TECHNIQUES OF INTEGRATION
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7TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION
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7.4Integration of Rational Functions
by Partial Fractions
TECHNIQUES OF INTEGRATION
In this section, we will learn:
How to integrate rational functions
by reducing them to a sum of simpler fractions.
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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.
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To illustrate the method, observe that,
by taking the fractions 2/(x – 1) and 1/(x – 2)
to a common denominator, we obtain:
INTEGRATION BY PARTIAL FRACTIONS
2
2 1 2( 2) ( 1)
1 2 ( 1)( 2)
5
2
x x
x x x x
x
x x
![Page 5: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/5.jpg)
If we now reverse the procedure, we see
how to integrate the function on the right side
of this equation:
INTEGRATION BY PARTIAL FRACTIONS
2
5 2 1
2 1 2
2ln | 1| ln | 2 |
xdx dx
x x x x
x x C
![Page 6: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/6.jpg)
To see how the method of partial fractions
works in general, let’s consider a rational
function
where P and Q are polynomials.
INTEGRATION BY PARTIAL FRACTIONS
( )( )
( )
P xf x
Q x
![Page 7: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/7.jpg)
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.
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Recall that, if
where an ≠ 0, then the degree of P is n
and we write deg(P) = n.
DEGREE OF P
1
11 0( ) n n
n nP x a x a x a x a
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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).
PARTIAL FRACTIONS
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The division statement is
where S and R are also polynomials.
PARTIAL FRACTIONS
( ) ( )( ) ( )
( ) ( )
P x R xf x S x
Q x Q x
Equation 1
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PARTIAL FRACTIONS
As the following example illustrates,
sometimes, this preliminary step is all
that is required.
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Find
The degree of the numerator is greater than that of the denominator.
So, we first perform the long division.
PARTIAL FRACTIONS Example 13
1
x xdx
x
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PARTIAL FRACTIONS
This enables us to write:
32
3 2
22
1 1
2 2ln | 1|3 2
x xdx x x dx
x x
x xx x C
Example 1
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The next step is to factor
the denominator Q(x) as far
as possible.
PARTIAL FRACTIONS
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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).
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FACTORISATION OF Q(x)
For instance, if Q(x) = x4 – 16, we could
factor it as:
2 2
2
( ) ( 4)( 4)
( 2)( 2)( 4)
Q x x x
x x x
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The third step is to express the proper rational
function R(x)/Q(x) as a sum of partial fractions
of the form:
FACTORISATION OF Q(x)
2or
( ) ( )
i j
A Ax B
ax b ax bx c
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A theorem in algebra guarantees that
it is always possible to do this.
We explain the details for the four cases that occur.
FACTORISATION OF Q(x)
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The denominator Q(x)
is a product of distinct linear
factors.
CASE 1
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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.
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In this case, the partial fraction theorem states
that there exist constants A1, A2, . . . , Ak such
that:
CASE 1
1 2
1 1 2 2
( )
( )k
k k
AA AR x
Q x a x b a x b a x b
Equation 2
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CASE 1
These constants can be
determined as in the following
example.
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Evaluate
The degree of the numerator is less than the degree of the denominator.
So, we don’t need to divide.
PARTIAL FRACTIONS Example 22
3 2
2 1
2 3 2
x xdx
x x x
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PARTIAL FRACTIONS
We factor the denominator as:
2x3 + 3x2 – 2x = x(2x2 + 3x – 2)
= x(2x – 1)(x + 2)
It has three distinct linear factors.
Example 2
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So, the partial fraction decomposition of
the integrand (Equation 2) has the form
PARTIAL FRACTIONS
2 2 1
(2 1)( 2) 2 1 2
x x A B C
x x x x x x
E. g. 2—Equation 3
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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),
obtaining:
x2 + 2x + 1 = A(2x – 1)(x + 2) + Bx(x + 2)
+ Cx(2x – 1)
PARTIAL FRACTIONS E. g. 2—Equation 4
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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
PARTIAL FRACTIONS E. g. 2—Equation 5
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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.
PARTIAL FRACTIONS Example 2
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This gives the following system of equations
for A, B, and C:
2A + B + 2C = 1
3A + 2B – C = 2
–2A = –1
PARTIAL FRACTIONS Example 2
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PARTIAL FRACTIONS
Solving, we get:
A = ½
B = 1/5
C = –1/10
Example 2
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Hence,
PARTIAL FRACTIONS
2
3 2
1 1 12 10 10
2 1
2 3 21 1 1 1 1 1
2 5 2 1 10 2
ln | | ln | 2 1| | 2 |
x xdx
x x x
dxx x x
x x x K
Example 2
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PARTIAL FRACTIONS
In integrating the middle term,
we have made the mental substitution
u = 2x – 1, which gives
du = 2 dx and dx = du/2.
Example 2
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We can use an alternative method
to find the coefficients A, B, and C
in Example 2.
NOTE
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Equation 4 is an identity.
It is true for every value of x.
Let’s choose values of x that simplify the equation.
NOTE
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NOTE
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.
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NOTE
Likewise, x = ½ gives 5B/4 = 1/4
and x = –2 gives 10C = –1.
Hence, B = 1/5 and C = –1/10.
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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?
NOTE
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NOTE
In fact, Equation 4 is true for all values
of x, even x = 0, ½, and –2 .
See Exercise 69 for the reason.
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Find , where a ≠ 0.
The method of partial fractions gives:
Therefore,
PARTIAL FRACTIONS Example 3
2 2
dx
x a
2 2
1 1
( )( )
A B
x a x a x a x a x a
( ) ( ) 1A x a B x a
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We use the method of the preceding
note.
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).
PARTIAL FRACTIONS Example 3
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PARTIAL FRACTIONS
Therefore,
2 2
1 1 1
2
1(ln | | ln | |)
2
dx
dxx a a x a x a
x a x a Ca
Example 3
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Since ln x – ln y = ln(x/y), we can write
the integral as:
See Exercises 55–56 for ways of using Formula 6.
PARTIAL FRACTIONS
2 2
1ln
2
dx x aC
x a a x a
E. g. 3—Formula 6
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Q(x) is a product of
linear factors, some of which
are repeated.
CASE 2
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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).
![Page 45: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/45.jpg)
Then, instead of the single term A1/(a1x + b1)
in Equation 2, we would use:
CASE 2
1 22
1 1 1 1 1 1 ( ) ( )r
r
A A A
a x b a x b a x b
Equation 7
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By way of illustration, we could write:
However, we prefer to work out in detail a simpler example, as follows.
CASE 2
3
2 3 2 2 3
1
( 1) 1 ( 1) ( 1)
x x A B C D E
x x x x x x x
![Page 47: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/47.jpg)
Find
The first step is to divide.
The result of long division is:
PARTIAL FRACTIONS Example 44 2
3 2
2 4 1
1
x x xdx
x x x
4 2
3 2
3 2
2 4 1
14
11
x x x
x x xx
xx x x
![Page 48: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/48.jpg)
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:
PARTIAL FRACTIONS
3 2 2
2
1 ( 1)( 1)
( 1)( 1)( 1)
( 1) ( 1)
x x x x x
x x x
x x
Example 4
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The linear factor x – 1 occurs twice.
So, the partial fraction decomposition is:
PARTIAL FRACTIONS
2 2
4
( 1) ( 1) 1 ( 1) 1
x A B C
x x x x x
Example 4
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Multiplying by the least common denominator,
(x – 1)2 (x + 1), we get:
PARTIAL FRACTIONS
2
2
4 ( 1)( 1) ( 1) ( 1)
( ) ( 2 ) ( )
x A x x B x C x
A C x B C x A B C
E. g. 4—Equation 8
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PARTIAL FRACTIONS
Now, we equate coefficients:
0
2 4
0
A C
B C
A B C
Example 4
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Solving, we obtain:
A = 1
B = 2
C = -1
PARTIAL FRACTIONS Example 4
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PARTIAL FRACTIONS
Thus, 4 2
3 2
2
2
2
2 4 1
1
1 2 11
1 ( 1) 1
2ln | 1| ln | 1|
2 1
2 1ln
2 1 1
x x xdx
x x x
x dxx x x
xx x x K
x
x xx Kx x
Example 4
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Q(x) contains irreducible
quadratic factors, none of which
is repeated.
CASE 3
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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
where A and B are constants to be
determined.
CASE 3 Formula 9
2
Ax B
ax bx c
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For instance, the function given by
f(x) = x/[(x – 2)(x2 + 1)(x2 + 4) has a partial
fraction decomposition of the form
CASE 3
2 2
2 2
( 2)( 1)( 4)
2 1 4
x
x x x
A Bx C Dx E
x x x
![Page 57: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/57.jpg)
The term in Formula 9 can be integrated
by completing the square and using
the formula
CASE 3
12 2
1tan
dx xC
x a a a
Formula 10
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Evaluate
As x3 + 4x = x(x2 + 4) can’t be factored further, we write:
PARTIAL FRACTIONS Example 52
3
2 4
4
x xdx
x x
2
2 2
2 4
( 4) 4
x x A Bx C
x x x x
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Multiplying by x(x2 + 4), we have:
PARTIAL FRACTIONS
2 2
2
2 4 ( 4) ( )
( ) 4
x x A x Bx C x
A B x Cx A
Example 5
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PARTIAL FRACTIONS
Equating coefficients, we obtain:
A + B = 2 C = –1 4A = 4
Thus, A = 1, B = 1, and C = –1.
Example 5
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Hence,
PARTIAL FRACTIONS
2
3 2
2 4 1 1
4 4
x x xdx dx
x x x x
Example 5
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In order to integrate the second term,
we split it into two parts:
PARTIAL FRACTIONS
2 2 2
1 1
4 4 4
x xdx dx dx
x x x
Example 5
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We make the substitution u = x2 + 4
in the first of these integrals so that
du = 2x dx.
PARTIAL FRACTIONS Example 5
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We evaluate the second integral by means
of Formula 10 with a = 2:
PARTIAL FRACTIONS
2
2
2 2
2 11 12 2
2 4
( 4)
1 1
4 4
ln | | ln( 4) tan ( / 2)
x xdx
x x
xdx dx dxx x x
x x x K
Example 5
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Evaluate
The degree of the numerator is not less than the degree of the denominator.
So, we first divide and obtain:
PARTIAL FRACTIONS2
2
4 3 2
4 4 3
x xdx
x x
2
2
2
4 3 2
4 4 31
14 4 3
x x
x xx
x x
Example 6
![Page 66: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/66.jpg)
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.
PARTIAL FRACTIONS Example 6
![Page 67: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/67.jpg)
To integrate the function, we complete
the square in the denominator:
This suggests we make the substitution u = 2x – 1.
Then, du = 2 dx, and x = ½(u + 1).
PARTIAL FRACTIONS
2 24 4 3 (2 1) 2 x x x
Example 6
![Page 68: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/68.jpg)
Thus,
PARTIAL FRACTIONS
2
2
2
121
2 2
14 2
4 3 2
4 4 31
14 4 3
( 1) 1
21
2
x xdx
x xx
dxx x
ux du
uu
x duu
Example 6
![Page 69: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/69.jpg)
PARTIAL FRACTIONS
1 14 42 2
2 118
2 118
1
2 2
1 1ln( 2) tan
4 2 2
1 2 1ln(4 4 3) tan
4 2 2
u
x du duu u
ux u C
xx x x C
Example 6
![Page 70: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/70.jpg)
Example 6 illustrates the general
procedure for integrating a partial fraction
of the form
NOTE
22
where 4 0Ax B
b acax bx c
![Page 71: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/71.jpg)
We complete the square in the denominator
and then make a substitution that brings
the integral into the form
Then, the first integral is a logarithm and the second is expressed in terms of tan-1.
NOTE
2 2 2 2 2 2
1Cu D udu C du D du
u a u a u a
![Page 72: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/72.jpg)
Q(x) contains
a repeated irreducible
quadratic factor.
CASE 4
![Page 73: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/73.jpg)
Suppose Q(x) has the factor
(ax2 + bx + c)r
where b2 – 4ac < 0.
CASE 4
![Page 74: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/74.jpg)
Then, instead of the single partial fraction
(Formula 9), the sum
occurs in the partial fraction decomposition
of R(x)/Q(x).
CASE 4
1 1 2 22 2 2 2 ( ) ( )
r r
r
A x B A x B A x B
ax bx c ax bx c ax bx c
Formula 11
![Page 75: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/75.jpg)
CASE 4
Each of the terms in Formula 11
can be integrated by first completing
the square.
![Page 76: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/76.jpg)
Write out the form of the partial fraction
decomposition of the function
PARTIAL FRACTIONS Example 7
3 2
2 2 3
1
( 1)( 1)( 1)
x x
x x x x x
![Page 77: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/77.jpg)
We have:
PARTIAL FRACTIONS
3 2
2 2 3
2 2
2 2 2 3
1
( 1)( 1)( 1)
1 1 1
( 1) ( 1)
x x
x x x x x
A B Cx D Ex F
x x x x xGx h Ix J
x x
Example 7
![Page 78: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/78.jpg)
Evaluate
The form of the partial fraction decomposition is:
PARTIAL FRACTIONS2 3
2 2
1 2
( 1)
x x xdx
x x
2 3
2 2 2 2 2
1 2
( 1) 1 ( 1)
x x x A Bx C Dx E
x x x x x
Example 8
![Page 79: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/79.jpg)
Multiplying by x(x2 + 1)2,
we have:
PARTIAL FRACTIONS
3 2
2 2 2
4 2 4 2 3 2
4 3 2
2 1
( 1) ( ) ( 1) ( )
( 2 1) ( ) ( )
( ) (2 ) ( )
x x x
A x Bx C x x Dx E x
A x x B x x C x x Dx Ex
A B x Cx A B D x C E x A
Example 8
![Page 80: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/80.jpg)
If we equate coefficients,
we get the system
This has the solution A = 1, B = –1, C = –1, D = 1, E =
0.
PARTIAL FRACTIONS
0
1
2 2
1
1
A B
C
A B D
C E
A
Example 8
![Page 81: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/81.jpg)
Thus,
PARTIAL FRACTIONS
2 3
2 2
2 2 2
2 2 2 2
2 112 2 2
1 2
( 1)
1 1
1 ( 1)
1 1 ( 1)
1ln | | ln( 1) tan
2( 1)
x x xdx
x x
x xdx
x x x
dx x dx x dxdx
x x x x
x x x Kx
Example 8
![Page 82: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/82.jpg)
We note that, sometimes,
partial fractions can be avoided
when integrating a rational function.
AVOIDING PARTIAL FRACTIONS
![Page 83: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/83.jpg)
For instance, the integral
could be evaluated by the method
of Case 3.
AVOIDING PARTIAL FRACTIONS
2
2
1
( 3)
xdx
x x
![Page 84: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/84.jpg)
However, it is much easier to observe that,
if u = x(x2 + 3) = x3 + 3x, then du = (3x2 + 3) dx
and so
AVOIDING PARTIAL FRACTIONS
231
32
1ln | 3 |
( 3)
xdx x x C
x x
![Page 85: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/85.jpg)
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.
RATIONALIZING SUBSTITUTIONS
![Page 86: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/86.jpg)
Evaluate
Let
Then, u2 = x + 4
So, x = u2 – 4 and dx = 2u du
RATIONALIZING SUBSTITUTIONS Example 9
4xdx
x
4u x
![Page 87: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/87.jpg)
Therefore,
RATIONALIZING SUBSTITUTIONS
2
2
2
2
42
4
244
2 1 4
x udx u du
x u
udu
u
duu
Example 9
![Page 88: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/88.jpg)
We can evaluate this integral
by factoring u2 – 4 as (u – 2)(u + 2)
and using partial fractions.
RATIONALIZING SUBSTITUTIONS Example 9
![Page 89: TECHNIQUES OF INTEGRATION](https://reader035.fdocuments.net/reader035/viewer/2022062217/56812bc6550346895d9010bf/html5/thumbnails/89.jpg)
Alternatively, we can use Formula 6
with a = 2:
RATIONALIZING SUBSTITUTIONS
2
4
2 84
1 22 8 ln
2 2 2
4 22 4 2ln
4 2
xdx
xdu
duu
uu C
u
xx C
x
Example 9