Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 3618-3621 | 2
1
9
6
5
x
+x
+
12 2
1
7
–6
x – x
13 (
)(
)
1
1
2
x –
x –
14 2
1
8
3x – x
+
15 |
1 | 3619-3622 | 2
1
7
–6
x – x
13 (
)(
)
1
1
2
x –
x –
14 2
1
8
3x – x
+
15 (
)(
)
1
x – a
x –b
16 |
1 | 3620-3623 | (
)(
)
1
1
2
x –
x –
14 2
1
8
3x – x
+
15 (
)(
)
1
x – a
x –b
16 42
1
2
3
x
x
x –
+
+
17 |
1 | 3621-3624 | 2
1
8
3x – x
+
15 (
)(
)
1
x – a
x –b
16 42
1
2
3
x
x
x –
+
+
17 2
2
1
x
x –
+
18 |
1 | 3622-3625 | (
)(
)
1
x – a
x –b
16 42
1
2
3
x
x
x –
+
+
17 2
2
1
x
x –
+
18 2
5
2
1 2
3
xx
x
−
+
+
19 |
1 | 3623-3626 | 42
1
2
3
x
x
x –
+
+
17 2
2
1
x
x –
+
18 2
5
2
1 2
3
xx
x
−
+
+
19 (
)(
)
6
7
5
4
x
x –
x –
+
20 |
1 | 3624-3627 | 2
2
1
x
x –
+
18 2
5
2
1 2
3
xx
x
−
+
+
19 (
)(
)
6
7
5
4
x
x –
x –
+
20 2
2
4
x
x – x
+
21 |
1 | 3625-3628 | 2
5
2
1 2
3
xx
x
−
+
+
19 (
)(
)
6
7
5
4
x
x –
x –
+
20 2
2
4
x
x – x
+
21 2
2
2
3
x
x
x
+
+
+
22 |
1 | 3626-3629 | (
)(
)
6
7
5
4
x
x –
x –
+
20 2
2
4
x
x – x
+
21 2
2
2
3
x
x
x
+
+
+
22 2
3
2
5
x
x – x
+
−
23 |
1 | 3627-3630 | 2
2
4
x
x – x
+
21 2
2
2
3
x
x
x
+
+
+
22 2
3
2
5
x
x – x
+
−
23 2
5
3
4
10
x
x
x
+
+
+ |
1 | 3628-3631 | 2
2
2
3
x
x
x
+
+
+
22 2
3
2
5
x
x – x
+
−
23 2
5
3
4
10
x
x
x
+
+
+ Choose the correct answer in Exercises 24 and 25 |
1 | 3629-3632 | 2
3
2
5
x
x – x
+
−
23 2
5
3
4
10
x
x
x
+
+
+ Choose the correct answer in Exercises 24 and 25 24 |
1 | 3630-3633 | 2
5
3
4
10
x
x
x
+
+
+ Choose the correct answer in Exercises 24 and 25 24 2
equals
2
2
dx
x
+x
+
∫
(A)
x tan–1 (x + 1) + C
(B)
tan–1 (x + 1) + C
(C)
(x + 1) tan–1x + C
(D)
tan–1x + C
25 |
1 | 3631-3634 | Choose the correct answer in Exercises 24 and 25 24 2
equals
2
2
dx
x
+x
+
∫
(A)
x tan–1 (x + 1) + C
(B)
tan–1 (x + 1) + C
(C)
(x + 1) tan–1x + C
(D)
tan–1x + C
25 2 equals
9
4
dx
x
x
−
∫
(A)
–1
1
9
8
sin
C
9
8
x −
+
(B)
–1
1
8
9
sin
C
2
9
x −
+
(C)
–1
1
9
8
sin
C
3
8
x −
+
(D)
–1
1
9
8
sin
C
2
9
x −
+
7 |
1 | 3632-3635 | 24 2
equals
2
2
dx
x
+x
+
∫
(A)
x tan–1 (x + 1) + C
(B)
tan–1 (x + 1) + C
(C)
(x + 1) tan–1x + C
(D)
tan–1x + C
25 2 equals
9
4
dx
x
x
−
∫
(A)
–1
1
9
8
sin
C
9
8
x −
+
(B)
–1
1
8
9
sin
C
2
9
x −
+
(C)
–1
1
9
8
sin
C
3
8
x −
+
(D)
–1
1
9
8
sin
C
2
9
x −
+
7 5 Integration by Partial Fractions
Recall that a rational function is defined as the ratio of two polynomials in the form
P( )
Q( )
x
x
, where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0 |
1 | 3633-3636 | 2
equals
2
2
dx
x
+x
+
∫
(A)
x tan–1 (x + 1) + C
(B)
tan–1 (x + 1) + C
(C)
(x + 1) tan–1x + C
(D)
tan–1x + C
25 2 equals
9
4
dx
x
x
−
∫
(A)
–1
1
9
8
sin
C
9
8
x −
+
(B)
–1
1
8
9
sin
C
2
9
x −
+
(C)
–1
1
9
8
sin
C
3
8
x −
+
(D)
–1
1
9
8
sin
C
2
9
x −
+
7 5 Integration by Partial Fractions
Recall that a rational function is defined as the ratio of two polynomials in the form
P( )
Q( )
x
x
, where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0 If the degree of P(x)
is less than the degree of Q(x), then the rational function is called proper, otherwise, it
is called improper |
1 | 3634-3637 | 2 equals
9
4
dx
x
x
−
∫
(A)
–1
1
9
8
sin
C
9
8
x −
+
(B)
–1
1
8
9
sin
C
2
9
x −
+
(C)
–1
1
9
8
sin
C
3
8
x −
+
(D)
–1
1
9
8
sin
C
2
9
x −
+
7 5 Integration by Partial Fractions
Recall that a rational function is defined as the ratio of two polynomials in the form
P( )
Q( )
x
x
, where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0 If the degree of P(x)
is less than the degree of Q(x), then the rational function is called proper, otherwise, it
is called improper The improper rational functions can be reduced to the proper rational
INTEGRALS 317
functions by long division process |
1 | 3635-3638 | 5 Integration by Partial Fractions
Recall that a rational function is defined as the ratio of two polynomials in the form
P( )
Q( )
x
x
, where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0 If the degree of P(x)
is less than the degree of Q(x), then the rational function is called proper, otherwise, it
is called improper The improper rational functions can be reduced to the proper rational
INTEGRALS 317
functions by long division process Thus, if P( )
Q( )
x
x is improper, then
1P ( )
P( )
T( )
Q( )
Q( )
x
x
x
x
x
=
+
,
where T(x) is a polynomial in x and
1P ( )
Q( )
x
x is a proper rational function |
1 | 3636-3639 | If the degree of P(x)
is less than the degree of Q(x), then the rational function is called proper, otherwise, it
is called improper The improper rational functions can be reduced to the proper rational
INTEGRALS 317
functions by long division process Thus, if P( )
Q( )
x
x is improper, then
1P ( )
P( )
T( )
Q( )
Q( )
x
x
x
x
x
=
+
,
where T(x) is a polynomial in x and
1P ( )
Q( )
x
x is a proper rational function As we know
how to integrate polynomials, the integration of any rational function is reduced to the
integration of a proper rational function |
1 | 3637-3640 | The improper rational functions can be reduced to the proper rational
INTEGRALS 317
functions by long division process Thus, if P( )
Q( )
x
x is improper, then
1P ( )
P( )
T( )
Q( )
Q( )
x
x
x
x
x
=
+
,
where T(x) is a polynomial in x and
1P ( )
Q( )
x
x is a proper rational function As we know
how to integrate polynomials, the integration of any rational function is reduced to the
integration of a proper rational function The rational functions which we shall consider
here for integration purposes will be those whose denominators can be factorised into
linear and quadratic factors |
1 | 3638-3641 | Thus, if P( )
Q( )
x
x is improper, then
1P ( )
P( )
T( )
Q( )
Q( )
x
x
x
x
x
=
+
,
where T(x) is a polynomial in x and
1P ( )
Q( )
x
x is a proper rational function As we know
how to integrate polynomials, the integration of any rational function is reduced to the
integration of a proper rational function The rational functions which we shall consider
here for integration purposes will be those whose denominators can be factorised into
linear and quadratic factors Assume that we want to evaluate
P( )
Q( )
∫xx dx
, where P( )
Q( )
x
x
is proper rational function |
1 | 3639-3642 | As we know
how to integrate polynomials, the integration of any rational function is reduced to the
integration of a proper rational function The rational functions which we shall consider
here for integration purposes will be those whose denominators can be factorised into
linear and quadratic factors Assume that we want to evaluate
P( )
Q( )
∫xx dx
, where P( )
Q( )
x
x
is proper rational function It is always possible to write the integrand as a sum of
simpler rational functions by a method called partial fraction decomposition |
1 | 3640-3643 | The rational functions which we shall consider
here for integration purposes will be those whose denominators can be factorised into
linear and quadratic factors Assume that we want to evaluate
P( )
Q( )
∫xx dx
, where P( )
Q( )
x
x
is proper rational function It is always possible to write the integrand as a sum of
simpler rational functions by a method called partial fraction decomposition After this,
the integration can be carried out easily using the already known methods |
1 | 3641-3644 | Assume that we want to evaluate
P( )
Q( )
∫xx dx
, where P( )
Q( )
x
x
is proper rational function It is always possible to write the integrand as a sum of
simpler rational functions by a method called partial fraction decomposition After this,
the integration can be carried out easily using the already known methods The following
Table 7 |
1 | 3642-3645 | It is always possible to write the integrand as a sum of
simpler rational functions by a method called partial fraction decomposition After this,
the integration can be carried out easily using the already known methods The following
Table 7 2 indicates the types of simpler partial fractions that are to be associated with
various kind of rational functions |
1 | 3643-3646 | After this,
the integration can be carried out easily using the already known methods The following
Table 7 2 indicates the types of simpler partial fractions that are to be associated with
various kind of rational functions Table 7 |
1 | 3644-3647 | The following
Table 7 2 indicates the types of simpler partial fractions that are to be associated with
various kind of rational functions Table 7 2
S |
1 | 3645-3648 | 2 indicates the types of simpler partial fractions that are to be associated with
various kind of rational functions Table 7 2
S No |
1 | 3646-3649 | Table 7 2
S No Form of the rational function
Form of the partial fraction
1 |
1 | 3647-3650 | 2
S No Form of the rational function
Form of the partial fraction
1 ( – ) ( – )
px
q
x a
+x b
, a ≠ b
A
B
x – a
x – b
+
2 |
1 | 3648-3651 | No Form of the rational function
Form of the partial fraction
1 ( – ) ( – )
px
q
x a
+x b
, a ≠ b
A
B
x – a
x – b
+
2 ( – )2
px
q
x
a
+
(
)
2
A
B
x – a
x – a
+
3 |
1 | 3649-3652 | Form of the rational function
Form of the partial fraction
1 ( – ) ( – )
px
q
x a
+x b
, a ≠ b
A
B
x – a
x – b
+
2 ( – )2
px
q
x
a
+
(
)
2
A
B
x – a
x – a
+
3 2
( – ) (
) (
)
px
qx
r
x
a
x – b
x – c
+
+
A
B
C
x – a
x – b
x –c
+
+
4 |
1 | 3650-3653 | ( – ) ( – )
px
q
x a
+x b
, a ≠ b
A
B
x – a
x – b
+
2 ( – )2
px
q
x
a
+
(
)
2
A
B
x – a
x – a
+
3 2
( – ) (
) (
)
px
qx
r
x
a
x – b
x – c
+
+
A
B
C
x – a
x – b
x –c
+
+
4 2
2
( – ) (
)
px
qx
r
x
a
x – b
+
+
2
A
B
C
(
)
x – a
x – b
+x – a
+
5 |
1 | 3651-3654 | ( – )2
px
q
x
a
+
(
)
2
A
B
x – a
x – a
+
3 2
( – ) (
) (
)
px
qx
r
x
a
x – b
x – c
+
+
A
B
C
x – a
x – b
x –c
+
+
4 2
2
( – ) (
)
px
qx
r
x
a
x – b
+
+
2
A
B
C
(
)
x – a
x – b
+x – a
+
5 2
2
( –
) (
)
px
qx
r
x
a
x
bx
c
+
+
+
+
2
A
B + C
x
x – a
x
bx
c
+
+
+
,
where x2 + bx + c cannot be factorised further
In the above table, A, B and C are real numbers to be determined suitably |
1 | 3652-3655 | 2
( – ) (
) (
)
px
qx
r
x
a
x – b
x – c
+
+
A
B
C
x – a
x – b
x –c
+
+
4 2
2
( – ) (
)
px
qx
r
x
a
x – b
+
+
2
A
B
C
(
)
x – a
x – b
+x – a
+
5 2
2
( –
) (
)
px
qx
r
x
a
x
bx
c
+
+
+
+
2
A
B + C
x
x – a
x
bx
c
+
+
+
,
where x2 + bx + c cannot be factorised further
In the above table, A, B and C are real numbers to be determined suitably 318
MATHEMATICS
Example 11 Find
(
1) (
2)
dx
x
x
+
+
∫
Solution The integrand is a proper rational function |
1 | 3653-3656 | 2
2
( – ) (
)
px
qx
r
x
a
x – b
+
+
2
A
B
C
(
)
x – a
x – b
+x – a
+
5 2
2
( –
) (
)
px
qx
r
x
a
x
bx
c
+
+
+
+
2
A
B + C
x
x – a
x
bx
c
+
+
+
,
where x2 + bx + c cannot be factorised further
In the above table, A, B and C are real numbers to be determined suitably 318
MATHEMATICS
Example 11 Find
(
1) (
2)
dx
x
x
+
+
∫
Solution The integrand is a proper rational function Therefore, by using the form of
partial fraction [Table 7 |
1 | 3654-3657 | 2
2
( –
) (
)
px
qx
r
x
a
x
bx
c
+
+
+
+
2
A
B + C
x
x – a
x
bx
c
+
+
+
,
where x2 + bx + c cannot be factorised further
In the above table, A, B and C are real numbers to be determined suitably 318
MATHEMATICS
Example 11 Find
(
1) (
2)
dx
x
x
+
+
∫
Solution The integrand is a proper rational function Therefore, by using the form of
partial fraction [Table 7 2 (i)], we write
1
(
1) (
2)
x
x
+
+
=
A
B
1
2
x
x
+
+
+ |
1 | 3655-3658 | 318
MATHEMATICS
Example 11 Find
(
1) (
2)
dx
x
x
+
+
∫
Solution The integrand is a proper rational function Therefore, by using the form of
partial fraction [Table 7 2 (i)], we write
1
(
1) (
2)
x
x
+
+
=
A
B
1
2
x
x
+
+
+ (1)
where, real numbers A and B are to be determined suitably |
1 | 3656-3659 | Therefore, by using the form of
partial fraction [Table 7 2 (i)], we write
1
(
1) (
2)
x
x
+
+
=
A
B
1
2
x
x
+
+
+ (1)
where, real numbers A and B are to be determined suitably This gives
1 = A (x + 2) + B (x + 1) |
1 | 3657-3660 | 2 (i)], we write
1
(
1) (
2)
x
x
+
+
=
A
B
1
2
x
x
+
+
+ (1)
where, real numbers A and B are to be determined suitably This gives
1 = A (x + 2) + B (x + 1) Equating the coefficients of x and the constant term, we get
A + B = 0
and
2A + B = 1
Solving these equations, we get A =1 and B = – 1 |
1 | 3658-3661 | (1)
where, real numbers A and B are to be determined suitably This gives
1 = A (x + 2) + B (x + 1) Equating the coefficients of x and the constant term, we get
A + B = 0
and
2A + B = 1
Solving these equations, we get A =1 and B = – 1 Thus, the integrand is given by
1
(
1) (
2)
x
x
+
+
=
1
–1
1
2
x
x
+
+
+
Therefore,
(
1) (
2)
dx
x
x
+
+
∫
=
1
2
dx
dx
–
x
x
+
+
∫
∫
= log
1
log
2
C
x
x
+
−
+
+
=
1
log
C
2
xx
+
+
+
Remark The equation (1) above is an identity, i |
1 | 3659-3662 | This gives
1 = A (x + 2) + B (x + 1) Equating the coefficients of x and the constant term, we get
A + B = 0
and
2A + B = 1
Solving these equations, we get A =1 and B = – 1 Thus, the integrand is given by
1
(
1) (
2)
x
x
+
+
=
1
–1
1
2
x
x
+
+
+
Therefore,
(
1) (
2)
dx
x
x
+
+
∫
=
1
2
dx
dx
–
x
x
+
+
∫
∫
= log
1
log
2
C
x
x
+
−
+
+
=
1
log
C
2
xx
+
+
+
Remark The equation (1) above is an identity, i e |
1 | 3660-3663 | Equating the coefficients of x and the constant term, we get
A + B = 0
and
2A + B = 1
Solving these equations, we get A =1 and B = – 1 Thus, the integrand is given by
1
(
1) (
2)
x
x
+
+
=
1
–1
1
2
x
x
+
+
+
Therefore,
(
1) (
2)
dx
x
x
+
+
∫
=
1
2
dx
dx
–
x
x
+
+
∫
∫
= log
1
log
2
C
x
x
+
−
+
+
=
1
log
C
2
xx
+
+
+
Remark The equation (1) above is an identity, i e a statement true for all (permissible)
values of x |
1 | 3661-3664 | Thus, the integrand is given by
1
(
1) (
2)
x
x
+
+
=
1
–1
1
2
x
x
+
+
+
Therefore,
(
1) (
2)
dx
x
x
+
+
∫
=
1
2
dx
dx
–
x
x
+
+
∫
∫
= log
1
log
2
C
x
x
+
−
+
+
=
1
log
C
2
xx
+
+
+
Remark The equation (1) above is an identity, i e a statement true for all (permissible)
values of x Some authors use the symbol ‘≡’ to indicate that the statement is an
identity and use the symbol ‘=’ to indicate that the statement is an equation, i |
1 | 3662-3665 | e a statement true for all (permissible)
values of x Some authors use the symbol ‘≡’ to indicate that the statement is an
identity and use the symbol ‘=’ to indicate that the statement is an equation, i e |
1 | 3663-3666 | a statement true for all (permissible)
values of x Some authors use the symbol ‘≡’ to indicate that the statement is an
identity and use the symbol ‘=’ to indicate that the statement is an equation, i e , to
indicate that the statement is true only for certain values of x |
1 | 3664-3667 | Some authors use the symbol ‘≡’ to indicate that the statement is an
identity and use the symbol ‘=’ to indicate that the statement is an equation, i e , to
indicate that the statement is true only for certain values of x Example 12 Find
2
2
1
5
6
x
dx
x
−+x
+
∫
Solution Here the integrand
2
2
1
5
6
x
x – x
+
+
is not proper rational function, so we divide
x2 + 1 by x2 – 5x + 6 and find that
INTEGRALS 319
2
2
1
5
6
x
x – x
+
+
=
2
5
5
5
5
1
1
(
2) (
3)
5
6
x –
x –
x –
x –
+x – x
= +
+
Let
5
5
(
2) (
3)
x –
x –
x –
=
A
B
2
3
x –
x –
+
So that
5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5
and 3A + 2B = 5 |
1 | 3665-3668 | e , to
indicate that the statement is true only for certain values of x Example 12 Find
2
2
1
5
6
x
dx
x
−+x
+
∫
Solution Here the integrand
2
2
1
5
6
x
x – x
+
+
is not proper rational function, so we divide
x2 + 1 by x2 – 5x + 6 and find that
INTEGRALS 319
2
2
1
5
6
x
x – x
+
+
=
2
5
5
5
5
1
1
(
2) (
3)
5
6
x –
x –
x –
x –
+x – x
= +
+
Let
5
5
(
2) (
3)
x –
x –
x –
=
A
B
2
3
x –
x –
+
So that
5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5
and 3A + 2B = 5 Solving these equations, we get A = – 5 and B = 10
Thus,
2
2
1
5
6
x
x – x
+
+
=
5
10
1
2
3
x –
x –
−
+
Therefore,
2
2
1
5
6
x
dx
x – x
+
+
∫
=
1
5
10
2
3
dx
dx
dx
x –
x –
−
+
∫
∫
∫
= x – 5 log |x – 2| + 10 log |x – 3| + C |
1 | 3666-3669 | , to
indicate that the statement is true only for certain values of x Example 12 Find
2
2
1
5
6
x
dx
x
−+x
+
∫
Solution Here the integrand
2
2
1
5
6
x
x – x
+
+
is not proper rational function, so we divide
x2 + 1 by x2 – 5x + 6 and find that
INTEGRALS 319
2
2
1
5
6
x
x – x
+
+
=
2
5
5
5
5
1
1
(
2) (
3)
5
6
x –
x –
x –
x –
+x – x
= +
+
Let
5
5
(
2) (
3)
x –
x –
x –
=
A
B
2
3
x –
x –
+
So that
5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5
and 3A + 2B = 5 Solving these equations, we get A = – 5 and B = 10
Thus,
2
2
1
5
6
x
x – x
+
+
=
5
10
1
2
3
x –
x –
−
+
Therefore,
2
2
1
5
6
x
dx
x – x
+
+
∫
=
1
5
10
2
3
dx
dx
dx
x –
x –
−
+
∫
∫
∫
= x – 5 log |x – 2| + 10 log |x – 3| + C Example 13 Find
32
2
(
1) (
3)
x
dx
x
−x
+
+
∫
Solution The integrand is of the type as given in Table 7 |
1 | 3667-3670 | Example 12 Find
2
2
1
5
6
x
dx
x
−+x
+
∫
Solution Here the integrand
2
2
1
5
6
x
x – x
+
+
is not proper rational function, so we divide
x2 + 1 by x2 – 5x + 6 and find that
INTEGRALS 319
2
2
1
5
6
x
x – x
+
+
=
2
5
5
5
5
1
1
(
2) (
3)
5
6
x –
x –
x –
x –
+x – x
= +
+
Let
5
5
(
2) (
3)
x –
x –
x –
=
A
B
2
3
x –
x –
+
So that
5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5
and 3A + 2B = 5 Solving these equations, we get A = – 5 and B = 10
Thus,
2
2
1
5
6
x
x – x
+
+
=
5
10
1
2
3
x –
x –
−
+
Therefore,
2
2
1
5
6
x
dx
x – x
+
+
∫
=
1
5
10
2
3
dx
dx
dx
x –
x –
−
+
∫
∫
∫
= x – 5 log |x – 2| + 10 log |x – 3| + C Example 13 Find
32
2
(
1) (
3)
x
dx
x
−x
+
+
∫
Solution The integrand is of the type as given in Table 7 2 (4) |
1 | 3668-3671 | Solving these equations, we get A = – 5 and B = 10
Thus,
2
2
1
5
6
x
x – x
+
+
=
5
10
1
2
3
x –
x –
−
+
Therefore,
2
2
1
5
6
x
dx
x – x
+
+
∫
=
1
5
10
2
3
dx
dx
dx
x –
x –
−
+
∫
∫
∫
= x – 5 log |x – 2| + 10 log |x – 3| + C Example 13 Find
32
2
(
1) (
3)
x
dx
x
−x
+
+
∫
Solution The integrand is of the type as given in Table 7 2 (4) We write
32
2
(
1) (
3)
x –
x
x
+
+
=
2
A
B
C
1
3
(
1)
x
x
+x
+
+
+
+
So that
3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 )
Comparing coefficient of x 2, x and constant term on both sides, we get
A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2 |
1 | 3669-3672 | Example 13 Find
32
2
(
1) (
3)
x
dx
x
−x
+
+
∫
Solution The integrand is of the type as given in Table 7 2 (4) We write
32
2
(
1) (
3)
x –
x
x
+
+
=
2
A
B
C
1
3
(
1)
x
x
+x
+
+
+
+
So that
3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 )
Comparing coefficient of x 2, x and constant term on both sides, we get
A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2 Solving these equations, we get
11
5
11
A
B
and C
4
2
4
–
–
,
=
=
= |
1 | 3670-3673 | 2 (4) We write
32
2
(
1) (
3)
x –
x
x
+
+
=
2
A
B
C
1
3
(
1)
x
x
+x
+
+
+
+
So that
3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 )
Comparing coefficient of x 2, x and constant term on both sides, we get
A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2 Solving these equations, we get
11
5
11
A
B
and C
4
2
4
–
–
,
=
=
= Thus the integrand is given by
32
2
(
1) (
3)
x
x
x
−
+
+
=
2
11
5
11
4(
1)
4(
3)
2(
1)
–
–
x
x
x
+
+
+
Therefore,
32
2
(
1) (
3)
x
x
x
−
+
+
∫
=
2
11
5
11
4
1
2
4
3
(
1)
dx
dx
dx
–
x
x
x
−
+
+
+
∫
∫
∫
=
11
5
11
log
+1
log
3
C
4
2( + 1)
4
x
x
x
+
−
+
+
= 11
+1
5
log
+C
4
+3
2 ( +1)
xx
x
+
320
MATHEMATICS
Example 14 Find
2
2
2
(
1) (
4)
x
dx
x
x
+
+
∫
Solution Consider
2
2
2
(
1) (
4)
x
x
x
+
+
and put x2 = y |
1 | 3671-3674 | We write
32
2
(
1) (
3)
x –
x
x
+
+
=
2
A
B
C
1
3
(
1)
x
x
+x
+
+
+
+
So that
3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 )
Comparing coefficient of x 2, x and constant term on both sides, we get
A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2 Solving these equations, we get
11
5
11
A
B
and C
4
2
4
–
–
,
=
=
= Thus the integrand is given by
32
2
(
1) (
3)
x
x
x
−
+
+
=
2
11
5
11
4(
1)
4(
3)
2(
1)
–
–
x
x
x
+
+
+
Therefore,
32
2
(
1) (
3)
x
x
x
−
+
+
∫
=
2
11
5
11
4
1
2
4
3
(
1)
dx
dx
dx
–
x
x
x
−
+
+
+
∫
∫
∫
=
11
5
11
log
+1
log
3
C
4
2( + 1)
4
x
x
x
+
−
+
+
= 11
+1
5
log
+C
4
+3
2 ( +1)
xx
x
+
320
MATHEMATICS
Example 14 Find
2
2
2
(
1) (
4)
x
dx
x
x
+
+
∫
Solution Consider
2
2
2
(
1) (
4)
x
x
x
+
+
and put x2 = y Then
2
2
2
(
1) (
4)
x
x
x
+
+
= (
1) (
4)
y
y
y
+
+
Write
(
1) (
4)
y
y
y
+
+
=
A
B
1
4
y
y
+
+
+
So that
y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
A =
1
4
and B
3
3
−
=
Thus,
2
2
2
(
1) (
4)
x
x
x
+
+
=
2
2
1
4
3(
1)
3 (
4)
–
x
x
+
+
+
Therefore,
2
2
2
(
1) (
4)
x dx
x
x
+
+
∫
=
2
2
1
4
3
3
1
4
dx
dx
–
x
x
+
+
+
∫
∫
=
1
1
1
4
1
tan
tan
C
3
3
2
2
–
– x
–
x +
×
+
=
1
1
1
2
tan
tan
C
3
3
2
–
– x
–
x +
+
In the above example, the substitution was made only for the partial fraction part
and not for the integration part |
1 | 3672-3675 | Solving these equations, we get
11
5
11
A
B
and C
4
2
4
–
–
,
=
=
= Thus the integrand is given by
32
2
(
1) (
3)
x
x
x
−
+
+
=
2
11
5
11
4(
1)
4(
3)
2(
1)
–
–
x
x
x
+
+
+
Therefore,
32
2
(
1) (
3)
x
x
x
−
+
+
∫
=
2
11
5
11
4
1
2
4
3
(
1)
dx
dx
dx
–
x
x
x
−
+
+
+
∫
∫
∫
=
11
5
11
log
+1
log
3
C
4
2( + 1)
4
x
x
x
+
−
+
+
= 11
+1
5
log
+C
4
+3
2 ( +1)
xx
x
+
320
MATHEMATICS
Example 14 Find
2
2
2
(
1) (
4)
x
dx
x
x
+
+
∫
Solution Consider
2
2
2
(
1) (
4)
x
x
x
+
+
and put x2 = y Then
2
2
2
(
1) (
4)
x
x
x
+
+
= (
1) (
4)
y
y
y
+
+
Write
(
1) (
4)
y
y
y
+
+
=
A
B
1
4
y
y
+
+
+
So that
y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
A =
1
4
and B
3
3
−
=
Thus,
2
2
2
(
1) (
4)
x
x
x
+
+
=
2
2
1
4
3(
1)
3 (
4)
–
x
x
+
+
+
Therefore,
2
2
2
(
1) (
4)
x dx
x
x
+
+
∫
=
2
2
1
4
3
3
1
4
dx
dx
–
x
x
+
+
+
∫
∫
=
1
1
1
4
1
tan
tan
C
3
3
2
2
–
– x
–
x +
×
+
=
1
1
1
2
tan
tan
C
3
3
2
–
– x
–
x +
+
In the above example, the substitution was made only for the partial fraction part
and not for the integration part Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method |
1 | 3673-3676 | Thus the integrand is given by
32
2
(
1) (
3)
x
x
x
−
+
+
=
2
11
5
11
4(
1)
4(
3)
2(
1)
–
–
x
x
x
+
+
+
Therefore,
32
2
(
1) (
3)
x
x
x
−
+
+
∫
=
2
11
5
11
4
1
2
4
3
(
1)
dx
dx
dx
–
x
x
x
−
+
+
+
∫
∫
∫
=
11
5
11
log
+1
log
3
C
4
2( + 1)
4
x
x
x
+
−
+
+
= 11
+1
5
log
+C
4
+3
2 ( +1)
xx
x
+
320
MATHEMATICS
Example 14 Find
2
2
2
(
1) (
4)
x
dx
x
x
+
+
∫
Solution Consider
2
2
2
(
1) (
4)
x
x
x
+
+
and put x2 = y Then
2
2
2
(
1) (
4)
x
x
x
+
+
= (
1) (
4)
y
y
y
+
+
Write
(
1) (
4)
y
y
y
+
+
=
A
B
1
4
y
y
+
+
+
So that
y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
A =
1
4
and B
3
3
−
=
Thus,
2
2
2
(
1) (
4)
x
x
x
+
+
=
2
2
1
4
3(
1)
3 (
4)
–
x
x
+
+
+
Therefore,
2
2
2
(
1) (
4)
x dx
x
x
+
+
∫
=
2
2
1
4
3
3
1
4
dx
dx
–
x
x
+
+
+
∫
∫
=
1
1
1
4
1
tan
tan
C
3
3
2
2
–
– x
–
x +
×
+
=
1
1
1
2
tan
tan
C
3
3
2
–
– x
–
x +
+
In the above example, the substitution was made only for the partial fraction part
and not for the integration part Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method Example 15 Find (
)
3 sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
Solution Let y = sinφ
Then
dy = cosφ dφ
INTEGRALS 321
Therefore,
(
)
3sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
=
2
(3 – 2)
5
(1
)
4
y
dy
–
– y
–
y
∫
=
32
2
4
4
y –
dy
y – y +
∫
=
(
)
2
3
2
I (say)
2
y –
y –
=
∫
Now, we write
(
)
2
3
2
2
y –
y –
=
2
A
B
2
(
2)
y
+y
−
−
[by Table 7 |
1 | 3674-3677 | Then
2
2
2
(
1) (
4)
x
x
x
+
+
= (
1) (
4)
y
y
y
+
+
Write
(
1) (
4)
y
y
y
+
+
=
A
B
1
4
y
y
+
+
+
So that
y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
A =
1
4
and B
3
3
−
=
Thus,
2
2
2
(
1) (
4)
x
x
x
+
+
=
2
2
1
4
3(
1)
3 (
4)
–
x
x
+
+
+
Therefore,
2
2
2
(
1) (
4)
x dx
x
x
+
+
∫
=
2
2
1
4
3
3
1
4
dx
dx
–
x
x
+
+
+
∫
∫
=
1
1
1
4
1
tan
tan
C
3
3
2
2
–
– x
–
x +
×
+
=
1
1
1
2
tan
tan
C
3
3
2
–
– x
–
x +
+
In the above example, the substitution was made only for the partial fraction part
and not for the integration part Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method Example 15 Find (
)
3 sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
Solution Let y = sinφ
Then
dy = cosφ dφ
INTEGRALS 321
Therefore,
(
)
3sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
=
2
(3 – 2)
5
(1
)
4
y
dy
–
– y
–
y
∫
=
32
2
4
4
y –
dy
y – y +
∫
=
(
)
2
3
2
I (say)
2
y –
y –
=
∫
Now, we write
(
)
2
3
2
2
y –
y –
=
2
A
B
2
(
2)
y
+y
−
−
[by Table 7 2 (2)]
Therefore,
3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4 |
1 | 3675-3678 | Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method Example 15 Find (
)
3 sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
Solution Let y = sinφ
Then
dy = cosφ dφ
INTEGRALS 321
Therefore,
(
)
3sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
=
2
(3 – 2)
5
(1
)
4
y
dy
–
– y
–
y
∫
=
32
2
4
4
y –
dy
y – y +
∫
=
(
)
2
3
2
I (say)
2
y –
y –
=
∫
Now, we write
(
)
2
3
2
2
y –
y –
=
2
A
B
2
(
2)
y
+y
−
−
[by Table 7 2 (2)]
Therefore,
3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4 Therefore, the required integral is given by
I =
2
3
4
[
+
]
2
(
2)
dy
y –
y –
∫
=
2
3
2+ 4
(
2)
dy
dy
y –
y –
∫
∫
=
1
3log
2
4
C
2
y
–
y
−
+
+
−
=
4
3log sin
2
C
2
sin
–
φ −
+
+
φ
=
4
3log (2
sin )
+ C
2
sin
−
φ +
−
φ
(since, 2 – sinφ is always positive)
Example 16 Find
2
2
1
(
2) (
1)
x
x
dx
x
x
+
+
+
+
∫
Solution The integrand is a proper rational function |
1 | 3676-3679 | Example 15 Find (
)
3 sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
Solution Let y = sinφ
Then
dy = cosφ dφ
INTEGRALS 321
Therefore,
(
)
3sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
=
2
(3 – 2)
5
(1
)
4
y
dy
–
– y
–
y
∫
=
32
2
4
4
y –
dy
y – y +
∫
=
(
)
2
3
2
I (say)
2
y –
y –
=
∫
Now, we write
(
)
2
3
2
2
y –
y –
=
2
A
B
2
(
2)
y
+y
−
−
[by Table 7 2 (2)]
Therefore,
3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4 Therefore, the required integral is given by
I =
2
3
4
[
+
]
2
(
2)
dy
y –
y –
∫
=
2
3
2+ 4
(
2)
dy
dy
y –
y –
∫
∫
=
1
3log
2
4
C
2
y
–
y
−
+
+
−
=
4
3log sin
2
C
2
sin
–
φ −
+
+
φ
=
4
3log (2
sin )
+ C
2
sin
−
φ +
−
φ
(since, 2 – sinφ is always positive)
Example 16 Find
2
2
1
(
2) (
1)
x
x
dx
x
x
+
+
+
+
∫
Solution The integrand is a proper rational function Decompose the rational function
into partial fraction [Table 2 |
1 | 3677-3680 | 2 (2)]
Therefore,
3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4 Therefore, the required integral is given by
I =
2
3
4
[
+
]
2
(
2)
dy
y –
y –
∫
=
2
3
2+ 4
(
2)
dy
dy
y –
y –
∫
∫
=
1
3log
2
4
C
2
y
–
y
−
+
+
−
=
4
3log sin
2
C
2
sin
–
φ −
+
+
φ
=
4
3log (2
sin )
+ C
2
sin
−
φ +
−
φ
(since, 2 – sinφ is always positive)
Example 16 Find
2
2
1
(
2) (
1)
x
x
dx
x
x
+
+
+
+
∫
Solution The integrand is a proper rational function Decompose the rational function
into partial fraction [Table 2 2(5)] |
1 | 3678-3681 | Therefore, the required integral is given by
I =
2
3
4
[
+
]
2
(
2)
dy
y –
y –
∫
=
2
3
2+ 4
(
2)
dy
dy
y –
y –
∫
∫
=
1
3log
2
4
C
2
y
–
y
−
+
+
−
=
4
3log sin
2
C
2
sin
–
φ −
+
+
φ
=
4
3log (2
sin )
+ C
2
sin
−
φ +
−
φ
(since, 2 – sinφ is always positive)
Example 16 Find
2
2
1
(
2) (
1)
x
x
dx
x
x
+
+
+
+
∫
Solution The integrand is a proper rational function Decompose the rational function
into partial fraction [Table 2 2(5)] Write
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
A
B + C
2
(
1)
x
x
+x
+
+
Therefore,
x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)
322
MATHEMATICS
Equating the coefficients of x2, x and of constant term of both sides, we get
A + B =1, 2B + C = 1 and A + 2C = 1 |
1 | 3679-3682 | Decompose the rational function
into partial fraction [Table 2 2(5)] Write
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
A
B + C
2
(
1)
x
x
+x
+
+
Therefore,
x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)
322
MATHEMATICS
Equating the coefficients of x2, x and of constant term of both sides, we get
A + B =1, 2B + C = 1 and A + 2C = 1 Solving these equations, we get
3
2
1
A
, B
and C
5
5
5
=
=
=
Thus, the integrand is given by
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
2
1
3
5
5
5 (
2)
1
x
x
x
+
+
+
+
=
2
3
1
2
1
5 (
2)
5
1
x
x
x
+
+
+
+
Therefore,
2
2
1
(
+1) (
2)
x
x
dx
x
+x
+
+
∫
=
2
2
3
1
2
1
1
5
2
5
5
1
1
dx
x
dx
dx
x
x
x
+
+
+
+
+
∫
∫
∫
=
2
1
3
1
1
log
2
log
1
tan
C
5
5
5
–
x
x
x
+
+
+
+
+
EXERCISE 7 |
1 | 3680-3683 | 2(5)] Write
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
A
B + C
2
(
1)
x
x
+x
+
+
Therefore,
x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)
322
MATHEMATICS
Equating the coefficients of x2, x and of constant term of both sides, we get
A + B =1, 2B + C = 1 and A + 2C = 1 Solving these equations, we get
3
2
1
A
, B
and C
5
5
5
=
=
=
Thus, the integrand is given by
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
2
1
3
5
5
5 (
2)
1
x
x
x
+
+
+
+
=
2
3
1
2
1
5 (
2)
5
1
x
x
x
+
+
+
+
Therefore,
2
2
1
(
+1) (
2)
x
x
dx
x
+x
+
+
∫
=
2
2
3
1
2
1
1
5
2
5
5
1
1
dx
x
dx
dx
x
x
x
+
+
+
+
+
∫
∫
∫
=
2
1
3
1
1
log
2
log
1
tan
C
5
5
5
–
x
x
x
+
+
+
+
+
EXERCISE 7 5
Integrate the rational functions in Exercises 1 to 21 |
1 | 3681-3684 | Write
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
A
B + C
2
(
1)
x
x
+x
+
+
Therefore,
x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)
322
MATHEMATICS
Equating the coefficients of x2, x and of constant term of both sides, we get
A + B =1, 2B + C = 1 and A + 2C = 1 Solving these equations, we get
3
2
1
A
, B
and C
5
5
5
=
=
=
Thus, the integrand is given by
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
2
1
3
5
5
5 (
2)
1
x
x
x
+
+
+
+
=
2
3
1
2
1
5 (
2)
5
1
x
x
x
+
+
+
+
Therefore,
2
2
1
(
+1) (
2)
x
x
dx
x
+x
+
+
∫
=
2
2
3
1
2
1
1
5
2
5
5
1
1
dx
x
dx
dx
x
x
x
+
+
+
+
+
∫
∫
∫
=
2
1
3
1
1
log
2
log
1
tan
C
5
5
5
–
x
x
x
+
+
+
+
+
EXERCISE 7 5
Integrate the rational functions in Exercises 1 to 21 1 |
1 | 3682-3685 | Solving these equations, we get
3
2
1
A
, B
and C
5
5
5
=
=
=
Thus, the integrand is given by
2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
=
2
2
1
3
5
5
5 (
2)
1
x
x
x
+
+
+
+
=
2
3
1
2
1
5 (
2)
5
1
x
x
x
+
+
+
+
Therefore,
2
2
1
(
+1) (
2)
x
x
dx
x
+x
+
+
∫
=
2
2
3
1
2
1
1
5
2
5
5
1
1
dx
x
dx
dx
x
x
x
+
+
+
+
+
∫
∫
∫
=
2
1
3
1
1
log
2
log
1
tan
C
5
5
5
–
x
x
x
+
+
+
+
+
EXERCISE 7 5
Integrate the rational functions in Exercises 1 to 21 1 (
1) (
2)
x
x
x
+
+
2 |
1 | 3683-3686 | 5
Integrate the rational functions in Exercises 1 to 21 1 (
1) (
2)
x
x
x
+
+
2 2
1
x –9
3 |
1 | 3684-3687 | 1 (
1) (
2)
x
x
x
+
+
2 2
1
x –9
3 3
1
(
1) (
2) (
3)
x –
x –
x –
x –
4 |
1 | 3685-3688 | (
1) (
2)
x
x
x
+
+
2 2
1
x –9
3 3
1
(
1) (
2) (
3)
x –
x –
x –
x –
4 (
1) (
2) (
3)
x
x –
x –
x –
5 |
1 | 3686-3689 | 2
1
x –9
3 3
1
(
1) (
2) (
3)
x –
x –
x –
x –
4 (
1) (
2) (
3)
x
x –
x –
x –
5 2
32
2
x
x
+x
+
6 |
1 | 3687-3690 | 3
1
(
1) (
2) (
3)
x –
x –
x –
x –
4 (
1) (
2) (
3)
x
x –
x –
x –
5 2
32
2
x
x
+x
+
6 2
(11
2 )
– x
x
– x
7 |
1 | 3688-3691 | (
1) (
2) (
3)
x
x –
x –
x –
5 2
32
2
x
x
+x
+
6 2
(11
2 )
– x
x
– x
7 (2
1) ( – 1)
x
x
x
+
8 |
1 | 3689-3692 | 2
32
2
x
x
+x
+
6 2
(11
2 )
– x
x
– x
7 (2
1) ( – 1)
x
x
x
+
8 2
(
1) (
2)
x
x –
x +
9 |
1 | 3690-3693 | 2
(11
2 )
– x
x
– x
7 (2
1) ( – 1)
x
x
x
+
8 2
(
1) (
2)
x
x –
x +
9 3
32
5
1
x – xx
+x
−
+
10 |
1 | 3691-3694 | (2
1) ( – 1)
x
x
x
+
8 2
(
1) (
2)
x
x –
x +
9 3
32
5
1
x – xx
+x
−
+
10 2
2
3
(
1) (2
3)
x –x
x
−
+
11 |
1 | 3692-3695 | 2
(
1) (
2)
x
x –
x +
9 3
32
5
1
x – xx
+x
−
+
10 2
2
3
(
1) (2
3)
x –x
x
−
+
11 2
5
(
1) (
4)
x
x
x
+
−
12 |
1 | 3693-3696 | 3
32
5
1
x – xx
+x
−
+
10 2
2
3
(
1) (2
3)
x –x
x
−
+
11 2
5
(
1) (
4)
x
x
x
+
−
12 3
2
1
1
x
x
x
+
+
−
13 |
1 | 3694-3697 | 2
2
3
(
1) (2
3)
x –x
x
−
+
11 2
5
(
1) (
4)
x
x
x
+
−
12 3
2
1
1
x
x
x
+
+
−
13 2
2
(1
) (1
)
x
x
−
+
14 |
1 | 3695-3698 | 2
5
(
1) (
4)
x
x
x
+
−
12 3
2
1
1
x
x
x
+
+
−
13 2
2
(1
) (1
)
x
x
−
+
14 2
3
1
(
2)
x –
x +
15 |
1 | 3696-3699 | 3
2
1
1
x
x
x
+
+
−
13 2
2
(1
) (1
)
x
x
−
+
14 2
3
1
(
2)
x –
x +
15 4
1
1
x −
16 |
1 | 3697-3700 | 2
2
(1
) (1
)
x
x
−
+
14 2
3
1
(
2)
x –
x +
15 4
1
1
x −
16 1
(
1)
n
x x +
[Hint: multiply numerator and denominator by x n – 1 and put xn = t ]
17 |
1 | 3698-3701 | 2
3
1
(
2)
x –
x +
15 4
1
1
x −
16 1
(
1)
n
x x +
[Hint: multiply numerator and denominator by x n – 1 and put xn = t ]
17 cos
(1– sin ) (2 – sin )
x
x
x
[Hint : Put sin x = t]
INTEGRALS 323
18 |
1 | 3699-3702 | 4
1
1
x −
16 1
(
1)
n
x x +
[Hint: multiply numerator and denominator by x n – 1 and put xn = t ]
17 cos
(1– sin ) (2 – sin )
x
x
x
[Hint : Put sin x = t]
INTEGRALS 323
18 2
2
2
2
(
1) (
2)
(
3)(
4)
x
x
x
x
+
+
+
+
19 |
1 | 3700-3703 | 1
(
1)
n
x x +
[Hint: multiply numerator and denominator by x n – 1 and put xn = t ]
17 cos
(1– sin ) (2 – sin )
x
x
x
[Hint : Put sin x = t]
INTEGRALS 323
18 2
2
2
2
(
1) (
2)
(
3)(
4)
x
x
x
x
+
+
+
+
19 2
2
2
(
1) (
3)
x
x
x
+
+
20 |
1 | 3701-3704 | cos
(1– sin ) (2 – sin )
x
x
x
[Hint : Put sin x = t]
INTEGRALS 323
18 2
2
2
2
(
1) (
2)
(
3)(
4)
x
x
x
x
+
+
+
+
19 2
2
2
(
1) (
3)
x
x
x
+
+
20 4
1
(
1)
x x –
21 |
1 | 3702-3705 | 2
2
2
2
(
1) (
2)
(
3)(
4)
x
x
x
x
+
+
+
+
19 2
2
2
(
1) (
3)
x
x
x
+
+
20 4
1
(
1)
x x –
21 1
(
1)
x
e –
[Hint : Put ex = t]
Choose the correct answer in each of the Exercises 22 and 23 |
1 | 3703-3706 | 2
2
2
(
1) (
3)
x
x
x
+
+
20 4
1
(
1)
x x –
21 1
(
1)
x
e –
[Hint : Put ex = t]
Choose the correct answer in each of the Exercises 22 and 23 22 |
1 | 3704-3707 | 4
1
(
1)
x x –
21 1
(
1)
x
e –
[Hint : Put ex = t]
Choose the correct answer in each of the Exercises 22 and 23 22 (
1) (
2)
x dx
x
x
−
−
∫
equals
(A)
2
(
1)
log
C
2
xx
−
+
−
(B)
2
(
2)
log
C
1
x
x
−
+
−
(C)
12
log
C
2
xx
−
+
−
(D) log (
1) (
2)
C
x
x
−
−
+
23 |
1 | 3705-3708 | 1
(
1)
x
e –
[Hint : Put ex = t]
Choose the correct answer in each of the Exercises 22 and 23 22 (
1) (
2)
x dx
x
x
−
−
∫
equals
(A)
2
(
1)
log
C
2
xx
−
+
−
(B)
2
(
2)
log
C
1
x
x
−
+
−
(C)
12
log
C
2
xx
−
+
−
(D) log (
1) (
2)
C
x
x
−
−
+
23 (2
1)
dx
∫x x +
equals
(A)
2
1
log
log (
+1) + C
2
x
x
−
(B)
2
1
log
log (
+1) + C
2
x
x
+
(C)
2
1
log
log (
+1) +C
2
x
x
−
+
(D)
2
1log
log (
+1) + C
2
x
x
+
7 |
1 | 3706-3709 | 22 (
1) (
2)
x dx
x
x
−
−
∫
equals
(A)
2
(
1)
log
C
2
xx
−
+
−
(B)
2
(
2)
log
C
1
x
x
−
+
−
(C)
12
log
C
2
xx
−
+
−
(D) log (
1) (
2)
C
x
x
−
−
+
23 (2
1)
dx
∫x x +
equals
(A)
2
1
log
log (
+1) + C
2
x
x
−
(B)
2
1
log
log (
+1) + C
2
x
x
+
(C)
2
1
log
log (
+1) +C
2
x
x
−
+
(D)
2
1log
log (
+1) + C
2
x
x
+
7 6 Integration by Parts
In this section, we describe one more method of integration, that is found quite useful in
integrating products of functions |
1 | 3707-3710 | (
1) (
2)
x dx
x
x
−
−
∫
equals
(A)
2
(
1)
log
C
2
xx
−
+
−
(B)
2
(
2)
log
C
1
x
x
−
+
−
(C)
12
log
C
2
xx
−
+
−
(D) log (
1) (
2)
C
x
x
−
−
+
23 (2
1)
dx
∫x x +
equals
(A)
2
1
log
log (
+1) + C
2
x
x
−
(B)
2
1
log
log (
+1) + C
2
x
x
+
(C)
2
1
log
log (
+1) +C
2
x
x
−
+
(D)
2
1log
log (
+1) + C
2
x
x
+
7 6 Integration by Parts
In this section, we describe one more method of integration, that is found quite useful in
integrating products of functions If u and v are any two differentiable functions of a single variable x (say) |
1 | 3708-3711 | (2
1)
dx
∫x x +
equals
(A)
2
1
log
log (
+1) + C
2
x
x
−
(B)
2
1
log
log (
+1) + C
2
x
x
+
(C)
2
1
log
log (
+1) +C
2
x
x
−
+
(D)
2
1log
log (
+1) + C
2
x
x
+
7 6 Integration by Parts
In this section, we describe one more method of integration, that is found quite useful in
integrating products of functions If u and v are any two differentiable functions of a single variable x (say) Then, by
the product rule of differentiation, we have
(
)
d
uv
dx
=
dv
du
u
v
dx
dx
+
Integrating both sides, we get
uv =
dv
du
u
dx
v
dx
dx
dx
+
∫
∫
or
udv
dx
∫dx
=
uv – vdu
dx
∫dx |
1 | 3709-3712 | 6 Integration by Parts
In this section, we describe one more method of integration, that is found quite useful in
integrating products of functions If u and v are any two differentiable functions of a single variable x (say) Then, by
the product rule of differentiation, we have
(
)
d
uv
dx
=
dv
du
u
v
dx
dx
+
Integrating both sides, we get
uv =
dv
du
u
dx
v
dx
dx
dx
+
∫
∫
or
udv
dx
∫dx
=
uv – vdu
dx
∫dx (1)
Let
u = f (x) and dv
dx = g(x) |
1 | 3710-3713 | If u and v are any two differentiable functions of a single variable x (say) Then, by
the product rule of differentiation, we have
(
)
d
uv
dx
=
dv
du
u
v
dx
dx
+
Integrating both sides, we get
uv =
dv
du
u
dx
v
dx
dx
dx
+
∫
∫
or
udv
dx
∫dx
=
uv – vdu
dx
∫dx (1)
Let
u = f (x) and dv
dx = g(x) Then
du
dx = f ′(x) and v =
( )
∫g x dx
324
MATHEMATICS
Therefore, expression (1) can be rewritten as
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
]
( )
f x
g x dx –
g x dx f
′x dx
∫
∫ ∫
i |
1 | 3711-3714 | Then, by
the product rule of differentiation, we have
(
)
d
uv
dx
=
dv
du
u
v
dx
dx
+
Integrating both sides, we get
uv =
dv
du
u
dx
v
dx
dx
dx
+
∫
∫
or
udv
dx
∫dx
=
uv – vdu
dx
∫dx (1)
Let
u = f (x) and dv
dx = g(x) Then
du
dx = f ′(x) and v =
( )
∫g x dx
324
MATHEMATICS
Therefore, expression (1) can be rewritten as
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
]
( )
f x
g x dx –
g x dx f
′x dx
∫
∫ ∫
i e |
1 | 3712-3715 | (1)
Let
u = f (x) and dv
dx = g(x) Then
du
dx = f ′(x) and v =
( )
∫g x dx
324
MATHEMATICS
Therefore, expression (1) can be rewritten as
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
]
( )
f x
g x dx –
g x dx f
′x dx
∫
∫ ∫
i e ,
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
( )
]
f x
g x dx –
f
x
g x dx dx
′
∫
∫
∫
If we take f as the first function and g as the second function, then this formula
may be stated as follows:
“The integral of the product of two functions = (first function) × (integral
of the second function) – Integral of [(differential coefficient of the first function)
× (integral of the second function)]”
Example 17 Find
xcos
x dx
∫
Solution Put f (x) = x (first function) and g (x) = cos x (second function) |
1 | 3713-3716 | Then
du
dx = f ′(x) and v =
( )
∫g x dx
324
MATHEMATICS
Therefore, expression (1) can be rewritten as
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
]
( )
f x
g x dx –
g x dx f
′x dx
∫
∫ ∫
i e ,
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
( )
]
f x
g x dx –
f
x
g x dx dx
′
∫
∫
∫
If we take f as the first function and g as the second function, then this formula
may be stated as follows:
“The integral of the product of two functions = (first function) × (integral
of the second function) – Integral of [(differential coefficient of the first function)
× (integral of the second function)]”
Example 17 Find
xcos
x dx
∫
Solution Put f (x) = x (first function) and g (x) = cos x (second function) Then, integration by parts gives
xcos
x dx
∫
=
cos
[
( ) cos
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
=
sin
sin
x
x –
x dx
∫
= x sin x + cos x + C
Suppose, we take
f (x) = cos x and g(x) = x |
1 | 3714-3717 | e ,
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
( )
]
f x
g x dx –
f
x
g x dx dx
′
∫
∫
∫
If we take f as the first function and g as the second function, then this formula
may be stated as follows:
“The integral of the product of two functions = (first function) × (integral
of the second function) – Integral of [(differential coefficient of the first function)
× (integral of the second function)]”
Example 17 Find
xcos
x dx
∫
Solution Put f (x) = x (first function) and g (x) = cos x (second function) Then, integration by parts gives
xcos
x dx
∫
=
cos
[
( ) cos
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
=
sin
sin
x
x –
x dx
∫
= x sin x + cos x + C
Suppose, we take
f (x) = cos x and g(x) = x Then
xcos
x dx
∫
= cos
[
(cos )
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
= (
)
2
2
cos
sin
2
2
x
x
x
x
dx
+∫
Thus, it shows that the integral
xcos
x dx
∫
is reduced to the comparatively more
complicated integral having more power of x |
1 | 3715-3718 | ,
( ) ( )
∫f x g x dx
=
( )
( )
[
( )
( )
]
f x
g x dx –
f
x
g x dx dx
′
∫
∫
∫
If we take f as the first function and g as the second function, then this formula
may be stated as follows:
“The integral of the product of two functions = (first function) × (integral
of the second function) – Integral of [(differential coefficient of the first function)
× (integral of the second function)]”
Example 17 Find
xcos
x dx
∫
Solution Put f (x) = x (first function) and g (x) = cos x (second function) Then, integration by parts gives
xcos
x dx
∫
=
cos
[
( ) cos
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
=
sin
sin
x
x –
x dx
∫
= x sin x + cos x + C
Suppose, we take
f (x) = cos x and g(x) = x Then
xcos
x dx
∫
= cos
[
(cos )
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
= (
)
2
2
cos
sin
2
2
x
x
x
x
dx
+∫
Thus, it shows that the integral
xcos
x dx
∫
is reduced to the comparatively more
complicated integral having more power of x Therefore, the proper choice of the first
function and the second function is significant |
1 | 3716-3719 | Then, integration by parts gives
xcos
x dx
∫
=
cos
[
( ) cos
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
=
sin
sin
x
x –
x dx
∫
= x sin x + cos x + C
Suppose, we take
f (x) = cos x and g(x) = x Then
xcos
x dx
∫
= cos
[
(cos )
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
= (
)
2
2
cos
sin
2
2
x
x
x
x
dx
+∫
Thus, it shows that the integral
xcos
x dx
∫
is reduced to the comparatively more
complicated integral having more power of x Therefore, the proper choice of the first
function and the second function is significant Remarks
(i)
It is worth mentioning that integration by parts is not applicable to product of
functions in all cases |
1 | 3717-3720 | Then
xcos
x dx
∫
= cos
[
(cos )
]
d
x
x dx –
x
x dx dx
dx
∫
∫
∫
= (
)
2
2
cos
sin
2
2
x
x
x
x
dx
+∫
Thus, it shows that the integral
xcos
x dx
∫
is reduced to the comparatively more
complicated integral having more power of x Therefore, the proper choice of the first
function and the second function is significant Remarks
(i)
It is worth mentioning that integration by parts is not applicable to product of
functions in all cases For instance, the method does not work for
xsin
x dx
∫ |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.