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In this section, we will learn: How to integrate rational functions by reducing them to a sum of simpler fractions.

1.
7

TECHNIQUES OF INTEGRATION

TECHNIQUES OF INTEGRATION

2.
TECHNIQUES OF INTEGRATION

7.4

Integration of Rational Functions

by Partial Fractions

In this section, we will learn:

How to integrate rational functions

by reducing them to a sum of simpler fractions.

7.4

Integration of Rational Functions

by Partial Fractions

In this section, we will learn:

How to integrate rational functions

by reducing them to a sum of simpler fractions.

3.
PARTIAL FRACTIONS

We show how to integrate any rational

function (a ratio of polynomials) by

expressing it as a sum of simpler fractions,

called partial fractions.

We already know how to integrate

partial functions.

We show how to integrate any rational

function (a ratio of polynomials) by

expressing it as a sum of simpler fractions,

called partial fractions.

We already know how to integrate

partial functions.

4.
INTEGRATION BY PARTIAL FRACTIONS

To illustrate the method, observe that,

by taking the fractions 2/(x – 1) and 1/(x – 2)

to a common denominator, we obtain:

2 1 2( x 2) ( x 1)

x 1 x2 ( x 1)( x 2)

x 5

2

x x 2

To illustrate the method, observe that,

by taking the fractions 2/(x – 1) and 1/(x – 2)

to a common denominator, we obtain:

2 1 2( x 2) ( x 1)

x 1 x2 ( x 1)( x 2)

x 5

2

x x 2

5.
INTEGRATION BY PARTIAL FRACTIONS

If we now reverse the procedure, we see

how to integrate the function on the right side

of this equation:

x 5 2 1

x 2 x 2 dx x 1 x 2 dx

2 ln | x 1| ln | x 2 | C

If we now reverse the procedure, we see

how to integrate the function on the right side

of this equation:

x 5 2 1

x 2 x 2 dx x 1 x 2 dx

2 ln | x 1| ln | x 2 | C

6.
INTEGRATION BY PARTIAL FRACTIONS

To see how the method of partial fractions

works in general, let’s consider a rational

P( x)

f ( x)

Q( x)

where P and Q are polynomials.

To see how the method of partial fractions

works in general, let’s consider a rational

P( x)

f ( x)

Q( x)

where P and Q are polynomials.

7.
PROPER FUNCTION

It’s possible to express f as a sum of

simpler fractions if the degree of P is less

than the degree of Q.

Such a rational function is called proper.

It’s possible to express f as a sum of

simpler fractions if the degree of P is less

than the degree of Q.

Such a rational function is called proper.

8.
DEGREE OF P

Recall that, if

n n 1

P( x) an x an 1 x a1 x a0

where an ≠ 0, then the degree of P is n

and we write deg(P) = n.

Recall that, if

n n 1

P( x) an x an 1 x a1 x a0

where an ≠ 0, then the degree of P is n

and we write deg(P) = n.

9.
PARTIAL FRACTIONS

If f is improper, that is, deg(P) ≥ deg(Q), then

we must take the preliminary step of dividing

Q into P (by long division).

This is done until a remainder R(x) is obtained

such that deg(R) < deg(Q).

If f is improper, that is, deg(P) ≥ deg(Q), then

we must take the preliminary step of dividing

Q into P (by long division).

This is done until a remainder R(x) is obtained

such that deg(R) < deg(Q).

10.
PARTIAL FRACTIONS Equation 1

The division statement is

P ( x) R( x)

f ( x) S ( x )

Q( x) Q( x)

where S and R are also polynomials.

The division statement is

P ( x) R( x)

f ( x) S ( x )

Q( x) Q( x)

where S and R are also polynomials.

11.
PARTIAL FRACTIONS

As the following example illustrates,

sometimes, this preliminary step is all

that is required.

As the following example illustrates,

sometimes, this preliminary step is all

that is required.

12.
PARTIAL FRACTIONS Example 1

3

x x

Find dx

x 1

The degree of the numerator is greater

than that of the denominator.

So, we first perform the long division.

3

x x

Find dx

x 1

The degree of the numerator is greater

than that of the denominator.

So, we first perform the long division.

13.
PARTIAL FRACTIONS Example 1

This enables us to write:

3

x x 2 2

x 1 dx

x x 2 dx

x 1

3 2

x x

2 x 2 ln | x 1| C

3 2

This enables us to write:

3

x x 2 2

x 1 dx

x x 2 dx

x 1

3 2

x x

2 x 2 ln | x 1| C

3 2

14.
PARTIAL FRACTIONS

The next step is to factor

the denominator Q(x) as far

as possible.

The next step is to factor

the denominator Q(x) as far

as possible.

15.
FACTORISATION OF Q(x)

It can be shown that any polynomial Q

can be factored as a product of:

Linear factors (of the form ax + b)

Irreducible quadratic factors (of the form ax2 + bx + c,

where b2 – 4ac < 0).

It can be shown that any polynomial Q

can be factored as a product of:

Linear factors (of the form ax + b)

Irreducible quadratic factors (of the form ax2 + bx + c,

where b2 – 4ac < 0).

16.
FACTORISATION OF Q(x)

For instance, if Q(x) = x4 – 16, we could

factor it as:

2 2

Q ( x) ( x 4)( x 4)

2

( x 2)( x 2)( x 4)

For instance, if Q(x) = x4 – 16, we could

factor it as:

2 2

Q ( x) ( x 4)( x 4)

2

( x 2)( x 2)( x 4)

17.
FACTORISATION OF Q(x)

The third step is to express the proper rational

function R(x)/Q(x) as a sum of partial fractions

of the form:

A Ax B

i

or 2 j

(ax b) (ax bx c)

The third step is to express the proper rational

function R(x)/Q(x) as a sum of partial fractions

of the form:

A Ax B

i

or 2 j

(ax b) (ax bx c)

18.
FACTORISATION OF Q(x)

A theorem in algebra guarantees that

it is always possible to do this.

We explain the details for the four cases

that occur.

A theorem in algebra guarantees that

it is always possible to do this.

We explain the details for the four cases

that occur.

19.
CASE 1

The denominator Q(x)

is a product of distinct linear

The denominator Q(x)

is a product of distinct linear

20.
CASE 1

This means that we can write

Q(x) = (a1x + b1) (a2x + b2)…(akx + bk)

where no factor is repeated (and no factor

is a constant multiple of another.

This means that we can write

Q(x) = (a1x + b1) (a2x + b2)…(akx + bk)

where no factor is repeated (and no factor

is a constant multiple of another.

21.
CASE 1 Equation 2

In this case, the partial fraction theorem states

that there exist constants A1, A2, . . . , Ak such

R( x) A1 A2 Ak

Q( x) a1 x b1 a2 x b2 ak x bk

In this case, the partial fraction theorem states

that there exist constants A1, A2, . . . , Ak such

R( x) A1 A2 Ak

Q( x) a1 x b1 a2 x b2 ak x bk

22.
CASE 1

These constants can be

determined as in the following

These constants can be

determined as in the following

23.
PARTIAL FRACTIONS Example 2

2

x 2x 1

Evaluate 3 2

dx

2 x 3x 2 x

The degree of the numerator is less than

the degree of the denominator.

So, we don’t need to divide.

2

x 2x 1

Evaluate 3 2

dx

2 x 3x 2 x

The degree of the numerator is less than

the degree of the denominator.

So, we don’t need to divide.

24.
PARTIAL FRACTIONS Example 2

We factor the denominator as:

2x3 + 3x2 – 2x = x(2x2 + 3x – 2)

= x(2x – 1)(x + 2)

It has three distinct linear factors.

We factor the denominator as:

2x3 + 3x2 – 2x = x(2x2 + 3x – 2)

= x(2x – 1)(x + 2)

It has three distinct linear factors.

25.
PARTIAL FRACTIONS E. g. 2—Equation 3

So, the partial fraction decomposition of

the integrand (Equation 2) has the form

2

x 2x 1 A B C

x(2 x 1)( x 2) x 2 x 1 x 2

So, the partial fraction decomposition of

the integrand (Equation 2) has the form

2

x 2x 1 A B C

x(2 x 1)( x 2) x 2 x 1 x 2

26.
PARTIAL FRACTIONS E. g. 2—Equation 4

To determine the values of A, B, and C, we

multiply both sides of the equation by the

product of the denominators, x(2x – 1)(x + 2),

x2 + 2x + 1 = A(2x – 1)(x + 2) + Bx(x + 2)

+ Cx(2x – 1)

To determine the values of A, B, and C, we

multiply both sides of the equation by the

product of the denominators, x(2x – 1)(x + 2),

x2 + 2x + 1 = A(2x – 1)(x + 2) + Bx(x + 2)

+ Cx(2x – 1)

27.
PARTIAL FRACTIONS E. g. 2—Equation 5

Expanding the right side of Equation 4 and

writing it in the standard form for polynomials,

we get:

x2 + 2x + 1 = (2A + B + 2C)x2

+ (3A + 2B – C) – 2A

Expanding the right side of Equation 4 and

writing it in the standard form for polynomials,

we get:

x2 + 2x + 1 = (2A + B + 2C)x2

+ (3A + 2B – C) – 2A

28.
PARTIAL FRACTIONS Example 2

The polynomials in Equation 5 are identical.

So, their coefficients must be equal.

The coefficient of x2 on the right side, 2A + B + 2C,

must equal that of x2 on the left side—namely, 1.

Likewise, the coefficients of x are equal and

the constant terms are equal.

The polynomials in Equation 5 are identical.

So, their coefficients must be equal.

The coefficient of x2 on the right side, 2A + B + 2C,

must equal that of x2 on the left side—namely, 1.

Likewise, the coefficients of x are equal and

the constant terms are equal.

29.
PARTIAL FRACTIONS Example 2

This gives the following system of equations

for A, B, and C:

2A + B + 2C = 1

3A + 2B – C = 2

–2A = –1

This gives the following system of equations

for A, B, and C:

2A + B + 2C = 1

3A + 2B – C = 2

–2A = –1

30.
PARTIAL FRACTIONS Example 2

Solving, we get:

A=½

B = 1/5

C = –1/10

Solving, we get:

A=½

B = 1/5

C = –1/10

31.
PARTIAL FRACTIONS Example 2

2

x 2x 1

2 x3 3x 2 2 x dx

11 1 1 1 1

dx

2 x 5 2 x 1 10 x 2

12 ln | x | 101 ln | 2 x 1| 101 | x 2 | K

2

x 2x 1

2 x3 3x 2 2 x dx

11 1 1 1 1

dx

2 x 5 2 x 1 10 x 2

12 ln | x | 101 ln | 2 x 1| 101 | x 2 | K

32.
PARTIAL FRACTIONS Example 2

In integrating the middle term,

we have made the mental substitution

u = 2x – 1, which gives

du = 2 dx and dx = du/2.

In integrating the middle term,

we have made the mental substitution

u = 2x – 1, which gives

du = 2 dx and dx = du/2.

33.
We can use an alternative method

to find the coefficients A, B, and C

in Example 2.

to find the coefficients A, B, and C

in Example 2.

34.
Equation 4 is an identity.

It is true for every value of x.

Let’s choose values of x that simplify

the equation.

It is true for every value of x.

Let’s choose values of x that simplify

the equation.

35.
If we put x = 0 in Equation 4, the second

and third terms on the right side vanish, and

the equation becomes –2A = –1.

Hence, A = ½.

and third terms on the right side vanish, and

the equation becomes –2A = –1.

Hence, A = ½.

36.
Likewise, x = ½ gives 5B/4 = 1/4

and x = –2 gives 10C = –1.

Hence, B = 1/5 and C = –1/10.

and x = –2 gives 10C = –1.

Hence, B = 1/5 and C = –1/10.

37.
You may object that Equation 3 is not

valid for x = 0, ½, or –2.

So, why should Equation 4 be valid for

those values?

valid for x = 0, ½, or –2.

So, why should Equation 4 be valid for

those values?

38.
In fact, Equation 4 is true for all values

of x, even x = 0, ½, and –2 .

See Exercise 69 for the reason.

of x, even x = 0, ½, and –2 .

See Exercise 69 for the reason.

39.
PARTIAL FRACTIONS Example 3

dx

Find 2 2

, where a ≠ 0.

x a

The method of partial fractions gives:

1 1 A B

2 2

x a ( x a)( x a ) x a x a

Therefore, A( x a) B ( x a) 1

dx

Find 2 2

, where a ≠ 0.

x a

The method of partial fractions gives:

1 1 A B

2 2

x a ( x a)( x a ) x a x a

Therefore, A( x a) B ( x a) 1

40.
PARTIAL FRACTIONS Example 3

We use the method of the preceding

We put x = a in the equation and get A(2a) = 1.

So, A = 1/(2a).

If we put x = –a, we get B(–2a) = 1.

So, B = –1/(2a).

We use the method of the preceding

We put x = a in the equation and get A(2a) = 1.

So, A = 1/(2a).

If we put x = –a, we get B(–2a) = 1.

So, B = –1/(2a).

41.
PARTIAL FRACTIONS Example 3

dx 1 1 1

2 2

x a 2a x a x a dx

1

(ln | x a | ln | x a |) C

2a

dx 1 1 1

2 2

x a 2a x a x a dx

1

(ln | x a | ln | x a |) C

2a

42.
PARTIAL FRACTIONS E. g. 3—Formula 6

Since ln x – ln y = ln(x/y), we can write

the integral as:

dx 1 x a

x 2 a 2 2a ln x a C

See Exercises 55–56 for ways of using Formula 6.

Since ln x – ln y = ln(x/y), we can write

the integral as:

dx 1 x a

x 2 a 2 2a ln x a C

See Exercises 55–56 for ways of using Formula 6.

43.
CASE 2

Q(x) is a product of

linear factors, some of which

are repeated.

Q(x) is a product of

linear factors, some of which

are repeated.

44.
CASE 2

Suppose the first linear factor (a1x + b1) is

repeated r times.

That is, (a1x + b1)r occurs in the factorization

of Q(x).

Suppose the first linear factor (a1x + b1) is

repeated r times.

That is, (a1x + b1)r occurs in the factorization

of Q(x).

45.
CASE 2 Equation 7

Then, instead of the single term A1/(a1x + b1)

in Equation 2, we would use:

A1 A2 Ar

2

r

a1 x b1 (a1 x b1 ) (a1 x b1 )

Then, instead of the single term A1/(a1x + b1)

in Equation 2, we would use:

A1 A2 Ar

2

r

a1 x b1 (a1 x b1 ) (a1 x b1 )

46.
CASE 2

By way of illustration, we could write:

3

x x 1 A B C D E

2 3

2 2

3

x ( x 1) x x x 1 ( x 1) ( x 1)

However, we prefer to work out in detail

a simpler example, as follows.

By way of illustration, we could write:

3

x x 1 A B C D E

2 3

2 2

3

x ( x 1) x x x 1 ( x 1) ( x 1)

However, we prefer to work out in detail

a simpler example, as follows.

47.
PARTIAL FRACTIONS Example 4

4 2

x 2 x 4 x 1

Find 3 2 dx

x x x 1

The first step is to divide.

4 2

The result of long division is:

x 2 x 4 x 1

x3 x 2 x 1

4x

x 1 3

x x2 x 1

4 2

x 2 x 4 x 1

Find 3 2 dx

x x x 1

The first step is to divide.

4 2

The result of long division is:

x 2 x 4 x 1

x3 x 2 x 1

4x

x 1 3

x x2 x 1

48.
PARTIAL FRACTIONS Example 4

The second step is to factor the

denominator Q(x) = x3 – x2 – x + 1.

Since Q(1) = 0, we know that x – 1 is a factor,

and we obtain:

x3 x 2 x 1 ( x 1)( x 2 1)

( x 1)( x 1)( x 1)

2

( x 1) ( x 1)

The second step is to factor the

denominator Q(x) = x3 – x2 – x + 1.

Since Q(1) = 0, we know that x – 1 is a factor,

and we obtain:

x3 x 2 x 1 ( x 1)( x 2 1)

( x 1)( x 1)( x 1)

2

( x 1) ( x 1)

49.
PARTIAL FRACTIONS Example 4

The linear factor x – 1 occurs twice.

So, the partial fraction decomposition is:

4x A B C

2

2

( x 1) ( x 1) x 1 ( x 1) x 1

The linear factor x – 1 occurs twice.

So, the partial fraction decomposition is:

4x A B C

2

2

( x 1) ( x 1) x 1 ( x 1) x 1

50.
PARTIAL FRACTIONS E. g. 4—Equation 8

Multiplying by the least common denominator,

(x – 1)2 (x + 1), we get:

2

4 x A( x 1)( x 1) B( x 1) C ( x 1)

2

( A C ) x ( B 2C ) x ( A B C )

Multiplying by the least common denominator,

(x – 1)2 (x + 1), we get:

2

4 x A( x 1)( x 1) B( x 1) C ( x 1)

2

( A C ) x ( B 2C ) x ( A B C )

51.
PARTIAL FRACTIONS Example 4

Now, we equate coefficients:

A C 0

B 2C 4

A B C 0

Now, we equate coefficients:

A C 0

B 2C 4

A B C 0

52.
PARTIAL FRACTIONS Example 4

Solving, we obtain:

A=1

B=2

C = -1

Solving, we obtain:

A=1

B=2

C = -1

53.
PARTIAL FRACTIONS Example 4

Thus, 4 2

x 2 x 4 x 1

x3 x 2 x 1 dx

1 2 1

x 1 2

dx

x 1 ( x 1) x 1

x2 2

x ln | x 1| ln | x 1| K

2 x 1

2

x 2 x 1

x ln K

2 x 1 x 1

Thus, 4 2

x 2 x 4 x 1

x3 x 2 x 1 dx

1 2 1

x 1 2

dx

x 1 ( x 1) x 1

x2 2

x ln | x 1| ln | x 1| K

2 x 1

2

x 2 x 1

x ln K

2 x 1 x 1

54.
CASE 3

Q(x) contains irreducible

quadratic factors, none of which

is repeated.

Q(x) contains irreducible

quadratic factors, none of which

is repeated.

55.
CASE 3 Formula 9

If Q(x) has the factor ax2 + bx + c, where

b2 – 4ac < 0, then, in addition to the partial

fractions in Equations 2 and 7, the expression

for R(x)/Q(x) will have a term of the form

Ax B

2

ax bx c

where A and B are constants to be

If Q(x) has the factor ax2 + bx + c, where

b2 – 4ac < 0, then, in addition to the partial

fractions in Equations 2 and 7, the expression

for R(x)/Q(x) will have a term of the form

Ax B

2

ax bx c

where A and B are constants to be

56.
CASE 3

For instance, the function given by

f(x) = x/[(x – 2)(x2 + 1)(x2 + 4) has a partial

fraction decomposition of the form

x

2 2

( x 2)( x 1)( x 4)

A Bx C Dx E

2 2

x 2 x 1 x 4

For instance, the function given by

f(x) = x/[(x – 2)(x2 + 1)(x2 + 4) has a partial

fraction decomposition of the form

x

2 2

( x 2)( x 1)( x 4)

A Bx C Dx E

2 2

x 2 x 1 x 4

57.
CASE 3 Formula 10

The term in Formula 9 can be integrated

by completing the square and using

the formula

dx 1 1 x

x 2 a 2 a tan a C

The term in Formula 9 can be integrated

by completing the square and using

the formula

dx 1 1 x

x 2 a 2 a tan a C

58.
PARTIAL FRACTIONS Example 5

2

2

Evaluate 3x x 4

dx

x 4x

As x3 + 4x = x(x2 + 4) can’t be factored further,

we write: 2

2 x x 4 A Bx C

2

2

x( x 4) x x 4

2

2

Evaluate 3x x 4

dx

x 4x

As x3 + 4x = x(x2 + 4) can’t be factored further,

we write: 2

2 x x 4 A Bx C

2

2

x( x 4) x x 4

59.
PARTIAL FRACTIONS Example 5

Multiplying by x(x2 + 4), we have:

2 2

2 x x 4 A( x 4) ( Bx C ) x

2

( A B) x Cx 4 A

Multiplying by x(x2 + 4), we have:

2 2

2 x x 4 A( x 4) ( Bx C ) x

2

( A B) x Cx 4 A

60.
PARTIAL FRACTIONS Example 5

Equating coefficients, we obtain:

A+B=2 C = –1 4A = 4

Thus, A = 1, B = 1, and C = –1.

Equating coefficients, we obtain:

A+B=2 C = –1 4A = 4

Thus, A = 1, B = 1, and C = –1.

61.
PARTIAL FRACTIONS Example 5

2

2x x 4 1 x 1

x3 4 x dx

x x 2 4

dx

2

2x x 4 1 x 1

x3 4 x dx

x x 2 4

dx

62.
PARTIAL FRACTIONS Example 5

In order to integrate the second term,

we split it into two parts:

x 1 x 1

x 2 4 dx x 2 4 dx x2 4 dx

In order to integrate the second term,

we split it into two parts:

x 1 x 1

x 2 4 dx x 2 4 dx x2 4 dx

63.
PARTIAL FRACTIONS Example 5

We make the substitution u = x2 + 4

in the first of these integrals so that

du = 2x dx.

We make the substitution u = x2 + 4

in the first of these integrals so that

du = 2x dx.

64.
PARTIAL FRACTIONS Example 5

We evaluate the second integral by means

of Formula 10 with a = 2:

2

2x x 4

x( x 2 4) dx

1 x 1

dx 2 dx 2 dx

x x 4 x 4

2 1

ln | x | 2 ln( x 4) 2 tan ( x / 2) K

1 1

We evaluate the second integral by means

of Formula 10 with a = 2:

2

2x x 4

x( x 2 4) dx

1 x 1

dx 2 dx 2 dx

x x 4 x 4

2 1

ln | x | 2 ln( x 4) 2 tan ( x / 2) K

1 1

65.
PARTIAL FRACTIONS Example 6

2

4 x 3x 2

Evaluate 2 dx

4x 4x 3

The degree of the numerator is not less than

the degree of the denominator.

4 x 2 3x 2

So, we first divide and obtain:

4x2 4x 3

x 1

1 2

4x 4x 3

2

4 x 3x 2

Evaluate 2 dx

4x 4x 3

The degree of the numerator is not less than

the degree of the denominator.

4 x 2 3x 2

So, we first divide and obtain:

4x2 4x 3

x 1

1 2

4x 4x 3

66.
PARTIAL FRACTIONS Example 6

Notice that the quadratic 4x2 – 4x + 3

is irreducible because its discriminant

is b2 – 4ac = –32 < 0.

This means it can’t be factored.

So, we don’t need to use the partial fraction technique.

Notice that the quadratic 4x2 – 4x + 3

is irreducible because its discriminant

is b2 – 4ac = –32 < 0.

This means it can’t be factored.

So, we don’t need to use the partial fraction technique.

67.
PARTIAL FRACTIONS Example 6

To integrate the function, we complete

the square in the denominator:

2 2

4 x 4 x 3 (2 x 1) 2

This suggests we make the substitution u = 2x – 1.

Then, du = 2 dx, and x = ½(u + 1).

To integrate the function, we complete

the square in the denominator:

2 2

4 x 4 x 3 (2 x 1) 2

This suggests we make the substitution u = 2x – 1.

Then, du = 2 dx, and x = ½(u + 1).

68.
PARTIAL FRACTIONS Example 6

2

4 x 3x 2

4 x 2 4 x 3 dx

x 1

1 2 dx

4x 4x 3

2 (u 1) 1

1

x 2 2

1

du

u 2

u 1

x 4 2

1

du

u 2

2

4 x 3x 2

4 x 2 4 x 3 dx

x 1

1 2 dx

4x 4x 3

2 (u 1) 1

1

x 2 2

1

du

u 2

u 1

x 4 2

1

du

u 2

69.
PARTIAL FRACTIONS Example 6

u 1

x 2

1

4 du 4 2

1

du

u 2 u 2

2 1 1 1 u

x 8 ln(u 2) tan

1

C

4 2 2

2 1 1 2x 1

x 8 ln(4 x 4 x 3)

1

tan C

4 2 2

u 1

x 2

1

4 du 4 2

1

du

u 2 u 2

2 1 1 1 u

x 8 ln(u 2) tan

1

C

4 2 2

2 1 1 2x 1

x 8 ln(4 x 4 x 3)

1

tan C

4 2 2

70.
Example 6 illustrates the general

procedure for integrating a partial fraction

of the form

Ax B 2

2

where b 4ac 0

ax bx c

procedure for integrating a partial fraction

of the form

Ax B 2

2

where b 4ac 0

ax bx c

71.
We complete the square in the denominator

and then make a substitution that brings

the integral into the form

Cu D u 1

u 2 a 2 du C u 2 a 2 du D u 2 a 2 du

Then, the first integral is a logarithm and

the second is expressed in terms of tan-1.

and then make a substitution that brings

the integral into the form

Cu D u 1

u 2 a 2 du C u 2 a 2 du D u 2 a 2 du

Then, the first integral is a logarithm and

the second is expressed in terms of tan-1.

72.
CASE 4

Q(x) contains

a repeated irreducible

quadratic factor.

Q(x) contains

a repeated irreducible

quadratic factor.

73.
CASE 4

Suppose Q(x) has the factor

(ax2 + bx + c)r

where b2 – 4ac < 0.

Suppose Q(x) has the factor

(ax2 + bx + c)r

where b2 – 4ac < 0.

74.
CASE 4 Formula 11

Then, instead of the single partial fraction

(Formula 9), the sum

A1 x B1 A2 x B2 Ar x Br

2

2 2

ax bx c (ax bx c) (ax 2 bx c )r

occurs in the partial fraction decomposition

of R(x)/Q(x).

Then, instead of the single partial fraction

(Formula 9), the sum

A1 x B1 A2 x B2 Ar x Br

2

2 2

ax bx c (ax bx c) (ax 2 bx c )r

occurs in the partial fraction decomposition

of R(x)/Q(x).

75.
CASE 4

Each of the terms in Formula 11

can be integrated by first completing

the square.

Each of the terms in Formula 11

can be integrated by first completing

the square.

76.
PARTIAL FRACTIONS Example 7

Write out the form of the partial fraction

decomposition of the function

3 2

x x 1

2 2 3

x( x 1)( x x 1)( x 1)

Write out the form of the partial fraction

decomposition of the function

3 2

x x 1

2 2 3

x( x 1)( x x 1)( x 1)

77.
PARTIAL FRACTIONS Example 7

We have:

3 2

x x 1

2 2 3

x( x 1)( x x 1)( x 1)

A B Cx D Ex F

2 2

x x 1 x x 1 x 1

Gx h Ix J

2 2

2 3

( x 1) ( x 1)

We have:

3 2

x x 1

2 2 3

x( x 1)( x x 1)( x 1)

A B Cx D Ex F

2 2

x x 1 x x 1 x 1

Gx h Ix J

2 2

2 3

( x 1) ( x 1)

78.
PARTIAL FRACTIONS Example 8

2 3

1 x 2x x

Evaluate 2 2

dx

x( x 1)

The form of the partial fraction decomposition is:

2 3

1 x 2x x A Bx C Dx E

2 2

2 2 2

x( x 1) x x 1 ( x 1)

2 3

1 x 2x x

Evaluate 2 2

dx

x( x 1)

The form of the partial fraction decomposition is:

2 3

1 x 2x x A Bx C Dx E

2 2

2 2 2

x( x 1) x x 1 ( x 1)

79.
PARTIAL FRACTIONS Example 8

Multiplying by x(x2 + 1)2,

we have:

x3 2 x 2 x 1

2 2 2

A( x 1) ( Bx C ) x( x 1) ( Dx E ) x

A( x 4 2 x 2 1) B ( x 4 x 2 ) C ( x 3 x) Dx 2 Ex

( A B ) x 4 Cx 3 (2 A B D) x 2 (C E ) x A

Multiplying by x(x2 + 1)2,

we have:

x3 2 x 2 x 1

2 2 2

A( x 1) ( Bx C ) x( x 1) ( Dx E ) x

A( x 4 2 x 2 1) B ( x 4 x 2 ) C ( x 3 x) Dx 2 Ex

( A B ) x 4 Cx 3 (2 A B D) x 2 (C E ) x A

80.
PARTIAL FRACTIONS Example 8

If we equate coefficients,

we get the system A B 0

C 1

2 A B D 2

C E 1

A 1

This has the solution

A = 1, B = –1, C = –1, D = 1, E =

0.

If we equate coefficients,

we get the system A B 0

C 1

2 A B D 2

C E 1

A 1

This has the solution

A = 1, B = –1, C = –1, D = 1, E =

0.

81.
PARTIAL FRACTIONS Example 8

2 3

1 x 2x x

x( x 2 1)2 dx

1 x 1 x

2 2 2

dx

x x 1 ( x 1)

dx x dx x dx

2 dx 2 2

x x 1 x 1 ( x 1) 2

2 1 1

ln | x | 2 ln( x 1) tan x

1

2 2

K

2( x 1)

2 3

1 x 2x x

x( x 2 1)2 dx

1 x 1 x

2 2 2

dx

x x 1 ( x 1)

dx x dx x dx

2 dx 2 2

x x 1 x 1 ( x 1) 2

2 1 1

ln | x | 2 ln( x 1) tan x

1

2 2

K

2( x 1)

82.
AVOIDING PARTIAL FRACTIONS

We note that, sometimes,

partial fractions can be avoided

when integrating a rational function.

We note that, sometimes,

partial fractions can be avoided

when integrating a rational function.

83.
AVOIDING PARTIAL FRACTIONS

For instance, the integral

2

x 1

x( x 2 3) dx

could be evaluated by the method

of Case 3.

For instance, the integral

2

x 1

x( x 2 3) dx

could be evaluated by the method

of Case 3.

84.
AVOIDING PARTIAL FRACTIONS

However, it is much easier to observe that,

if u = x(x2 + 3) = x3 + 3x, then du = (3x2 + 3) dx

and so

2

x 1 3

x( x 2 3) dx 3 ln | x 3x | C

1

However, it is much easier to observe that,

if u = x(x2 + 3) = x3 + 3x, then du = (3x2 + 3) dx

and so

2

x 1 3

x( x 2 3) dx 3 ln | x 3x | C

1

85.
RATIONALIZING SUBSTITUTIONS

Some nonrational functions can be

changed into rational functions by means

of appropriate substitutions.

In particular, when an integrand contains

an expression of the form n√g(x), then

the substitution u = n√g(x) may be effective.

Some nonrational functions can be

changed into rational functions by means

of appropriate substitutions.

In particular, when an integrand contains

an expression of the form n√g(x), then

the substitution u = n√g(x) may be effective.

86.
RATIONALIZING SUBSTITUTIONS Example 9

Evaluate x4

dx

x

Let u x4

Then, u2 = x + 4

So, x = u2 – 4 and dx = 2u du

Evaluate x4

dx

x

Let u x4

Then, u2 = x + 4

So, x = u2 – 4 and dx = 2u du

87.
RATIONALIZING SUBSTITUTIONS Example 9

Therefore,

x4 u

dx 2 2u du

x u 4

2

u

2 2 du

u 4

4

2 1 2 4 du

u

Therefore,

x4 u

dx 2 2u du

x u 4

2

u

2 2 du

u 4

4

2 1 2 4 du

u

88.
RATIONALIZING SUBSTITUTIONS Example 9

We can evaluate this integral

by factoring u2 – 4 as (u – 2)(u + 2)

and using partial fractions.

We can evaluate this integral

by factoring u2 – 4 as (u – 2)(u + 2)

and using partial fractions.

89.
RATIONALIZING SUBSTITUTIONS Example 9

Alternatively, we can use Formula 6

with a = 2: x4

x dx

du

2du 8 2

u 4

1 u 2

2u 8 ln C

2 2 u 2

x4 2

2 x 4 2 ln C

x4 2

Alternatively, we can use Formula 6

with a = 2: x4

x dx

du

2du 8 2

u 4

1 u 2

2u 8 ln C

2 2 u 2

x4 2

2 x 4 2 ln C

x4 2