LESSON 10 INTEGRATION OF RATIONAL
FUNCTIONS BY PARTIAL FRACTIONS
Recall that a rational function is a quotient of polynomial functions.
2
Definition The quadratic a x  b x  c is called irreducible if the quadratic
2
2
equation a x  b x  c  0 has no real solutions. Thus, b  4 a c  0 .
A theorem from algebra states that a polynomial with real coefficients can be
factored into a product of linear factors, a product of irreducible quadratic factors or
a product of linear and irreducible quadratic factors.
Given a rational function, if the degree of the polynomial in the numerator is less
than the degree of the polynomial in the denominator, then the rational function can
be expressed as the sum of rational functions whose denominators are powers of
linear polynomials and powers of irreducible quadratic functions. This sum is
called the partial fraction decomposition of the rational function.
p( x)
.
q( x )
Given the rational function r ( x ) 
If the factorization of the polynomial q ( x ) contains m identical linear factors
a x  b , then the partial fraction decomposition of r ( x ) contains a sum of the form
A1
ax  b

A2
(a x  b) 2

A3
(a x  b) 3
  
Am
(a x  b) m
,
where A1 , A 2 , A3 , . . . . , A m are constants to be determined.
If the factorization of the polynomial q ( x ) contains m identical irreducible
2
quadratic factors a x  b x  c , then the partial fraction decomposition of r ( x )
contains a sum of the form
A1 x  B 1
ax2  bx  c

A2 x  B 2
(a x 2  b x  c) 2

A3 x  B 3
(a x 2  b x  c)3
  
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Am x  B m
(a x 2  b x  c) m
,
where A1 , A 2 , A3 , . . . . , A m and B 1 , B 2 , B 3 , . . . . , B m are constants to be
determined.
Example Find the partial fraction decomposition for the following rational
expression. Do NOT solve for the constants.
4 x 3  3 x 2  5 x  12
( 3 x  5) ( x  2 ) 3 ( x 2  4 x  6 ) ( 6 x 2  2 x  7 ) 2
B
A
4 x 3  3 x 2  5 x  12
=
+
+
x  2
3x  5
( 3 x  5) ( x  2 ) 3 ( x 2  4 x  6 ) ( 6 x 2  2 x  7 ) 2
Ex  F
Gx  H
C
D
Ix  J
+
+
,
2
2
2 +
3 +
2
x  4x  6
6x  2x  7
( x  2)
( x  2)
(6 x  2 x  7) 2
where A, B, C, D, E, F, G, H, I, and J are constants to be determined.
NOTE: If we were asked to integrate the rational function given by the expression
on the left-hand side of the equation above, we could integrate the rational
functions given by the expressions on the right-hand side of the equation. In order
to integrate the rational functions with quadratic denominators, we might need to
complete the square for the quadratic.
Examples Evaluate the following integrals.
1.

6
dx
x2  9x
6
6
A
B

=
=
. We need to solve for A and B.
x( x  9)
x2  9x
x
x 9
Multiplying both sides of the equation
6
A
B

=
by
x( x  9)
x
x 9
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x ( x  9 ) , we obtain the equation 6  A ( x  9 )  B x . From here, there are
two methods for solving for A and B.
Method 1
The first method is to solve a system of equations obtained from equating the
coefficients of the terms on each side of the equation 6  A ( x  9 )  B x .
6  A( x  9 )  B x  6  A x  9 A  B x  6  ( A  B ) x  9 A
The coefficient of the x term on the right side of the equation
6  ( A  B ) x  9 A is A  B . Since there is not an x term on the left side of
the equation, then its coefficient is zero. Equating the coefficients of the x
terms on each side of the equation, we obtain that A  B  0 . The constant
term on the right side of the equation 6  ( A  B ) x  9 A is  9 A . The
constant term on the left side of the equation is 6. Equating the constant
terms on each side of the equation, we obtain that  9 A  6 . Thus, to solve
for A and B, we will solve the system of equations A  B  0 and  9 A  6 .
2
The second equation gives us that A   . The first equation gives us that
3
2
B  A .
3
Method 2
The second method involves choosing a value for x that will leave A but will
eliminate B in the equation 6  A ( x  9 )  B x and choosing a value for x
that will leave B but will eliminate A.
To solve for A, choose x  0 : 6  A ( 0  9 )  B ( 0 )  6   9 A 
2
A 
3
To solve for B, choose x  9 : 6  A ( 9  9 )  B ( 9 )  6  9 B  B 
Copyrighted by James D. Anderson, The University of Toledo
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2
3
6
A
B

Thus,
=
=
x( x  9)
x
x 9
6
dx =
x2  9x


2
2
2 1
1
3  3

 . Thus,

= 
3  x  9 x 
x
x 9

2
6
dx =
x( x  9)
3
 1
1


  x  9 x  dx =
2
x 9
2
 c
( ln x  9  ln x )  c = ln
3
x
3
Answer:
2.

5
1
2
x 9
ln
 c
3
x
4x  7
dx
3 x  16 x  12
2
2
Since 3 x  16 x  12  ( x  6 ) ( 3 x  2 ) , then the function
4x  7
2
f ( x) 
x


is
defined
for
all
real
numbers
x
such
that
3 x 2  16 x  12
3
and x  6 . The function f is continuous on its domain of definition. Thus,
the function f is continuous on the closed interval [1, 5 ] . Thus, the
Fundamental Theorem of Calculus can be applied.
4x  7
 3 x 2  16 x  12 dx using the technique of
Partial Fraction Decomposition. Then we will apply the Fundamental
Theorem of Calculus with one of these antiderivatives.
We will first evaluate the integral
4x  7
A
B
4x  7

=
=
( x  6)(3x  2)
x  6 3x  2
3 x  16 x  12
2
Multiplying both sides of the equation
4x  7
A
B

=
( x  6)(3x  2)
x  6 3x  2
Copyrighted by James D. Anderson, The University of Toledo
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by ( x  6 ) ( 3 x  2 ) , we obtain the equation 4 x  7  A ( 3 x  2 )  B ( x  6 ) .
Method 1
4 x  7  A( 3 x  2 )  B ( x  6 )  4 x  7  3 A x  2 A  B x  6 B 
4 x  7  (3A  B) x  2 A  6B
Equating the coefficients of x on both sides of the equation, we obtain that
3A  B  4.
Equating the constant terms on both sides of the equation, we obtain that
2 A  6B   7 .
3A  B  4
We can solve the system of equations 2 A  6 B   7 by the addition
method.
18 A  6 B  24
Multiplying the first equation by 6, we obtain 2 A  6 B   7
20 A
 17
17
A

Thus,
.
20
Using the first equation of 3 A  B  4 to solve for B, we have that
80 51 29
B  4  3A 


.
20 20 20
Method 2
4 x  7  A( 3 x  2 )  B ( x  6 )
To solve for A, choose x  6 : 17  20 A  A 
To solve for B, choose x  
17
20
8 21
2
 2 18 
 B   
:  
3
3
3
3
 3
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29
20
29
 
B  B 
3
3
20

17
29
4x  7
A
B
20  20

Thus,
=
=
=
( x  6)(3x  2)
x  6 3x  2
x  6 3x  2
1  17
29 

 . Thus,

20  x  6 3 x  2 
 3x
2
1  17
29 
4x  7
dx =

 dx =


 16 x  12
20  x  6 3 x  2 
1 
1

 17 ln x  6  29  ln 3 x  2   c =
20 
3

1 
29

ln 3 x  2   c =
 17 ln x  6 
20 
3

1
( 51 ln x  6  29 ln 3 x  2 )  c
60
Now, that we have all the antiderivatives of the integrand
4x  7
f ( x) 
, we can apply the Fundamental Theorem of Calculus
2
3 x  16 x  12
5
4x  7
dx .
to evaluate the definite integral  1
3 x 2  16 x  12

5
1
5
4x  7
1

dx
(
51
ln
x

6

29
ln
3
x

2
)
=
 =
3 x 2  16 x  12
60
1
1
1
[ 51 ln 1  29 ln 17  ( 51 ln 5  29 ln 5 ) ] =
( 0  29 ln 17  80 ln 5 ) =
60
60
1
( 29 ln 17  80 ln 5 )
60
Copyrighted by James D. Anderson, The University of Toledo
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Answer:
3.

1
( 29 ln 17  80 ln 5 )
60
t 2  3t  2
dt
10 t 4  6 t 3
t 2  3t  2
t 2  3t  2
=
. Thus,
10 t 4  6 t 3
2 t 3 ( 5t  3)
1
2


t 2  3t  2
dt =
10 t 4  6 t 3

t 2  3t  2
dt =
2 t 3 ( 5t  3)
t 2  3t  2
dt
t 3 ( 5t  3)
t 2  3t  2
Thus, we will find the partial fraction decomposition for 3
. Thus,
t ( 5t  3)
A
B
C
D
t 2  3t  2
 2  3 
=
. Multiplying both sides of this
3
t
t
t
5t  3
t ( 5t  3)
3
equation by t ( 5 t  3 ) , we obtain the following equation.
t 2  3t  2  A t 2 ( 5 t  3 )  B t ( 5 t  3 )  C ( 5 t  3 )  D t 3
To solve for C, choose t  0 :  2  3C  C  
2
3
3 9
9
27
 2 
D 
To solve for D, choose t   :
5 25 5
125
20
45  225  250   27 D  20   27 D  D  
27
In order to solve for A and B, we will need to equate coefficients.
2
3
Note that the coefficient of the t term in the product A t ( 5 t  3 ) is 5 A .
The products of B t ( 5 t  3 ) and C ( 5 t  3 ) do not have a t 3 term. Thus, the
3
coefficient of the t term on the right side of the equation is 5 A  D . The
Copyrighted by James D. Anderson, The University of Toledo
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coefficient of the t 3 term on the left side of the equation is 0. Equating the
coefficients of t 3 on both sides of the equation, we obtain that 5 A  D  0 .
20
20
20
 0  5A 

Since D  
from above, then we have that 5 A 
27
27
27
4
A
.
27
t 2  3t  2  A t 2 ( 5 t  3 )  B t ( 5 t  3 )  C ( 5 t  3 )  D t 3
Note that the coefficient of the t term in the product B t ( 5 t  3 ) is 3 B and
the coefficient of the t term in the product C ( 5 t  3 ) is 5 C . The product of
A t 2 ( 5 t  3 ) does not have a t term. Thus, the coefficient of the t term on
the right side of the equation is 3 B  5 C . The coefficient of the t term on
the left side of the equation is  3 . Equating the coefficients of t on both
2
sides of the equation, we obtain that 3 B  5 C   3 . Since C   from
3
10
 3 
above, then we have that 3 B  5C   3  3B 
3
1
9 B  10   9  B  .
9
4
1
2
20
A
B
C
D
t  3t  2
27  9  3  27
 2  3 
Thus, 3
=
=
=
t
t
t
5t  3
t
t 2 t 3 5t  3
t ( 5t  3)
4
3
18
20
1  4
3 18
20 
27  27  27  27

 . Thus,



=
27  t
t2
t 3 5 t  3 
t
t2
t3
5t  3
2

1 1
t 2  3t  2
1 t 2  3t  2
dt =  3
dt = 
4
3
2 27
10 t  6 t
2 t ( 5t  3)
1
54
 4
3 18
20 




  t t 2 t 3 5t  3  dt =
 4
20 
 dt =
 3t  2  18 t  3 
t
5 t  3 
 
Copyrighted by James D. Anderson, The University of Toledo
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
1 
3t  1 18 t  2
1
 4 ln t 

 20  ln 5 t  3   c =
54 
1
2
5

1 
3
9

 4 ln t   2  4 ln 5 t  3   c
54 
t
t

Answer:
4.

1 
3
9

 4 ln t   2  4 ln 5 t  3   c
54 
t
t

3x  5
dx
( x  4 ) ( 4 x 2  12 x  9 )
2
2
NOTE: 4 x  12 x  9  ( 2 x  3 ) . Thus,

3x  5
dx =
( x  4 ) ( 4 x 2  12 x  9 )

3x  5
dx
( x  4 ) ( 2 x  3) 2
Thus, we will find the partial fraction decomposition for
3x  5
.
( x  4 ) ( 2 x  3) 2
Thus,
3x  5
A
B
C


=
. Multiplying both sides
( x  4 ) ( 2 x  3) 2
x  4 2 x  3 ( 2 x  3) 2
2
of this equation by ( x  4 ) ( 2 x  3 ) , we obtain the following equation.
3 x  5  A( 2 x  3) 2  B ( x  4 ) ( 2 x  3)  C ( x  4 )
To solve for A, choose x   4 :  7  121 A  A  
To solve for C, choose x 
19 11

C 
2
2
C 
7
121
3 9 10
 3 8 

 C   
:
2
2 2
 2 2 
19
11
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In order to solve for B, we will need to equate coefficients.
2
Note that the coefficient of the x 2 term in the product A ( 2 x  3 ) is 4 A .
The coefficient of the x 2 term in the product B ( x  4 ) ( 2 x  3 ) is 2 B . The
product of C ( x  4 ) does not have a x 2 term. Thus, the coefficient of the x 2
term on the right side of the equation is 4 A  2 B . The coefficient of the x 2
term on the left side of the equation is 0. Equating the coefficients of x 2 on
both sides of the equation, we obtain that 4 A  2 B  0  2 A  B  0 .
7
Since A  
from above, then we have that 2 A  B  0  B   2 A =
121
14
.
121
3x  5
A
B
C


Thus,
=
=
( x  4 ) ( 2 x  3) 2
x  4 2 x  3 ( 2 x  3) 2
7
14
19
7
14
209

121  121 
11
121  121 
121
=
2 =
x  4 2 x  3 ( 2 x  3)
x  4 2 x  3 ( 2 x  3) 2

1 
7
14
209
 


121  x  4 2 x  3 ( 2 x  3 ) 2

3x  5
dx =
( x  4 ) ( 4 x 2  12 x  9 )
1
121
1
121


 . Thus,

3x  5
dx =
( x  4 ) ( 2 x  3) 2


7
14
209
 


2
 x  4 2 x  3 ( 2 x  3)

 dx =



7
14
 

 209 ( 2 x  3 )  2
 x  4 2x  3

 dx =

1 
1
1 ( 2 x  3)  1
  7 ln x  4  14  ln 2 x  3  209  
121 
2
2
1
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
  c =


1 
209
  7 ln x  4  7 ln 2 x  3 
c
121 
2 ( 2 x  3 ) 
Answer:

1 
209
  7 ln x  4  7 ln 2 x  3 
  c OR
121 
2 ( 2 x  3 ) 

5.


1 
209
 7 ln x  4  7 ln 2 x  3 
c
121 
2 ( 2 x  3 ) 
6t 2  8t  5
dt
( t  3 ) ( 2 t  1) ( t  1)
6t 2  8t  5
We will find the partial fraction decomposition for
.
( t  3 ) ( 2 t  1) ( t  1)
Thus,
A
B
C
6t 2  8t  5


=
. Multiplying both sides of
t  3 2t  1 t  1
( t  3 ) ( 2 t  1) ( t  1)
this equation by ( t  3 ) ( 2 t  1) ( t  1) , we obtain the following equation.
6 t 2  8 t  5  A ( 2 t  1) ( t  1)  B ( t  3 ) ( t  1)  C ( t  3 ) ( 2 t  1)
To solve for A, choose t  3 : 54  24  5  ( 7 ) ( 4 ) A  73  28 A 
73
A
28
1 3
 1 6  1 2 
 4  5        B 
To solve for B, choose t   :
2 2
 2 2  2 2 
3 18  7   1 
15
7
30

   B  
  B  B 
2
2
2
4
7
 2 2
To solve for C, choose t   1 : 6  8  5  (  4 ) (  1) C   7  4 C 
Copyrighted by James D. Anderson, The University of Toledo
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C  
7
4
A
B
C
6t 2  8t  5


Thus,
=
=
t  3 2t  1 t  1
( t  3 ) ( 2 t  1) ( t  1)
73
30
73
120
49
7
28  7  4
28  28  28
=
=
t  3 2t  1 t  1
t  3 2t  1 t  1
1  73
120
49



28  t  3 2 t  1 t  1


 . Thus,

1
6t 2  8t  5
dt =
28 
( t  3 ) ( 2 t  1) ( t  1)
 73
120
49



 t  3 2t  1 t  1

 dt =

1 
1

 73 ln t  3  120  ln 2 t  1  49 ln t  1   c =
28 
2

1
 73 ln t  3  60 ln 2 t  1  49 ln t  1
28
Answer:
6.

 c
1
 73 ln t  3  60 ln 2 t  1  49 ln t  1
28
 c
 y 3  2 y  12
dy
y 4  5 y 2  36
4
2
2
Note that y  5 y  36 is quadratic in y and we have that
y 4  5 y 2  36 = ( y 2  4 ) ( y 2  9) = ( y  2 ) ( y  2 ) ( y 2  9) . Thus,

 y 3  2 y  12
dy =
y 4  5 y 2  36

 y 3  2 y  12
dy
( y  2 ) ( y  2 ) ( y 2  9)
Copyrighted by James D. Anderson, The University of Toledo
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 y 3  2 y  12
We will find the partial fraction decomposition for
.
( y  2 ) ( y  2 ) ( y 2  9)
Thus,
A
B
Cy  D
 y 3  2 y  12


=
. Multiplying both
y  2
y 2
y2  9
( y  2 ) ( y  2 ) ( y 2  9)
2
sides of this equation by ( y  2 ) ( y  2 ) ( y  9) , we obtain the following
equation.
 y 3  2 y  12  A( y  2 ) ( y 2  9)  B ( y  2 ) ( y 2  9)  ( C y  D) ( y 2  4)
2
NOTE: ( y  2 ) ( y  2 )  y  4
To solve for A, choose y   2 : 8  4  12  (  4 ) (13 ) A 
2
 8  (  4 ) (13 ) A  2  13 A  A 
13
To solve for B, choose y  2 :  8  4  12  ( 4 ) (13 ) B 
4
 16  ( 4 ) (13 ) B   4  13 B  B  
13
In order to solve for C and D, we will need to equate coefficients.
2
3
Note that the coefficient of the y term in the product A ( y  2 ) ( y  9) is
A . The coefficient of the y 3 term in the product B ( y  2 ) ( y 2  9) is B .
2
3
The coefficient of the y term in the product ( C y  D) ( y  4) is C . Thus,
3
the coefficient of the y term on the right side of the equation is A  B  C .
3
The coefficient of the y term on the left side of the equation is  1 .
3
Equating the coefficients of y on both sides of the equation, we obtain that
2
4
A  B  C   1 . Since A 
and B  
from above, then we have
13
13
2
4
2
11

 C  1  
 C  1  C   .
that
13 13
13
13
 y 3  2 y  12  A( y  2 ) ( y 2  9)  B ( y  2 ) ( y 2  9)  ( C y  D) ( y 2  4)
Copyrighted by James D. Anderson, The University of Toledo
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2
2
Note that the coefficient of the y term in the product A ( y  2 ) ( y  9) is
 2 A . The coefficient of the y 2 term in the product B ( y  2 ) ( y 2  9) is
2 B . The coefficient of the y 2 term in the product ( C y  D) ( y 2  4) is D .
2
Thus, the coefficient of the y term on the right side of the equation is
 2 A  2 B  D . The coefficient of the y 2 term on the left side of the
2
equation is 0. Equating the coefficients of y on both sides of the equation,
2
4
we obtain that  2 A  2 B  D  0 . Since A 
and B  
from
13
13
4
8
12

 D 0  D  .
above, then we have that 
13 13
13
A
B
Cy  D
 y 3  2 y  12

 2
Thus,
=
=
2
y  2
y 2
y 9
( y  2 ) ( y  2 ) ( y  9)
2
4
11
12
 y 
1  2
4
 11 y  12
13  13  13
13



=
2
13  y  2 y  2
y2  9
y 2 y 2
y 9

 =

1  2
4
11 y
12 

 . Thus,

 2
 2
13  y  2 y  2 y  9 y  9 

 y 3  2 y  12
dy =
y 4  5 y 2  36
1
13 

 y 3  2 y  12
dy =
( y  2 ) ( y  2 ) ( y 2  9)
 2
4
11 y
12


 2
 2
 y 2 y 2 y 9 y 9

 dy =

1 
1
1
2
1 y 
 2 ln y  2  4 ln y  2  11  ln ( y  9 )  12  tan
 c =
13 
2
3
3
1 
11
2
1 y 
 2 ln y  2  4 ln y  2  ln ( y  9 )  4 tan
 c =
13 
2
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1 
y
2
4
2
11 / 2
 4 tan  1   c =
 ln ( y  2 )  ln ( y  2 )  ln ( y  9 )
13 
3
1 
( y  2) 2
y
 ln
 4 tan  1
4
2
11 / 2
13  ( y  2 ) ( y  9 )
3
Answer:

  c

1 
11
2
1 y 
 2 ln y  2  4 ln y  2  ln ( y  9 )  4 tan
  c OR
13 
2
3
1 
( y  2) 2
1 y
 ln

4
tan
13  ( y  2 ) 4 ( y 2  9 ) 11/ 2
3
7.


  c

3x 3  6 x 2  8
dx
9 x 4  24 x 2  16
4
2
Note that 9 x  24 x  16 is quadratic in x 2 and we have that
9 x 4  24 x 2  16 = ( 3 x 2  4 ) 2 . Thus,

3x 3  6 x 2  8
dx =
9 x 4  24 x 2  16

3x 3  6 x 2  8
dx
(3x 2  4) 2
3x 3  6 x 2  8
We will find the partial fraction decomposition for
. Thus,
(3x 2  4) 2
Ax  B
Cx  D
3x 3  6 x 2  8

=
. Multiplying both sides of this
2
3x  4 (3x 2  4) 2
(3x 2  4) 2
2
2
equation by ( 3 x  4 ) , we obtain the following equation.
3 x 3  6 x 2  8  ( A x  B ) ( 3 x 2  4)  C x  D
If we let x  0 , we obtain the equation  8  4 B  D .
Copyrighted by James D. Anderson, The University of Toledo
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2
Note that the coefficient of the x 3 term in the product ( A x  B ) ( 3 x  4) is
3 A . Thus, the coefficient of the x 3 term on the right side of the equation is
3 A . The coefficient of the x 3 term on the left side of the equation is 3.
Equating the coefficients of x 3 on both sides of the equation, we obtain that
3 A  3  A  1.
3 x 3  6 x 2  8  ( A x  B ) ( 3 x 2  4)  C x  D
2
Note that the coefficient of the x 2 term in the product ( A x  B ) ( 3 x  4) is
3 B . Thus, the coefficient of the x 2 term on the right side of the equation is
3 B . The coefficient of the x 2 term on the left side of the equation is  6 .
Equating the coefficients of x 2 on both sides of the equation, we obtain that
3B   6  B   2 .
Since  8  4 B  D from above and B   2 , then  8   8  D 
D  0.
2
Note that the coefficient of the x term in the product ( A x  B ) ( 3 x  4) is
4 A . Thus, the coefficient of the x term on the right side of the equation is
4 A  C . The coefficient of the x term on the left side of the equation is 0.
Equating the coefficients of x on both sides of the equation, we obtain that
4 A  C  0 . Since A  1 , then C   4 .
x 2
 4x
3x 3  6 x 2  8
Ax  B
Cx  D


Thus,
=
=
=
3x 2  4 (3x 2  4) 2
3x 2  4 (3x 2  4) 2
(3x 2  4) 2
x
3x 2  4
Thus,



2
3x 2  4

4x
.
(3x 2  4) 2
3x 3  6 x 2  8
dx =
9 x 4  24 x 2  16

3x 3  6 x 2  8
dx =
(3x 2  4) 2

x
2
4x



2
2
2
2
 3x  4 3x  4 (3x  4)

 dx =

Copyrighted by James D. Anderson, The University of Toledo
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


x
1
2

 
 4 x(3x 2  4)  2
2
 3 x  4 3 x 2  4
3



 dx =


3 x
1
2 3
1 (3x 2  4)  1
2
1
ln ( 3 x  4 )  
tan
 4
 c =
6
3 2
2
6
1
3
3 x
1
2
ln ( 3 x 2  4 ) 
tan  1

 c
6
3
2
3( 3 x 2  4 )
3
3 x
1
2
2
1
ln
(
3
x

4
)

tan

 c
Answer:
6
3
2
3( 3 x 2  4 )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860