know how to integrate the tangent and
secant and their squares. To integrate higher powers we use the identities tan2 x = sec2 x − 1
and sec2 x = tan2 x + 1, and integrate by parts when necessary to reduce the higher powers to
lower powers.

3.3. Products of Sines and Cosines. To evaluate the integrals of the form
ˆ
ˆ
ˆ
sin mx sin nx dx,
sin mx cos nx dx, and
cos mx cos nx dx,
we may apply the method of integration by parts, but two such integrations are required in
each case. So, it is better to use the identities
1
sin mx sin nx = (cos(m − n)x − cos(m + n)x)
2
1
sin mx cos nx = (sin(m − n)x + cos(m + n)x)
2
1
cos mx cos nx = (cos(m − n)x + cos(m + n)x)
2
ˆ
(1)
cos4 x sin x dx
ˆ
(2)
cos4 x sin3 x dx
ˆ √
(3)
sin x cos5 x dx
ˆ
(4)
cos3 x dx
ˆ
(5)
sin3 x cos7 x dx
ˆ
(6)
sin2 x dx
ˆ
(7)
cos4 x dx
ˆ
(8)
sin2 x cos2 x dx
ˆ
(9)
tan3 x sec3 x dx
ˆ
(10)
tan2 x dx
ˆ
(11)
tan2 x sec4 x dx
ˆ
(12)
tan3 x dx
ˆ
(13)
tan3 x sec5 x dx


5

ˆ
(14)

sec x dx
ˆ √
tan x sec4 x dx
(15)
ˆ
(16)
cot4 x dx
ˆ
(17)
csc x dx
ˆ
(18)
cot3 x csc3 x dx
ˆ
sin5 x
√
(19)
dx
cos x
3.4. Trigonometric Substitutions. Trigonometric substitutions can be eﬀective in transforming complicated integrals involving a2 − x2 , a2 + x2 and x2 − a2 into integrals we can
evaluate directly.
The following table gives the suitable trigonometric substitution in each case and the trigonometric identity to be used.
Expression Trigonometric Substitution Related Trigonometric Identity
a2 + x 2
x = a tan θ
1 + tan2 θ = sec2 θ
2
2
a −x
x = a sin θ
1 − sin2 θ = cos2 θ
x 2 − a2
x = a sec θ
sec2 θ − 1 = tan2 θ
Remark. We want any substitution we use in an integration to be reversible so that we can
change back to the original variable afterward. For example, if x = a tan θ, we want to be able
x
after the integration takes place. As we know, the functions in these
to set θ = tan−1
a
substitution have inverses only for selected values of θ. For reversibility,
x
π
π
x = a tan θ requires θ = tan−1
with − < θ < ,
a
2
2
π
π
≤θ≤ ,
2
2

 0 ≤ θ < π if x ≥ 1,


2
a
−1 x
x = a sec θ requires θ = sec
with

a
 π < θ ≤ π if x ≤ −1,

2
a
To simplify calculations with the substitution x = a sec θ, we will restrict its use to integrals in
√
x
π
which ≥ 1. This will place θ in 0,
and make tan θ ≥ 0. We will then have x2 − a2 =
a
2
√
a2 tan2 θ = |a tan θ| = a tan θ, free of absolute values, provided that a > 0.
x = a sin θ

requires θ = sin−1

x
a

with −

Example 3.2.
(1) Evaluate the following integrals.
ˆ
dx
√
(a)
2
2
ˆ x 4−x
dx
√
(b)
2
ˆ √9 2+ x
x − 25
(c)
dx, for x ≥ 5.
x

6


ˆ
(d)

ˆ

(e)
ˆ
(f)

ˆ

(g)
ˆ
(h)
(i)
(j)

ˆ
ˆ
ˆ

dx
(1 + x2 )2
√
4 − x2 dx
√
x2 − 1
dx
x
dx
√
2 x2 − 2
x
√
9x2 − 4
dx
x
dx
2 + 2x + 10
x
√
ex 1 − e2x dx
√

cos x

dx
1 + sin2 x
dx
(l)
2 + 4x + 5)2
ˆ (4x
2x + 2
(m)
dx
2 − 4x + 8
ˆ x
5x + 3
dx
(n)
2 − 4x + 8
ˆ x
x2
(o)
5 dx
(x2 − 1) 2
(2) Find the volume of the solid generated by revolving about the x-axis the region in the
2
ﬁrst quadrant enclosed by the coordinate axes, the curve y =
, and the line x = 1.
1 + x2
(k)

ˆ

4. Integration of Rational Functions Using Partial Fractions
The method based on the fact that “every rational function can be written as a sum of simpler
fractions that we can integrate with the techniques we already know”.
General Description of the Method of Partial Fractions. To write a rational function
P (x)
as a sum of partial fractions, do the following:
Q(x)
(1) The degree of P (x) must be less than the degree of Q(x). (Otherwise, divide P (x) by
Q(x) and work on the remainder.)
(2) Factor out the polynomial Q(x).
• For each linear factor, assign the sum of m partial fractions to this factor where m
is the exponent of that linear factor as follows:
A1
A2
Am
+
+ ··· +
2
x − r (x − r)
(x − r)m
• For each irreducible quadratic factor (a polynomial of second degree that cannot
be written as a product of linear factors), assign the sum of n partial fractions to


this factor where n is the exponent of that quadratic factor as:
B1 x + C1
B2 x + C2
Bn x + Cn
+ 2
+ ··· + 2
2 + px + q
2
x
(x + px + q)
(x + px + q)n
Several Examples. We assume that the degree of P (x) is less than the degree of Q(x).
(1) Distinct Linear Factors:
P (x)
A
B
C
P (x)
=
=
+
+
.
Q(x)
(x − r1 )(x − r2 )(x − r3 )
x − r1 x − r2 x − r3
(2) Repeated Linear Factors:
P (x)
P (x)
B
A
C
D
=
+
=
+
+
.
4
2
3
Q(x)
(x − r)
x − r (x − r)
(x − r)
(x − r)4
(3) Some distinct, some repeated linear factors:
P (x)
A B
C
D
E
F
P (x)
+
+
= 2
= + 2+
+
.
3
2
Q(x)
x (x − r1 )(x − r2 )
x x
x − r1 x − r2 (x − r2 )
(x − r3 )3
(4) Distinct irreducible quadratic factors:
P (x)
P (x)
Ax + B
Cx + D
= 2
= 2
+ 2
.
2+p x+q )
Q(x)
(x + p1 x + q1 )(x
x + p1 x + q 1 x + p2 x + q2
2
2
(5) Repeated irreducible quadratic factors:
P (x)
P (x)
Ax + B
Cx + D
Ex + F
= 2
= 2
+ 2
+ 2
.
3
2
Q(x)
(x + p1 x + q1 )
x + p1 x + q1 (x + p1 x + q1 )
(x + p1 x + q1 )3
(6) General Case:
P (x)
P (x)
=
2 (x2 + p x + q )2 (x2 + p x + q )
Q(x)
x(x − r)
1
1
2
2
A
B
C
Dx + E
Fx + G
Hx + I
= +
+
+ 2
+ 2
+ 2
.
2
2
x x − r (x − r)
x + p1 x + q1 (x + p1 x + q1 )
x + p2 x + q 2
ˆ
1
(1)
dx
2+x−2
ˆ x 2
3x − 7x − 2
(2)
dx
x3 − x
ˆ
2x3 − 4x2 − 15x + 5
(3)
dx
x2 − 2x − 8
ˆ
5x2 + 20x + 6
(4)
dx
3
2
ˆ x 2 + 2x + x
2x − 5x + 2
(5)
dx
x3 + x
ˆ
5x2 + 6x + 2
dx
(6)
2
ˆ (x + 2)(x + 2x + 5)
3x − 1
(7)
dx
(x + 2)(x − 3)

7

8


ˆ
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(20)
(21)
(22)
(23)
(24)

ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ

x2 + 1
dx
x(x − 1)(x − 2)(x − 3)
x
dx
(x − 3)2
x−1
dx
(x + 1)3
x2 + 1
dx
(x − 1)2 (x + 1)
x+1
dx
2 + 1)(x − 2)
(x
6x2 − 15x + 22
dx
(x + 3)(x2 + 2)2
2x + 1
dx
3x + 1
x5 − x4 − 3x + 5
dx
x4 − 2x3 + 2x2 − 2x + 1
x3 + 1
dx
x2 − x
sec x dx
x3
dx
x2 + 6x + 5
x4
dx
x4 − 1
dx
3−1
x
dx
x − e2x
e
sin x
dx
cos x + cos 2x
sin4 x
dx
cos3 x
dx
3

2 2
ˆ x(1 − x )
3
2x + 5x2 + 8x + 4
(25)
dx
(x2 + 2x + 2)2


9

Brief Summary of Integration Techniques.
ˆ
ˆ
• Integration by Substitution:
f (u(x))u (x) dx = f (u) du
What to look for
– Compositions of the form f (u(x)), where the integrand also contains u (x); for
example,
ˆ
ˆ
ˆ
2
2
2x cos(x ) dx = cos(x )2x dx = cos udu.
– Compositions of the form f (ax + b); for example,
ˆ
ˆ
x
u−1
√
√ du.
dx =
u
x+1
´
´
• Integration by Parts: u dv = uv − v du
What to look for: products of diﬀerent types of functions: xn , cos x, ex ; for example,
ˆ
ˆ
2x cos x dx = x sin x − sin x dx.
• Trigonometric Substitutions:
What to look √
for:
– Terms like a2 − x2 : Let x = a sin θ, − π ≤ θ ≤ π , so that dx = a cos θ dθ and
2
2
√
a2 − x2 = a2 − a2 sin2 θ = a cos θ; for example,
ˆ
ˆ
x2
√
dx = sin2 θ dθ.
2
1−x
√
– Terms like x2 + a2 : Let x = a tan θ, − π < θ < π , so that dx = a sec2 θ dθ and
2
2
√
√
x2 + a2 = a2 tan2 θ + a2 = a sec θ; for example,
ˆ
ˆ
x3
√
dx = 27 tan3 θ sec θ dθ.
2+9
x
√
2 − a2 : Let x = a sec θ, for θ ∈ [0, π ) ∪ ( π , π], so that dx =
– Terms like x
2
2
√
√
a sec θ tan θ dθ and x2 − a2 = a2 sec2 θ − a2 = a tan θ; for example,
ˆ
ˆ
√
3
2 − 4 dx = 32
x x
sec4 θ tan2 θ dθ.
• Partial Fractions:
What to look for: rational functions, for example,
ˆ
ˆ
ˆ
x+2
x+2
A
B
dx =
dx =
+
2 − 4x + 3
x
(x − 1)(x − 3)
x−1 x−3

dx.

10


5. Improper Integrals

ˆ

2

Example 5.1. Evaluate
−1

1
dx.
x2

When studying the definite integrals, we required two things. First, the domain of integration
(from a to b) [a, b], be finite. Second, function is finite on the domain of integration.
Question: What happens if the domain of integration is infinite? What if function becomes
infinite in the domain of integration?
Answer: Improper integration!
The integrals of type described above are called improper integrals. If the limits exist, we
evaluate them with the following definitions
(1) If f is continuous on [a, ∞), then
ˆ ∞
ˆ b
f (x) dx = lim
f (x) dx.
b→∞

a

a

(2) If f is continuous on (−∞, b], then
ˆ b
ˆ b
f (x) dx = lim
f (x) dx.
a→−∞

−∞

a

(3) If f is continuous on [a, b) then
ˆ b
ˆ c
f (x) dx = lim
f (x) dx.
−
c→b

a

a

(4) If f is continuous on (a, b] then
ˆ b
ˆ b
f (x) dx = lim
f (x) dx.
+
c→a

a

c

In each case, if the limit exists and is finite we say that the improper integral converges and
the limit is the value of the improper integral. Otherwise the improper integral diverges.
Similarly, if f becomes infinite at an interior point d ∈ [a, b], then
ˆ b
ˆ d
ˆ b
f (x) dx =
f (x) dx +
f (x) dx.
a

a

d

This integral (on [a, b]) converges if both integrals (on [a, d] and on [d, b]) converges. Otherwise,
the integral from a to b diverges.
ˆ
ˆ
∞

a

Finally, if f is continuous on (−∞, ∞) and if
ˆ ∞
f (x) dx converges and its value is
−∞

ˆ

ˆ

∞

−∞

f (x) dx +
−∞

f (x) dx both converge, then
a

ˆ

a

f (x) dx =
−∞

f (x) dx and

∞

f (x) dx.
a

If either one or both of the integrals on the right-hand side of this equation diverge, the integral
diverges.
Example 5.2. In each part, determine whether the improper integral converges or diverges,
and find its value if it converges.


ˆ
(1)
(2)
(3)
(4)
(5)
(6)
(7)

1

dx
√
.
1−x
0
ˆ 0
1
dx.
2
−1 x
ˆ 1
1
√ dx.
ˆ0 2 x
1
dx.
1 x−1
ˆ 2
1
dx.
2
ˆ−1 x
∞
1
dx.
2
ˆ1 ∞ x
1
√ dx.
x
ˆ1
∞

(8)
(9)

e−x dx.

ˆ1 ∞
ˆ0

sin x dx.
∞

xe−x dx.

(10)
ˆ0 1
(11)
ˆ−1
2

(12)
(13)
(14)
(15)
(16)
(17)
(18)

1

(20)

dx.

1
√
dx.
2
ˆ0 3 4 − x
dx
2 .
0 (x − 1) 3
ˆ 1
dx
.
x
ˆ−1
1
dx
.
p
0 x
ˆ −1
1
dx.
−∞ x
ˆ 0
1
dx.
2
ˆ−∞ (x − 1)
∞
1
dx.
2
ˆ−∞ 1 + x
∞

(19)

1

x3

ˆ−∞
∞
ˆ−∞
∞

(21)
0

2

xe−x dx.
e−x dx.
1
dx.
(x − 1)2

11

12


ˆ
(22)
(23)

∞

ˆ1 ∞
ˆ−∞
∞

1
dx.
xp
e−|x| dx.
cos x dx.

(24)
−∞

Exercises 5.3. In each part, determine whether the improper integral converges or diverges,
and ﬁnd its value if it converges.
ˆ 1
dx
(1)
.
2
−1 x
ˆ 16
dx
√ .
(2)
4
x
ˆ0 1
dx
√
(3)
.
2
ˆ0 ∞ 1 − x
ln x
dx.
(4)
ˆ1 ∞ x
dx
(5)
.
1.001
ˆ1 1 x
dx
(6)
1 .
−8 x 3
ˆ 1
dx
(7)
.
0.999
ˆ0 ∞x
x dx
(8)
3 .
2
−∞ (x + 4) 2
ˆ ∞
16 tan−1 x
dx.
(9)
2
ˆ0 ∞ 1 + x
2e−θ sin θ dθ.
(10)
ˆ0 1
(11)
− ln x dx.
ˆ0 ∞
dx
(12)
.
2
ˆ0 1 (x + 1)(x + 1)
(13)
−x ln |x| dx.
ˆ−1
∞
dx
(14)
.
x + e−x
ˆ−∞ e
∞
(15)
ln(ln x).
ee

A comparison test.
Theorem 5.1 (Direct Comparison Test). Let f and g be continuous on [a, ∞) and suppose
that 0 ≤ f (x) ≤ g(x) for all x ≥ a.
ˆ ∞
ˆ ∞
(i) If
g(x) dx converges then
f (x) dx converges.
a

a


ˆ
(ii) If

ˆ

∞

f (x) dx diverges then
a

13

∞

g(x) dx diverges.
a

Example 5.4. In each part, determine whether the improper integral converges or diverges.
ˆ ∞
1
(1)
dx.
x
ˆ0 ∞ x + e
2
e−x dx.
(2)
ˆ0 ∞
2 + sin x
√
(3)
dx.
x
1
ˆ ∞
sin2 x
(4)
dx.
2
ˆ1 ∞ x
dx
√
.
(5)
4
ˆ1 ∞ x + 5
ln x
√ dx.
(6)
ˆ1 ∞ x
dx
(7)
.
5x + 2x
1
ˆ ∞
dx
(8)
.
x
ˆ0 ∞ 1 + e
1 + sin x
dx.
(9)
x2
ˆπ ∞
dx
(10)
.
ˆ2 1 ln x
ln x
√ dx.
(11)
x
0

14


Answers
cos(ax)
+ c.
a
x
1
Answers 1.2. tan−1
+ c.
a
a
x5 10x3
−
+ 25x + c.
Answers 1.3.
5
3
x−3
Answers 1.4. sin−1
+ c.
2
Answers 1.1. −

Answers 1.5. ln x2 + 2 x + 5 −

3
tan−1
4

x+1
2

Answers 2.1.
(1) sin x − x cos x + c.
(2) x2 ex − 2xex + 2ex + c.
ex (cos x + sin x)
(3)
+ c.
2
2
2
x ln x x
(4)
−
+ c.
2
4
(5) x ln x − x + c.
√
(6) x sin−1 x + 1 − x2 + c.
2
ex (x2 − 1)
+ c.
(7)
2
(8) x tan x + ln(cos x) + c.
3x
x3x
− 2 + c.
(9)
ln 3 ln 3
x(cos(ln x) + sin(ln x))
(10)
+ c.
2
4 x
3 x
2 x
(11) x e − 4x e + 12x e − 24xex + 24ex + c.
x ln x x
− + c.
(12)
2
√ √x 2 √x
(13) 2 xe − 2e + c.
(14) −x2 cos x + 2x sin x + 2 cos x + c.
xe2x e2x
(15)
−
+ c.
2 −x 4 −x
(16) −xe − e + c.
2e2x cos(3x) 3e2x sin(3x)
(17)
+
+ c.
13
13
1
(18) x tan−1 x − ln(1 + x2 ) + c.
2
2
x4 e−x
2
2
(19) −
− x2 e−x − e−x + c.
2 √
−1
esin x ( 1 − x2 + x)
(20)
+ c.
2
5
3
2
2
(21) (1 − x) 2 − (1 − x) 2 + c.
5
3
15
(22) 4 ln 2 −
+ c.
16

+ c.


Answers 3.1.

(1) −

cos5 x
+ c.
5

cos7 x cos5 x
−
+ c.
7
5
3
3
11
4
2
2
(3) sin 2 x − sin 2 x +
sin 2 x + c.
3
7
11
sin3 x
(4) sin x −
+ c.
3
cos10 x cos8
(5)
−
+ c.
10
8
x sin(2x)
+ c.
(6) −
2
4
3x sin(2x) sin(4x)
(7)
+
+
c.
8
4
32
x sin(4x)
+ c.
(8) −
8
32
sec5 x sec3 x
(9)
−
+ c.
5
3
(10) tan x − x + c.
tan5 x tan3 x
(11)
+
+ c.
5
3
tan2 x
+ ln | cos x| + c.
(12)
2
7
sec x sec5 x
(13)
−
+ c.
7
5
(14) ln | sec x + tan x| + c.
7
3
2
2
(15) tan 2 + tan 2 +c.
7
3
1
3
(16) − cot x + cot x + x + c.
3
(17) − ln | csc x + cot x| + c.
csc3 x csc5 x
−
+ c.
(18)
3
5
√
5
9
4
2
(19) − cos x + cos 2 x − cos 2 x + c.
5
9
√
4 − x2
Answers 3.2.
(1) (a) −
+ c.
4x
√
x + 9 + x2
+ c.
(b) ln
3
√
x
(c) x2 − 25 − 5 sec−1 + c.
5
1
x
−1
(d) tan x +
+ c.
2
2(1 + x2 )
√
x x 4 − x2
(e) 2 sin−1 +
+ c.
2
2
√
(f) √x2 − 1 − sec−1 x + c.
x2 − 2
(g)
+ c.
2x
(2)

15

16


√

3x
9x2 − 4 − 2 sec−1
+ c.
2
1
x+1
(i) tan−1
+ c.
3
3
√
ex 1 − e2x
1 −1 x
+ c.
(j) sin (e ) +
2
2
(k) ln 1 + sin2 x + sin x + c.
1
2x + 1
2x + 1
(l)
tan−1
+
+ c.
32
2
16(4x2 + 4x + 5)
x−2
(m) ln x2 − 4 x + 8 + 3 tan−1
+ c.
2
5
13
x−2
(n) ln x2 − 4 x + 8 +
tan−1
+ c.
2
2
2
x3
(o) −
3 + c.
3(x2 − 1) 2
π(π + 2)
(2)
.
2
1
1
Answers 4.1.
(1) ln |x − 1| − ln |x + 2| + c.
3
3
(2) 2 ln |x| + 4 ln |x + 1| − 3 ln |x − 1| + c.
3
1
(3) x2 + ln |x − 4| − ln |x + 2| + c.
2
2
9
(4) −
+ 6 ln |x| − ln |x + 1| + c.
x+1
(5) −5 tan−1 x + 2 ln |x| + c.
3
7
x+1
(6) ln x2 + 2 x + 5 − tan−1
+ 2 ln |x + 2| + c.
2
2
2
7
8
(7) ln |x − 3| + ln |x + 2| + c.
5
5
5
1
5
(8) − ln |x − 2| − ln |x| + ln |x − 1| + ln |x − 3| + c.
2
6
3
3
(9) −
+ ln |x − 3| + c.
x−3
1
1
(10) −
+
+ c.
x + 1 (x + 1)2
1
1
1
(11) ln |x + 1| + ln |x − 1| −
+ c.
2
2
x−1
3
1
3
(12) −
ln x2 + 1 − tan−1 x + ln |x − 2| + c.
10
5
5
√
√
5
1
3 2
x 2
2
−1
(13) ln |3 + x| +
− ln x + 2 +
tan
2(x2 + 2) 2
2
2
2
1
(14) x + ln |3 x + 1| + c.
3
9
x2
1
(15) x +
+ ln x2 + 1 + tan−1 x − 2 ln |x − 1| −
+ c.
2
x−1
x2
(16) x +
− ln |x| + 2 ln |x − 1| + c.
2
(h)

+ c.


(17) ln |sec x + tan x| + c.
1
125
x2
− 6 x − ln |x + 1| +
ln |5 + x| + c.
(18)
2
4
4
1
1
1
(19) x + ln |x − 1| − ln |x + 1| − tan−1 x + c.
4
4 √
2
√
1
3
(2 x + 1) 3
1
2
−1
(20) − ln x + x + 1 −
tan
+ ln |x − 1| + c.
6
3
3
3
−x
x
(21) −e + x − ln |e − 1| + c.
1
1
(22) − ln |2 cos x − 1| + ln |1 + cos x| + c.
3
3
1
3
3
1
(23) sin x −
− ln |sin x + 1| + ln |sin x − 1| −
+ c.
4(sin x + 1) 4
4
4(sin x − 1)
√
√
1
1
1
(24) √
− ln 1 − x2 + 1 + ln 1 − x2 − 1 + c.
2
1 − x2 2
1
(25) ln x2 + 2 x + 2 − tan−1 (x + 1) − 2
+ c.
x + 2x + 2
Answers 5.1. See, Answers 5.2, (5).
Answers 5.2.
(1) Convergent, 2.
(2) Divergent.
(3) Convergent, 2.
(4) Divergent.
(5) Divergent.
(6) Convergent, 1.
(7) Divergent.
1
(8) Convergent, .
e
(9) Divergent.
(10) Convergent, 1.
(11) Convergent, 0.
π
(12) Convergent, .
2
√
3
(13) Convergent, 3 + 3 2.
(14) Divergent.
(15) Divergent if p ≥ 1. Convergent if p < 1,
(16)
(17)
(18)
(19)
(20)
(21)

1
.
1−p

Divergent.
Convergent, 1.
Convergent, π.
Convergent, 0.
Divergent.
Divergent.

(22) Divergent if p ≤ 1. Convergent if p > 1,
(23) Convergent, 2.
(24) Divergent.
Answers 5.3.

(1) Divergent.

1
.
p−1

17

18


32
.
3
π
(3) Convergent, .
2
(4) Divergent.
(5) Convergent, 1000.
9
(6) Convergent, − .
2
(7) Convergent, 1000.
(8) Convergent, 0.
(9) Convergent, 2π 2 .
(10) Convergent, 1.
(11) Convergent, 1.
π
(12) Convergent, .
4
(13) Convergent, 0.
π
(14) Convergent, .
2
Answers 5.4.
(1) Convergent.
(2) Convergent.
(3) Divergent.
(4) Convergent.
(5) Convergent.
(6) Divergent.
(7) Convergent.
(8) Convergent.
(9) Convergent.
(10) Divergent.
(11) Convergent.
(12) Divergent.
(2) Convergent,

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