Integrating Rational functions by the Method of Partial fraction Decomposition
By
Antony L. Foster
At times, especially in calculus, it is necessary, it is necessary to express a fraction as the sum of two or more others that
are simpler in form than the original. The simpler fractions thus obtained are called partial fractions. In this discussion,
we shall consider the problem of expressing a given fraction as the sum of partial fractions.
A rational function 𝑟𝑟(𝑥𝑥) = 𝑁𝑁(𝑥𝑥)/𝐷𝐷(𝑥𝑥) is the quotient of two polynomials 𝑁𝑁(𝑥𝑥) 𝑎𝑎𝑎𝑎𝑎𝑎 𝐷𝐷(𝑥𝑥). We shall deal exclusively
with rational functions in this discussion, and we shall develop methods that apply only to proper rational fractions, that
is, those rational functions in which the numerator is of lower degree than the denominator.
In algebra, it is known that we can express any polynomial as the product of integral powers of linear and irreducible
quadratic factors. Consequently, every rational fraction belongs to one of the following four cases:
Case 1: All factors of the denominator are linear and none of them are repeated.
Case 2: All factors of the denominator are linear and some of them are repeated.
Case 3: The denominator contains irreducible quadratic factors none of which are repeated.
Case 4: The denominator contains irreducible quadratic factors some of which are repeated.
We shall employ the following theorem in the next four paragraphs, each of which concerns one of the four cases. The
proof of the theorem is omitted because it is beyond the scope of this discussions.
𝑁𝑁(𝑥𝑥 )
THEOREM: If a proper rational function 𝑟𝑟(𝑥𝑥) = 𝐷𝐷(𝑥𝑥 ) in lowest terms is expressed as the sum of partial fractions, then:
1.
To every linear factor 𝑎𝑎𝑎𝑎 + 𝑏𝑏 of the denominator 𝐷𝐷(𝑥𝑥) that appears without repetition, there corresponds
a partial fractions
𝐴𝐴
𝐴𝐴
= 𝑎𝑎
𝑏𝑏
𝑎𝑎𝑎𝑎 + 𝑏𝑏
𝑥𝑥 + 𝑎𝑎
where 𝐴𝐴 is a constant.
2. To every factor (𝑎𝑎𝑎𝑎 + 𝑏𝑏)𝑘𝑘 of the denominator 𝐷𝐷(𝑥𝑥), there correspond the partial fractions
𝐴𝐴1
,
𝑎𝑎𝑎𝑎 + 𝑏𝑏
𝐴𝐴2
,
(𝑎𝑎𝑎𝑎 + 𝑏𝑏)2
𝐴𝐴3
,
(𝑎𝑎𝑎𝑎 + 𝑏𝑏)3
⋯ ,
𝐴𝐴𝑘𝑘
(𝑎𝑎𝑎𝑎 + 𝑏𝑏)𝑘𝑘
where 𝐴𝐴1 , 𝐴𝐴2 , … , 𝐴𝐴𝑘𝑘 are constants.
3. To every irreducible quadratic factor 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 of the denominator 𝐷𝐷(𝑥𝑥)that appears without
repetition, there corresponds the partial fraction
𝐴𝐴𝐴𝐴 + 𝐵𝐵
+ 𝑏𝑏𝑏𝑏 + 𝑐𝑐
𝑎𝑎𝑥𝑥 2
where 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵 are constants.
4. If 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 is irreducible, then to every factor (𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐)𝑘𝑘 of the denominator 𝐷𝐷(𝑥𝑥), there
correspond the partial fractions
𝐴𝐴1 𝑥𝑥 + 𝐵𝐵1
,
𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐
𝐴𝐴2 𝑥𝑥 + 𝐵𝐵2
𝐴𝐴𝑘𝑘 𝑥𝑥 + 𝐵𝐵𝑘𝑘
,⋯,
2
2
(𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐)
(𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐)𝑘𝑘
where 𝐴𝐴1 , 𝐴𝐴2 , … , 𝐴𝐴𝑘𝑘 , 𝐵𝐵1 , 𝐵𝐵2 , … , 𝐵𝐵𝑘𝑘 are constants.
CASE 1
The Denominator 𝐷𝐷 (𝑥𝑥) of the proper rational function contains only distinct linear factors:
Example: Express the function
as the sum of partial fractions.
𝑟𝑟(𝑥𝑥 ) =
𝑁𝑁(𝑥𝑥)
𝐷𝐷(𝑥𝑥)
(2𝑥𝑥 2 +𝑥𝑥+1)
= (𝑥𝑥+2)(3𝑥𝑥+1)(𝑥𝑥+3)
(1)
Solution: There are two methods of solution, the first of them is as follows:
METHOD I: Each factor of the denominator of 𝑟𝑟(𝑥𝑥) is linear and appears only once. Hence, by part 1 of the theorem
above, the partial fractions are
𝐴𝐴
,
𝑥𝑥 + 2
Thus, we have
𝑟𝑟(𝑥𝑥) = (𝑥𝑥
2𝑥𝑥 2 +𝑥𝑥+1
+2)(3𝑥𝑥 +1)(𝑥𝑥 +3)
𝐵𝐵
,
3𝑥𝑥 + 1
𝐴𝐴
𝐶𝐶
𝑥𝑥 + 3
𝐵𝐵
𝐶𝐶
= 𝑥𝑥+2 + 3𝑥𝑥 +1 + 𝑥𝑥 +3
(2)
in which 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 are to be determined so that (2) holds for every value of 𝑥𝑥 except possibly for those that cause one
of the denominators to vanish. If we multiply each member of (2) by the LCD (𝑥𝑥 + 2)(3𝑥𝑥 + 1)(𝑥𝑥 + 3), we get
2𝑥𝑥 2 + 𝑥𝑥 + 1 = 𝐴𝐴(3𝑥𝑥 + 1)(𝑥𝑥 + 3) + 𝐵𝐵(𝑥𝑥 + 2)(𝑥𝑥 + 3) + 𝐶𝐶(𝑥𝑥 + 2)(3𝑥𝑥 + 1)
(3)
By performing the indicated multiplications and collecting the terms in (3), we have
2𝑥𝑥 2 + 𝑥𝑥 + 1 = (3𝐴𝐴 + 𝐵𝐵 + 3𝐶𝐶 )𝑥𝑥 2 + (10𝐴𝐴 + 5𝐵𝐵 + 7𝐶𝐶 )𝑥𝑥 + (3𝐴𝐴 + 6𝐵𝐵 + 2𝐶𝐶)
If 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 are determined so that
(4)
3𝐴𝐴 + 𝐵𝐵 + 3𝐶𝐶 = 2
10𝐴𝐴 + 5𝐵𝐵 + 7𝐶𝐶 = 1
3𝐴𝐴 + 6𝐵𝐵 + 2𝐶𝐶 = 1
then both sides of (3) are identical, and, therefore, the equation is satisfied for every value of 𝑥𝑥. Consequently, for the
values of 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 thus determined, (2) is true for every value of 𝑥𝑥, with the possible exception of those for which a
denominator vanishes.
We can solve this system of three linear equations in 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 by methods learned in your elementary algebra
7
1
courses (make sure you know how to solve linear systems like that above) and obtain that 𝐴𝐴 = − 5 , 𝐵𝐵 = 5 , 𝐶𝐶 = 2.
Therefore,
−7
1
2
2𝑥𝑥 2 + 𝑥𝑥 + 1
=
+
+
(𝑥𝑥 + 2)(3𝑥𝑥 + 1)(𝑥𝑥 + 3) 5 (𝑥𝑥 + 2) 5 (3𝑥𝑥 + 1) 𝑥𝑥 + 3
(5)
METHOD II: We shall present a second method for determining 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 so that both sides of (2) are equal for all
1
values of 𝑥𝑥 except possibly for 𝑥𝑥 = −2, 𝑥𝑥 = −3, 𝑥𝑥 = − 3 for which one of the denominators vanishes.
If the members of (2) are equal for all values of 𝑥𝑥 with the possible exception of the three mentioned above, then the
members of (4) are equal for all values of 𝑥𝑥 including these three. Since the right members of (3) and (4) are two forms
of the same polynomial, the two members of (3) are equal for all values of 𝑥𝑥.
If we let 𝑥𝑥 = −2, the coefficients of 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 in (3) vanish, and we have
1
Similarly, when 𝑥𝑥 = − 3, (3) becomes
Finally, if we let 𝑥𝑥 = −3, we have
8 − 2 + 1 = (−5)(1)𝐴𝐴
−5𝐴𝐴 = 7
7
𝐴𝐴 = −
5
5 8
2 1
− + 1 = � � � � 𝐵𝐵
3 3
9 3
2 − 3 + 9 = 40𝐵𝐵
40𝐵𝐵 = 8
1
𝐵𝐵 =
5
18 − 3 + 1 = 8𝐶𝐶
8𝐶𝐶 = 16
𝐶𝐶 = 2
Hence, we have (5).
The use of the second method in case 1 enables us to avoid solving the system of equations which involves 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶
as unknowns above and thus usually saves time and labor. It can frequently be employed to an advantage in other
cases; in particular, if a linear factor appears in the denominator of 𝑟𝑟(𝑥𝑥)
FORMULA 1
�
𝐴𝐴
𝐴𝐴
𝑑𝑑𝑑𝑑 = ln |𝛼𝛼𝛼𝛼 + 𝑏𝑏| + 𝐶𝐶
𝛼𝛼
𝛼𝛼𝛼𝛼 + 𝛽𝛽
Illustration 1: Now in calculus, if we were asked to evaluated the indefinite integral
�
2𝑥𝑥 2 + 𝑥𝑥 + 1
𝑑𝑑𝑑𝑑
(𝑥𝑥 + 2)(3𝑥𝑥 + 1)(𝑥𝑥 + 3)
Then expressing the integrand as a sum of partial fractions as shown in (5) above, we can then write
�
−7
1
2
2𝑥𝑥 2 + 𝑥𝑥 + 1
𝑑𝑑𝑑𝑑 = � �
+
+
� 𝑑𝑑𝑑𝑑
(𝑥𝑥 + 2)(3𝑥𝑥 + 1)(𝑥𝑥 + 3)
5 (𝑥𝑥 + 2) 5 (3𝑥𝑥 + 1) 𝑥𝑥 + 3
7
1
1
1
3
= − �
𝑑𝑑𝑑𝑑 + �
𝑑𝑑𝑑𝑑 + 2 �
𝑑𝑑𝑑𝑑
5 𝑥𝑥 + 2
15 3𝑥𝑥 + 1
𝑥𝑥 + 3
7
1
= − ln |𝑥𝑥 + 2| + ln |3𝑥𝑥 + 1 | + 2 ln |𝑥𝑥 + 3| + 𝐶𝐶
5
15
EXERCISES FOR DISTINCT LINEAR FACTORS (CASE 1)
Evaluate the indefinite integrals below by decomposing each of the integrands as a sum of partial fractions. These will
be Collected!
5𝑥𝑥+4
1. ∫ (𝑥𝑥−1)(𝑥𝑥
𝑑𝑑𝑑𝑑
+2)
2𝑥𝑥+5
2. ∫ (𝑥𝑥+1)(𝑥𝑥
𝑑𝑑𝑑𝑑
+3)
8𝑡𝑡−7
3. ∫ (2𝑡𝑡−1)(𝑡𝑡−2) 𝑑𝑑𝑑𝑑
5𝑦𝑦
4. ∫ (𝑦𝑦+1)(2𝑦𝑦
𝑑𝑑𝑑𝑑
−3)
𝑠𝑠−8
5. ∫ (3𝑠𝑠+2)(2𝑠𝑠−3) 𝑑𝑑𝑑𝑑
8𝑥𝑥−31
6. ∫ (𝑥𝑥+3)(2𝑥𝑥
𝑑𝑑𝑑𝑑
−5)
2𝑤𝑤 +7
7. ∫ (2𝑤𝑤
𝑑𝑑𝑑𝑑
+5)(𝑤𝑤 +3)
5𝑦𝑦 −3
8. ∫ (𝑦𝑦−5)(2𝑦𝑦
𝑑𝑑𝑑𝑑
+1)
16𝑞𝑞+12
9. ∫ 4𝑞𝑞 2 +4𝑞𝑞−3 𝑑𝑑𝑑𝑑
𝑥𝑥 +8
10. ∫ 15𝑥𝑥 2 −7𝑥𝑥 −2 𝑑𝑑𝑑𝑑
19𝑥𝑥 −46
11. ∫ 6𝑥𝑥 2 −29𝑥𝑥 +35 𝑑𝑑𝑑𝑑
23𝑥𝑥 −10
12. ∫ 15𝑥𝑥 2 +𝑥𝑥 −6 𝑑𝑑𝑑𝑑
2𝑥𝑥 2 +4𝑥𝑥 −6
13. ∫ 2𝑥𝑥 2 −3𝑥𝑥 −2 𝑑𝑑𝑑𝑑
6𝑥𝑥 2 +9𝑥𝑥 −1
14. ∫ 3𝑥𝑥 2
𝑑𝑑𝑑𝑑
+2𝑥𝑥 −1
15. ∫
6𝑥𝑥 2 +3𝑥𝑥 −11
2𝑥𝑥 2 +𝑥𝑥 −6
𝑑𝑑𝑑𝑑
6𝑥𝑥 2 −14𝑥𝑥 +5
16. ∫ 6𝑥𝑥 2 −13𝑥𝑥 +6 𝑑𝑑𝑑𝑑
5𝑦𝑦 2 +5𝑦𝑦 −4
17. ∫ (𝑦𝑦−1)(𝑦𝑦+1)(𝑦𝑦
𝑑𝑑𝑑𝑑
+2)
4𝑠𝑠 2 +2𝑠𝑠−18
18. ∫ (𝑠𝑠+2)(𝑠𝑠−1)(𝑠𝑠+3) 𝑑𝑑𝑑𝑑
3𝑡𝑡 2 −13𝑡𝑡−28
19. ∫ (𝑡𝑡+1)(𝑡𝑡+2)(𝑡𝑡−3) 𝑑𝑑𝑑𝑑
4𝜃𝜃 2 +7𝜃𝜃 −6
20. ∫ (𝜃𝜃−2)(𝜃𝜃
𝑑𝑑𝑑𝑑
+2)(𝜃𝜃 +1)
12𝑥𝑥 2 −7
21. ∫ (𝑋𝑋−1)(2𝑥𝑥
𝑑𝑑𝑑𝑑
−1)(2𝑥𝑥 +3)
17𝑤𝑤 2 +18𝑤𝑤 −72
22. ∫ (𝑤𝑤+1)(2𝑤𝑤
𝑑𝑑𝑑𝑑
+5)(3𝑤𝑤 −1)
6𝑠𝑠 2 +25𝑠𝑠+6
23. ∫ (𝑠𝑠−3)(2𝑠𝑠−1)(2𝑠𝑠+3) 𝑑𝑑𝑑𝑑
7𝑌𝑌 2 −47𝑌𝑌−72
24. ∫ (𝑌𝑌+5)(3𝑌𝑌+2)(2𝑌𝑌−3) 𝑑𝑑𝑑𝑑
CASE 2
The Denominator 𝐷𝐷(𝑥𝑥) of the proper rational function contains only linear factors some of which repeat.
If the denominator of a proper rational function in factored form contains only linear factors, but one or more are
repeated, we use the method illustrated in the example below for expressing it as the sum of partial fractions.
Example: Decompose
into partial fractions.
𝑅𝑅(𝑥𝑥) =
3𝑥𝑥 2 + 5𝑥𝑥 + 1
(𝑥𝑥 − 1)(𝑥𝑥 + 2)2
Solution: Again we have available two methods of solution; the following is the first one. By Part 1 of the theorem, we
must have the partial fraction
𝐴𝐴
𝑥𝑥 − 1
corresponding to the linear factor 𝑥𝑥 − 1 of the denominator 𝐷𝐷 (𝑥𝑥) = (𝑥𝑥 − 1)(𝑥𝑥 + 2)2 , and by part 2 of the theorem, we
must also have the partial fractions
corresponding to the factor (𝑥𝑥 + 2)2 . Hence we have
𝐵𝐵
,
𝑥𝑥 + 2
𝐶𝐶
(𝑥𝑥 + 2)2
3𝑥𝑥 2 + 5𝑥𝑥 + 1
𝐴𝐴
𝐵𝐵
𝐶𝐶
=
+
+
2
(𝑥𝑥 − 1)(𝑥𝑥 + 2)
𝑥𝑥 − 1 𝑥𝑥 + 2 (𝑥𝑥 + 2)2
(1)
and we must find the values of 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶so that the member of (1) are equal for all values of 𝑥𝑥, with the possible
exception of 𝑥𝑥 = 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 = −2. The first step in this process is to multiply each member of (1) by (𝑥𝑥 − 1)(𝑥𝑥 + 2)2
and get
3𝑥𝑥 2 + 5𝑥𝑥 + 1 = 𝐴𝐴(𝑥𝑥 + 2)2 + 𝐵𝐵(𝑥𝑥 − 1)(𝑥𝑥 + 2) + 𝐶𝐶(𝑥𝑥 − 1)
3𝑥𝑥 2 + 5𝑥𝑥 + 1 = (𝐴𝐴 + 𝐵𝐵)𝑥𝑥 2 + (4𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶 )𝑥𝑥 + 4𝐴𝐴 − 2𝐵𝐵 − 𝐶𝐶
(2)
(3)
The members of (3) will be equal for all values of 𝑥𝑥 if 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 are determined so that the coefficients of the like
powers of 𝑥𝑥 in the right and left members are equal. By equating the coefficients of 𝑥𝑥 2 , 𝑥𝑥, and the two constant terms,
we obtain the following system of three linear equations in 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶:
𝐴𝐴 + 𝐵𝐵 = 3
4𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶 = 5
4𝐴𝐴 − 2𝐵𝐵 − 𝐶𝐶 = 1
The solutions of this system of equations is 𝐴𝐴 = 1, 𝐵𝐵 = 2, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 = −1. Therefore,
1
2
1
3𝑥𝑥 2 + 5𝑥𝑥 + 1
=
+
−
2
(𝑥𝑥 − 1)(𝑥𝑥 + 2)
𝑥𝑥 − 1 𝑥𝑥 + 2 (𝑥𝑥 + 2)2
The second method of evaluating 𝐴𝐴, 𝐵𝐵, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 is similar to the second method applicable to case 1. If the members of (2)
are equal for all values of 𝑥𝑥, with the possible exception of 𝑥𝑥 = 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 = −2, then the members of (2) are equal for
all values of 𝑥𝑥.
If we let 𝑥𝑥 = 1, in (2), the coefficients of 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 vanish, and we have
3 + 5 + 1 = 32 𝐴𝐴
9𝐴𝐴 = 9
𝐴𝐴 = 1
If 𝑥𝑥 = −2, the coefficients of 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵 vanish in (2), and we have
12 − 10 + 1 = −3𝐶𝐶
−3𝐶𝐶 = 3
𝐶𝐶 = −1
Since there is no single value of 𝑥𝑥 for which the coefficients of 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 vanish simultaneously in (2), we must resort to
another method to evaluate 𝐵𝐵. Since the members of (2) hold for all values of 𝑥𝑥, we may substitute any convenient
value for this variable together with the values of 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 above in (2) and obtain an equation containing only 𝐵𝐵. So if
we let 𝑥𝑥 = 0, 𝐴𝐴 = 1, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 = −1, (2) becomes
3(0)2 + 5(0) + 1 = (1 × 4) + (−1)(2)𝐵𝐵 + (−1)(−1)
1 = 5 − 2𝐵𝐵
2𝐵𝐵 = 4
𝐵𝐵 = 2
FORMULA 2
𝐴𝐴
𝐴𝐴
�
𝑑𝑑𝑑𝑑 =
+ 𝐶𝐶,
(
)(
(𝛼𝛼𝛼𝛼 + 𝛽𝛽)𝑘𝑘
𝛼𝛼 1 − 𝑘𝑘 𝛼𝛼𝛼𝛼 + 𝛽𝛽)𝑘𝑘−1
𝑘𝑘 ≥ 2
Illustration 2: Thus, in calculus, if we were asked to evaluate the indefinite integral
3𝑥𝑥 2 + 5𝑥𝑥 + 1
�
𝑑𝑑𝑑𝑑
(𝑥𝑥 − 1)(𝑥𝑥 + 2)2
Then expressing the integrand as a sum of partial fractions would give us
�
1
2
1
3𝑥𝑥 2 + 5𝑥𝑥 + 1
𝑑𝑑𝑑𝑑 = � �
+
−
� 𝑑𝑑𝑑𝑑
2
(𝑥𝑥 − 1)(𝑥𝑥 + 2)
𝑥𝑥 − 1 𝑥𝑥 + 2 (𝑥𝑥 + 2)2
1
1
1
𝑑𝑑𝑑𝑑 + 2 �
𝑑𝑑𝑑𝑑 − �
𝑑𝑑𝑑𝑑
= �
(𝑥𝑥 + 2)2
𝑥𝑥 + 2
𝑥𝑥 − 1
= ln |𝑥𝑥 − 1| + 2 ln |𝑥𝑥 + 2| +
1
+ 𝐶𝐶
𝑥𝑥 + 2
EXERCISES FOR REPEATED LINEAR FACTORS (CASE 2)
Evaluate the indefinite integrals below by decomposing each of the integrands as a sum of partial fractions. These will
be Collected!
2𝑡𝑡+5
1. ∫ (𝑡𝑡+1)2 𝑑𝑑𝑑𝑑
2. ∫
−2𝑥𝑥 +11
(𝑥𝑥−4)2
3𝑥𝑥+1
𝑑𝑑𝑑𝑑
3. ∫ (3𝑥𝑥−1)2 𝑑𝑑𝑑𝑑
4𝑥𝑥−5
4. ∫ (2𝑥𝑥−1)2 𝑑𝑑𝑑𝑑
5. ∫
27𝑥𝑥 2 +21𝑥𝑥 +8
(3𝑥𝑥 +2)3
6. ∫
−4𝑥𝑥 2 −5𝑥𝑥 +1
(𝑥𝑥+1)3
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
7. ∫
−16𝑧𝑧 2 +54𝑧𝑧−40
(2𝑧𝑧−3)3
8. ∫
20𝑠𝑠 2 −144𝑠𝑠+265
(2𝑠𝑠−7)3
𝑑𝑑𝑑𝑑
12𝑤𝑤 2 −9𝑤𝑤 +20
𝑑𝑑𝑑𝑑
9. ∫ (2𝑤𝑤
𝑑𝑑𝑑𝑑
+3)(3𝑤𝑤 −1)2
−8𝑥𝑥 2 +35𝑥𝑥 +9
10. ∫ (2𝑥𝑥−1)(𝑥𝑥−4)2 𝑑𝑑𝑑𝑑
24𝑥𝑥 2 −94𝑥𝑥 +88
11. ∫ (2𝑥𝑥−3)(3𝑥𝑥−5)2 𝑑𝑑𝑑𝑑
12. ∫
−22𝑇𝑇 2 +32𝑇𝑇+138
(5𝑇𝑇+2)(2𝑇𝑇−7)2
𝑑𝑑𝑑𝑑
−3𝑦𝑦 3 −9𝑦𝑦 2 +49𝑦𝑦 −25
13. ∫ (2𝑦𝑦
𝑑𝑑𝑑𝑑
−1)(𝑦𝑦 +2)(𝑦𝑦 −1)2
14. ∫
3𝑥𝑥 3 −11𝑥𝑥 2 −18𝑥𝑥 +46
(𝑥𝑥+3)(𝑥𝑥−2)(𝑥𝑥−1)2
−14𝑠𝑠 3 +14𝑠𝑠 2 +𝑠𝑠−7
𝑑𝑑𝑑𝑑
15. ∫ (2𝑠𝑠+1)(𝑠𝑠−4)(𝑠𝑠+1)2 𝑑𝑑𝑑𝑑
2𝑦𝑦 3 −5𝑦𝑦 2 −27𝑦𝑦 −24
16. ∫ (𝑦𝑦+1)(𝑦𝑦−1)(𝑦𝑦 2 𝑑𝑑𝑑𝑑
+2)
17. ∫
2𝑥𝑥 2 +5𝑥𝑥 +1
(𝑥𝑥+1) 2
18. ∫
3𝑡𝑡 2 −16𝑡𝑡+20
(𝑡𝑡−3)2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
19. ∫
2𝑧𝑧 3 −10𝑧𝑧 2 +4𝑧𝑧+11
(2𝑧𝑧+1)(𝑧𝑧−2)2
𝑑𝑑𝑑𝑑
20. ∫
3𝑥𝑥 3 −12𝑥𝑥 2 −3𝑥𝑥 +10
(3𝑥𝑥−2)(𝑥𝑥−2)2
𝑑𝑑𝑑𝑑
CASE 3
The Denominator 𝐷𝐷(𝑥𝑥) of the proper rational function contains distinct irreducible quadratic factors .
If the first power of an irreducible quadratic function appears among the factors of the denominator of a rational
function that is to be decomposed into partial fractions, it must appear as the denominator of one of the partial
fractions. The numerator of the partial fraction that has the quadratic denominator must be a linear function. The
linear factors of the denominator enter exactly as in the previous two cases (case 1 and case 2).
Example 1: Decompose
𝐹𝐹(𝑥𝑥) =
14𝑥𝑥 3 + 14𝑥𝑥 2 − 4𝑥𝑥 + 3
(3𝑥𝑥 2 − 𝑥𝑥 + 1)(𝑥𝑥 − 1)(𝑥𝑥 + 2)
into partial fractions.
Solution: The quadratic factor 3𝑥𝑥 2 − 𝑥𝑥 + 1 is irreducible in the real numbers; hence it must be used as the denominator
of a partial fraction that has a linear function 𝐴𝐴𝐴𝐴 + 𝐵𝐵 for its numerator. Each of the linear factors will enter as a
denominator of a partial fraction with a constant numerator. Thus,
14𝑥𝑥 3 + 14𝑥𝑥 2 − 4𝑥𝑥 + 3
𝐴𝐴𝐴𝐴 + 𝐵𝐵
𝐶𝐶
𝐷𝐷
=
+
+
2
2
(3𝑥𝑥 − 𝑥𝑥 + 1)(𝑥𝑥 − 1)(𝑥𝑥 + 2)
3𝑥𝑥 − 𝑥𝑥 + 1 𝑥𝑥 − 1 𝑥𝑥 + 2
(1)
In order to find the values of 𝐴𝐴, 𝐵𝐵, 𝐶𝐶, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐷𝐷 so that the members of (1) are equal, we first multiply each member of (1)
by (3𝑥𝑥 2 − 𝑥𝑥 + 1)(𝑥𝑥 − 1)(𝑥𝑥 + 2) and get
14𝑥𝑥 3 + 14𝑥𝑥 2 − 4𝑥𝑥 + 3 = (𝐴𝐴𝐴𝐴 + 𝐵𝐵)(𝑥𝑥 − 1)(𝑥𝑥 + 2) + 𝐶𝐶 (3𝑥𝑥 2 − 𝑥𝑥 + 1)(𝑥𝑥 + 2) + 𝐷𝐷 (3𝑥𝑥 2 − 𝑥𝑥 + 1)(𝑥𝑥 − 1) (2)
14𝑥𝑥 3 + 14𝑥𝑥 2 − 4𝑥𝑥 + 3 = (𝐴𝐴 + 3𝐶𝐶 + 3𝐷𝐷)𝑥𝑥 3 + (𝐴𝐴 + 𝐵𝐵 + 5𝐶𝐶 − 4𝐷𝐷)𝑥𝑥 2 + (−2𝐴𝐴 + 𝐵𝐵 − 𝐶𝐶 + 2𝐷𝐷)𝑥𝑥 + (−2𝐵𝐵 +
2𝐶𝐶 − 𝐷𝐷)
(3)
We may now obtain, by equating the coefficients of equal powers of 𝑥𝑥, the four linear equations in 𝐴𝐴, 𝐵𝐵, 𝐶𝐶, and 𝐷𝐷 which
follow:
𝐴𝐴 + 3𝐶𝐶 + 3𝐷𝐷 = 14
𝐴𝐴 + 𝐵𝐵 + 5𝐶𝐶 − 4𝐷𝐷 = 14
−2𝐴𝐴 + 𝐵𝐵 − 𝐶𝐶 + 2𝐷𝐷 = −4
−2𝐵𝐵 + 2𝐶𝐶 − 𝐷𝐷 = 3
The solution of this system can be obtained by Gaussian Elimination (which can be found in Precalculus and linear
algebra textbooks) or by solving the last equation for 𝐷𝐷 in terms of 𝐵𝐵 and 𝐶𝐶, replacing each 𝐷𝐷 in the other three
equations by this value, solving the resulting system of three linear equations in three unknowns, and finally, obtaining
the value of 𝐷𝐷 from the last equation. The solution thus obtained is 𝐴𝐴 = 2, 𝐵𝐵 = 1, 𝐶𝐶 = 3, 𝐷𝐷 = 1. Hence,
𝐹𝐹 (𝑥𝑥) =
14𝑥𝑥 3 + 14𝑥𝑥 2 − 4𝑥𝑥 + 3
2𝑥𝑥 + 1
3
1
=
+
+
2
2
(3𝑥𝑥 − 𝑥𝑥 + 1)(𝑥𝑥 − 1)(𝑥𝑥 + 2)
3𝑥𝑥 − 𝑥𝑥 + 1 𝑥𝑥 − 1 𝑥𝑥 + 2
Example 2: Decompose
𝑟𝑟(𝑥𝑥) =
4𝑥𝑥 4 + 4𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 + 1
(𝑥𝑥 2 + 𝑥𝑥 + 1)(𝑥𝑥 2 − 𝑥𝑥 − 3)(𝑥𝑥 + 1)
into partial fractions.
Solution
4𝑥𝑥 4 + 4𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 + 1
𝐴𝐴𝐴𝐴 + 𝐵𝐵
𝐶𝐶𝐶𝐶 + 𝐷𝐷
𝐸𝐸
=
+
+
(𝑥𝑥 2 + 𝑥𝑥 + 1)(𝑥𝑥 2 − 𝑥𝑥 − 3)(𝑥𝑥 + 1) 𝑥𝑥 2 + 𝑥𝑥 + 1 𝑥𝑥 2 − 𝑥𝑥 − 3 𝑥𝑥 + 1
(1)
After multiplying each member of (1) by (𝑥𝑥 2 + 𝑥𝑥 + 1)(𝑥𝑥 2 − 𝑥𝑥 − 3)(𝑥𝑥 + 1), we have
4𝑥𝑥 4 + 4𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 + 1 = (𝐴𝐴𝐴𝐴 + 𝐵𝐵)(𝑥𝑥 2 − 𝑥𝑥 − 3)(𝑥𝑥 + 1) + (𝐶𝐶𝐶𝐶 + 𝐷𝐷)(𝑥𝑥 2 + 𝑥𝑥 + 1)(𝑥𝑥 + 1) + 𝐸𝐸(𝑥𝑥 2 + 𝑥𝑥 + 1)(𝑥𝑥 2 − 𝑥𝑥 −
3)
(2)
= (𝐴𝐴 + 𝐶𝐶 + 𝐸𝐸 )𝑥𝑥 4 + (𝐵𝐵 + 2𝐶𝐶 + 𝐷𝐷)𝑥𝑥 3 + (−4𝐴𝐴 + 2𝐶𝐶 + 2𝐷𝐷 − 3𝐸𝐸 )𝑥𝑥 2 + (−3𝐴𝐴 − 4𝐵𝐵 + 𝐶𝐶 + 2𝐷𝐷 − 4𝐸𝐸 )𝑥𝑥 + (−3𝐵𝐵 + 𝐷𝐷 −
3𝐸𝐸)
(3)
We can obtain the following system of equations by equating the coefficients of the equal powers of 𝑥𝑥:
𝐴𝐴 + 𝐶𝐶 + 𝐸𝐸 = 4
𝐵𝐵 + 2𝐶𝐶 + 𝐷𝐷 = 4
−4𝐴𝐴 + 2𝐶𝐶 + 2𝐷𝐷 − 3𝐸𝐸 = −1
−3𝐴𝐴 − 4𝐵𝐵 + 𝐶𝐶 + 2𝐷𝐷 − 4𝐸𝐸 = 1
−3𝐵𝐵 + 𝐷𝐷 − 3𝐸𝐸 = 1
The solution of this system can be obtained by Gaussian elimination or by obtaining an expression for 𝐷𝐷 in terms of
𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝐸𝐸 from the last equation, substituting this for 𝐷𝐷 in the other four equations, thus getting a system of four linear
equations in four unknowns, and solving the system by any of the available methods. The method suggested in Example
1 can be used. The solution is 𝐴𝐴 = 1, 𝐵𝐵 = −1, 𝐶𝐶 = 2, 𝐷𝐷 = 1, 𝐸𝐸 = 1. Therefore,
𝑥𝑥 − 1
2𝑥𝑥 + 1
1
4𝑥𝑥 4 + 4𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 + 1
= 2
+ 2
+
2
2
(𝑥𝑥 + 𝑥𝑥 + 1)(𝑥𝑥 − 𝑥𝑥 − 3)(𝑥𝑥 + 1) 𝑥𝑥 + 𝑥𝑥 + 1 𝑥𝑥 − 𝑥𝑥 − 3 𝑥𝑥 + 1
FORMULA 3
�
𝛼𝛼𝑥𝑥 2
2
⎧
⎪
⎪
�4𝛼𝛼𝛼𝛼
− 𝛽𝛽2
2𝛼𝛼𝛼𝛼 + 𝛽𝛽
tan−1 �
� + 𝐶𝐶 𝑖𝑖𝑖𝑖 𝛽𝛽2 − 4𝛼𝛼𝛼𝛼 < 0
�4𝛼𝛼𝛼𝛼 − 𝛽𝛽2
1
𝑑𝑑𝑑𝑑 =
+ 𝛽𝛽𝛽𝛽 + 𝛾𝛾
⎨
2𝛼𝛼𝛼𝛼 + 𝛽𝛽 − �𝛽𝛽2 − 4𝛼𝛼𝛼𝛼
1
⎪
� + 𝐶𝐶 𝑖𝑖𝑖𝑖 𝛽𝛽2 − 4𝛼𝛼𝛼𝛼 > 0
ln �
⎪ 2
2
2𝛼𝛼𝛼𝛼 + 𝛽𝛽 + �𝛽𝛽 − 4𝛼𝛼𝛼𝛼
⎩�𝛽𝛽 − 4𝛼𝛼𝛼𝛼
FORMULA 4
�
𝛼𝛼𝑥𝑥 2
𝑥𝑥
1
𝛽𝛽
1
� 2
𝑑𝑑𝑑𝑑 =
ln |𝛼𝛼𝑥𝑥 2 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾| −
𝑑𝑑𝑑𝑑
+ 𝛽𝛽𝛽𝛽 + 𝛾𝛾
2𝛼𝛼
2𝛼𝛼 𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾
Illustration 3: Find the indefinite integral
�
14𝑥𝑥 3 + 14𝑥𝑥 2 − 4𝑥𝑥 + 3
𝑑𝑑𝑑𝑑
(3𝑥𝑥 2 − 𝑥𝑥 + 1)(𝑥𝑥 − 1)(𝑥𝑥 + 2)
Solution: By expressing the integrand as a sum of partial fractions as seen in Example 1 just above, we have
�
2𝑥𝑥 + 1
3
1
14𝑥𝑥 3 + 14𝑥𝑥 2 − 4𝑥𝑥 + 3
𝑑𝑑𝑑𝑑 = � � 2
+
+
� 𝑑𝑑𝑑𝑑
2
(3𝑥𝑥 − 𝑥𝑥 + 1)(𝑥𝑥 − 1)(𝑥𝑥 + 2)
3𝑥𝑥 − 𝑥𝑥 + 1 𝑥𝑥 − 1 𝑥𝑥 + 2
=�
2𝑥𝑥 + 1
1
1
𝑑𝑑𝑑𝑑 + 3 �
𝑑𝑑𝑑𝑑 + �
𝑑𝑑𝑑𝑑
2
3𝑥𝑥 − 𝑥𝑥 + 1
𝑥𝑥 − 1
𝑥𝑥 + 2
= �
= 2�
3𝑥𝑥 2
2𝑥𝑥 + 1
𝑑𝑑𝑑𝑑 + 3 ln|𝑥𝑥 − 1| + ln|𝑥𝑥 + 2| + 𝐶𝐶
− 𝑥𝑥 + 1
3𝑥𝑥 2
1
𝑥𝑥
𝑑𝑑𝑑𝑑 + � 2
𝑑𝑑𝑑𝑑 + 3 ln|𝑥𝑥 − 1| + ln|𝑥𝑥 + 2| + 𝐶𝐶
− 𝑥𝑥 + 1
3𝑥𝑥 − 𝑥𝑥 + 1
4
1
1
𝑑𝑑𝑑𝑑 + 3 ln|𝑥𝑥 − 1| + ln|𝑥𝑥 + 2| + 𝐶𝐶
= ln|3𝑥𝑥 2 − 𝑥𝑥 + 1| + � 2
3 3𝑥𝑥 − 𝑥𝑥 + 1
3
=
8
6𝑥𝑥 − 1
1
ln|3𝑥𝑥 2 − 𝑥𝑥 + 1| +
tan−1 �
� + 3 ln|𝑥𝑥 − 1| + ln|𝑥𝑥 + 2| + 𝐶𝐶
3
3√11
√11
Illustration 4 : Suppose that in calculus we are asked to find
�
4𝑥𝑥 4 + 4𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 + 1
𝑑𝑑𝑑𝑑
(𝑥𝑥 2 + 𝑥𝑥 + 1)(𝑥𝑥 2 − 𝑥𝑥 − 3)(𝑥𝑥 + 1)
Solution: By decomposing the integrand as a sum of partial fractions (using case 1, case 3) as shown in Example 2 above,
we have
�
𝑥𝑥 − 1
2𝑥𝑥 + 1
1
4𝑥𝑥 4 + 4𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 + 1
𝑑𝑑𝑑𝑑 = � � 2
+ 2
+
� 𝑑𝑑𝑑𝑑
(𝑥𝑥 2 + 𝑥𝑥 + 1)(𝑥𝑥 2 − 𝑥𝑥 − 3)(𝑥𝑥 + 1)
𝑥𝑥 + 𝑥𝑥 + 1 𝑥𝑥 − 𝑥𝑥 − 3 𝑥𝑥 + 1
=�
𝑥𝑥 2
2𝑥𝑥 + 1
1
𝑥𝑥 − 1
𝑑𝑑𝑑𝑑 + � 2
𝑑𝑑𝑑𝑑 + �
𝑑𝑑𝑑𝑑
+ 𝑥𝑥 + 1
𝑥𝑥 − 𝑥𝑥 − 3
𝑥𝑥 + 1
2𝑥𝑥 + 1
1
1
� + ln|𝑥𝑥 2 − 𝑥𝑥 − 3| + 2 � 2
𝑑𝑑𝑑𝑑 + ln|𝑥𝑥 + 1| + 𝐶𝐶
= ln|𝑥𝑥 2 + 𝑥𝑥 + 1| − √3 tan−1 �
𝑥𝑥 − 𝑥𝑥 − 3
2
√3
1
2𝑥𝑥 + 1
2
2𝑥𝑥 − 1 − √13
� + ln |𝑥𝑥 + 1| + 𝐶𝐶
= ln |𝑥𝑥 2 + 𝑥𝑥 + 1| − √3 tan−1 �
� + ln|𝑥𝑥 2 − 𝑥𝑥 − 3| +
ln �
2
√3
√13 2𝑥𝑥 − 1 + √13
EXERCISES FOR DISTINCT IRREDUCIBLE QUADRATIC FACTORS (CASE 3)
Find the following indefinite integral by decomposing the integrand into partial fractions
3𝑥𝑥 2 −10𝑥𝑥 +16
1. ∫ (𝑥𝑥−3)(𝑥𝑥 2 +𝑥𝑥 +1) 𝑑𝑑𝑑𝑑
−2𝑥𝑥 +5
2. ∫ (𝑥𝑥−1)(𝑥𝑥 2 +2) 𝑑𝑑𝑑𝑑
5𝑥𝑥 2 +𝑥𝑥+2
3. ∫ (𝑥𝑥+1)(𝑥𝑥 2 +1) 𝑑𝑑𝑑𝑑
𝑥𝑥 2 −4𝑥𝑥 −3
4. ∫ (2𝑥𝑥−3)(𝑥𝑥 2
𝑑𝑑𝑑𝑑
−𝑥𝑥−3)
5𝑥𝑥 3 +14𝑥𝑥 2 +71𝑥𝑥 +14
5. ∫ (𝑥𝑥+5)(2𝑥𝑥
𝑑𝑑𝑑𝑑
−1)(𝑥𝑥 2 −3)
𝑥𝑥 3 +11𝑥𝑥 2 +13𝑥𝑥 −5
6. ∫ (𝑥𝑥−3)(3𝑥𝑥
𝑑𝑑𝑑𝑑
+1)(𝑥𝑥 2 −𝑥𝑥 +2)
−9𝑥𝑥 3 +19𝑥𝑥 2 +2𝑥𝑥 +3
7. ∫ (2𝑥𝑥−3)(𝑥𝑥+2)(𝑥𝑥 2 +3) 𝑑𝑑𝑑𝑑
5𝑥𝑥 3 −31𝑥𝑥 2 −3𝑥𝑥−23
8. ∫ (3𝑥𝑥−1)(𝑥𝑥−3)(𝑥𝑥 2 +3𝑥𝑥 +4) 𝑑𝑑𝑑𝑑
9. ∫
3𝑥𝑥 3 −10𝑥𝑥 2 +9𝑥𝑥 −6
(𝑥𝑥−1)2 (𝑥𝑥 2 +1)
𝑑𝑑𝑑𝑑
10. ∫
–𝑥𝑥 3 +8𝑥𝑥 2 +2𝑥𝑥 −19
(𝑥𝑥−2) 2 (𝑥𝑥 2 +5)
11. ∫
4𝑥𝑥 3 −11𝑥𝑥 2 +12𝑥𝑥 −12
(𝑥𝑥−2)2 (𝑥𝑥 2 +𝑥𝑥+1)
12. ∫
𝑑𝑑𝑑𝑑
10𝑥𝑥 3 +3𝑥𝑥 2 −19𝑥𝑥 +12
(2𝑥𝑥−1)2 (𝑥𝑥 2 +𝑥𝑥−3)
2𝑥𝑥 3 −8𝑥𝑥 +4
13. ∫ (𝑥𝑥 2 +2)(𝑥𝑥 2 −2) 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
3𝑥𝑥 3 −3𝑥𝑥 2 −4𝑥𝑥 −14
14. ∫ (𝑥𝑥 2 −5)(𝑥𝑥 2 −𝑥𝑥+3) 𝑑𝑑𝑑𝑑
–𝑥𝑥 3 −13𝑥𝑥 2 +9
15. ∫ (𝑥𝑥 2 +3)(𝑥𝑥 2 −3𝑥𝑥−3) 𝑑𝑑𝑑𝑑
3𝑥𝑥 3 −10𝑥𝑥 2 +7𝑥𝑥 −3
16. ∫ (𝑥𝑥 2 +1)(𝑥𝑥 2 −3𝑥𝑥−1) 𝑑𝑑𝑑𝑑
17. ∫
6𝑥𝑥 3 −9𝑥𝑥 2 +33𝑥𝑥 −64
(2𝑥𝑥 −3)(𝑥𝑥 2 +5)
𝑑𝑑𝑑𝑑
18. ∫
2𝑥𝑥 3 −10𝑥𝑥 2 +11𝑥𝑥 +19
(𝑥𝑥−5)(2𝑥𝑥 2 −3𝑥𝑥 +2)
19. ∫
𝑥𝑥 3 +4𝑥𝑥 2 −7𝑥𝑥 +6
(𝑥𝑥−1)(𝑥𝑥 2 +1)
20. ∫
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
2𝑥𝑥 3 +4𝑥𝑥 2 −11𝑥𝑥 −19
(𝑥𝑥+4)(𝑥𝑥 2 −3)
𝑑𝑑𝑑𝑑
CASE 4
The Denominator 𝐷𝐷(𝑥𝑥) of the proper rational function contains repeated irreducible quadratic factors .
The final case to be considered (case 4) is that in which the factors of the denominator of the given proper rational
function contains powers of one or more irreducible quadratic factors. By part 4 of the theorem in the beginning of this
discussion, to every factor of the type (𝛼𝛼𝑥𝑥 2 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾)𝑘𝑘 appearing in the denominator of a rational function there
corresponds the partial fractions
𝐴𝐴1 𝑥𝑥 + 𝐵𝐵1
,
𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐
where 𝐴𝐴1 , 𝐴𝐴2 , … , 𝐴𝐴𝑘𝑘 , 𝐵𝐵1 , 𝐵𝐵2 , … , 𝐵𝐵𝑘𝑘 are constants.
𝐴𝐴2 𝑥𝑥 + 𝐵𝐵2
𝐴𝐴𝑘𝑘 𝑥𝑥 + 𝐵𝐵𝑘𝑘
,⋯,
2
2
(𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐)
(𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐)𝑘𝑘
The linear and nonrepeated quadratic factors of the given denominator enter in the same manner as in cases 1 to 3. The
numerator of each partial fraction whose denominator contains a quadratic factor should be a linear function.
Example: Decompose the following fraction
into partial fractions.
6𝑥𝑥 4 + 11𝑥𝑥 3 + 18𝑥𝑥 2 + 14𝑥𝑥 + 6
(𝑥𝑥 + 1)(𝑥𝑥 2 + 𝑥𝑥 + 1)2
Solution: The denominator contains a linear function and the square of an irreducible quadratic function as factors;
hence we let
𝐴𝐴
𝐵𝐵𝐵𝐵 + 𝐶𝐶
𝐷𝐷𝐷𝐷 + 𝐸𝐸
6𝑥𝑥 4 + 11𝑥𝑥 3 + 18𝑥𝑥 2 + 14𝑥𝑥 + 6
=
+ 2
+ 2
2
2
(𝑥𝑥 + 1)(𝑥𝑥 + 𝑥𝑥 + 1)
𝑥𝑥 + 1 𝑥𝑥 + 𝑥𝑥 + 1 (𝑥𝑥 + 𝑥𝑥 + 1)2
and evaluate the undetermined constants after multiplying each member by (𝑥𝑥 + 1)(𝑥𝑥 2 + 𝑥𝑥 + 1)2 . Thus
6𝑥𝑥 4 + 11𝑥𝑥 3 + 18𝑥𝑥 2 + 14𝑥𝑥 + 6 = 𝐴𝐴(𝑥𝑥 2 + 𝑥𝑥 + 1)2 + (𝐵𝐵𝐵𝐵 + 𝐶𝐶 )(𝑥𝑥 + 1)(𝑥𝑥 2 + 𝑥𝑥 + 1) + (𝐷𝐷𝐷𝐷 + 𝐸𝐸 )(𝑥𝑥 + 1)
= (𝐴𝐴 + 𝐵𝐵)𝑥𝑥 4 + (2𝐴𝐴 + 2𝐵𝐵 + 𝐶𝐶 )𝑥𝑥 3 + (3𝐴𝐴 + 2𝐵𝐵 + 2𝐶𝐶 + 𝐷𝐷)𝑥𝑥 2 + (2𝐴𝐴 + 𝐵𝐵 + 2𝐶𝐶 + 𝐷𝐷 + 𝐸𝐸 )𝑥𝑥 + (𝐴𝐴 + 𝐶𝐶 + 𝐸𝐸 ).
Therefore, by equating the coefficients of equal powers of 𝑥𝑥, we obtain
𝐴𝐴 + 𝐵𝐵 = 6
2𝐴𝐴 + 2𝐵𝐵 + 𝐶𝐶 = 11
3𝐴𝐴 + 2𝐵𝐵 + 2𝐶𝐶 + 𝐷𝐷 = 18
2𝐴𝐴 + 𝐵𝐵 + 2𝐶𝐶 + 𝐷𝐷 + 𝐸𝐸 = 14
𝐴𝐴 + 𝐶𝐶 + 𝐸𝐸 = 6
We can solve this system by means of the method suggested in the solution of Example 2 in Case 3. The solution to this
system of equations is 𝐴𝐴 = 5, 𝐵𝐵 = 1, 𝐶𝐶 = −1, 𝐷𝐷 = 3, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐸𝐸 = 2. Therefore,
5
𝑥𝑥 − 1
3𝑥𝑥 + 2
6𝑥𝑥 4 + 11𝑥𝑥 3 + 18𝑥𝑥 2 + 14𝑥𝑥 + 6
=
+ 2
+ 2
2
2
(𝑥𝑥 + 1)(𝑥𝑥 + 𝑥𝑥 + 1)
𝑥𝑥 + 1 𝑥𝑥 + 𝑥𝑥 + 1 (𝑥𝑥 + 𝑥𝑥 + 1)2
FORMULA 5
�
(𝛼𝛼𝑥𝑥 2
1
2𝛼𝛼𝛼𝛼 + 𝛽𝛽
2𝛼𝛼
1
� 2
𝑑𝑑𝑑𝑑 =
+
𝑑𝑑𝑑𝑑
2
2
2
2
(4𝛼𝛼𝛼𝛼 − 𝛽𝛽 )(𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾) 4𝛼𝛼𝛼𝛼 − 𝛽𝛽
+ 𝛽𝛽𝛽𝛽 + 𝛾𝛾)
𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾
FORMULA 6
�
(𝛼𝛼𝑥𝑥 2
𝑥𝑥
𝛽𝛽𝛽𝛽 + 2𝛾𝛾
𝛽𝛽
1
� 2
𝑑𝑑𝑑𝑑 = −
−
𝑑𝑑𝑑𝑑
2
2
2
2
(4𝛼𝛼𝛼𝛼 − 𝛽𝛽 )(𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾) 4𝛼𝛼𝛼𝛼 − 𝛽𝛽
+ 𝛽𝛽𝛽𝛽 + 𝛾𝛾)
𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾
FORMULA 7
�
2𝛼𝛼𝛼𝛼 + 𝛽𝛽
6𝛼𝛼 2 𝑥𝑥 + 3𝛼𝛼𝛼𝛼
6𝛼𝛼 2
1
1
� 2
𝑑𝑑𝑑𝑑
=
+
+
𝑑𝑑𝑑𝑑
2
3
2
2
2
2
2
2
2
2
(4𝛼𝛼𝛼𝛼 − 𝛽𝛽 ) (𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾) (4𝛼𝛼𝛾𝛾 − 𝛽𝛽 )
(𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾)
2(4𝛼𝛼𝛾𝛾 − 𝛽𝛽 )(𝛼𝛼𝑥𝑥 + 𝛽𝛽𝛽𝛽 + 𝛾𝛾)
𝛼𝛼𝑥𝑥 + 𝛽𝛽𝑥𝑥 + 𝛾𝛾
FORMULA 8
�
𝛽𝛽𝑥𝑥 + 2𝛾𝛾
6𝛼𝛼𝛽𝛽𝑥𝑥 + 3𝛽𝛽2
3𝛼𝛼𝛽𝛽
𝑑𝑑𝑑𝑑
𝑥𝑥
� 2
𝑑𝑑𝑑𝑑
=
−
−
−
2
3
2
2
2
2
2
2
2
2
(𝛼𝛼𝑥𝑥 + 𝛽𝛽𝑥𝑥 + 𝛾𝛾)
2(4𝛼𝛼𝛾𝛾 − 𝛽𝛽 ) (𝛼𝛼𝑥𝑥 + 𝛽𝛽𝑥𝑥 + 𝛾𝛾) (4𝛼𝛼𝛾𝛾 − 𝛽𝛽 )
2(4𝛼𝛼𝛾𝛾 − 𝛽𝛽 )(𝛼𝛼𝑥𝑥 + 𝛽𝛽𝑥𝑥 + 𝛾𝛾)
𝛼𝛼𝑥𝑥 + 𝛽𝛽𝑥𝑥 + 𝛾𝛾
Illustration 5: Find the indefinite integral
�
Solution
�
6𝑥𝑥 4 + 11𝑥𝑥 3 + 18𝑥𝑥 2 + 14𝑥𝑥 + 6
𝑑𝑑𝑑𝑑
(𝑥𝑥 + 1)(𝑥𝑥 2 + 𝑥𝑥 + 1)2
5
𝑥𝑥 − 1
3𝑥𝑥 + 2
6𝑥𝑥 4 + 11𝑥𝑥 3 + 18𝑥𝑥 2 + 14𝑥𝑥 + 6
𝑑𝑑𝑑𝑑 = � �
+ 2
+ 2
� 𝑑𝑑𝑑𝑑
2
2
(𝑥𝑥 + 1)(𝑥𝑥 + 𝑥𝑥 + 1)
𝑥𝑥 + 1 𝑥𝑥 + 𝑥𝑥 + 1 (𝑥𝑥 + 𝑥𝑥 + 1)2
= 5�
= 5�
1
𝑥𝑥 − 1
3𝑥𝑥 + 2
𝑑𝑑𝑑𝑑 + � 2
𝑑𝑑𝑑𝑑 + � 2
𝑑𝑑𝑑𝑑
(𝑥𝑥 + 𝑥𝑥 + 1)2
𝑥𝑥 + 1
𝑥𝑥 + 𝑥𝑥 + 1
1
𝑥𝑥 − 1
𝑥𝑥
1
�
𝑑𝑑𝑑𝑑 + � 2
𝑑𝑑𝑑𝑑 + 3 � 2
𝑑𝑑𝑑𝑑
+
2
𝑑𝑑𝑑𝑑
(𝑥𝑥 + 𝑥𝑥 + 1)2
(𝑥𝑥 2 + 𝑥𝑥 + 1)2
𝑥𝑥 + 1
𝑥𝑥 + 𝑥𝑥 + 1
1
2𝑥𝑥 + 1
3
2𝑥𝑥 + 1
1
1
= 5 ln|𝑥𝑥 + 1| + ln|𝑥𝑥 2 + 𝑥𝑥 + 1| − √3 tan−1 �
�+ � 2
𝑑𝑑𝑑𝑑 + � 2
𝑑𝑑𝑑𝑑
2
2
2 (𝑥𝑥 + 𝑥𝑥 + 1)
2 (𝑥𝑥 + 𝑥𝑥 + 1)2
√3
1
2𝑥𝑥 + 1
3
1
1
= 5 ln|𝑥𝑥 + 1| + ln|𝑥𝑥 2 + 𝑥𝑥 + 1| − √3 tan−1 �
�−
+ � 2
𝑑𝑑𝑑𝑑
2
)
(
(
2
2 𝑥𝑥 + 𝑥𝑥 + 1
2
𝑥𝑥 + 𝑥𝑥 + 1)2
√3
1
2𝑥𝑥 + 1
3
1
4
2𝑥𝑥 + 1
2𝑥𝑥 + 1
−1
�
= 5 ln|𝑥𝑥 + 1| + ln|𝑥𝑥 2 + 𝑥𝑥 + 1| − √3 tan−1 �
�−
+
+
tan
�
��
2
2(𝑥𝑥 2 + 𝑥𝑥 + 1) 2 3(𝑥𝑥 2 + 𝑥𝑥 + 1) 3√3
√3
√3
+ 𝐶𝐶
1
𝑥𝑥 − 4
7
2𝑥𝑥 + 1
= 5 ln|𝑥𝑥 + 1| + ln|𝑥𝑥 2 + 𝑥𝑥 + 1| +
−
tan−1 �
� + 𝐶𝐶
2
2
3(𝑥𝑥 + 𝑥𝑥 + 1) 3√3
√3
EXERCISE FOR REPEATED QUADRATIC FACTORS (CASE 4)
Find the value of each of the following integrals by decomposing the integrand in partial fractions. (These Problems will
be Collected!)
1. ∫
3𝑥𝑥 3 +4𝑥𝑥 −5
(𝑥𝑥 2 +2)2
𝑑𝑑𝑑𝑑
2. ∫
𝑥𝑥 3 +2𝑥𝑥 2 −3
(𝑥𝑥 2 −2)2
3. ∫
2𝑥𝑥 3 +𝑥𝑥 2 +4𝑥𝑥 +1
(𝑥𝑥 2 +𝑥𝑥+1)2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
4. ∫
2𝑥𝑥 3 −𝑥𝑥 2 −6𝑥𝑥 −8
(𝑥𝑥 2 −𝑥𝑥−3)2
5. ∫
𝑥𝑥 5 +2𝑥𝑥 4 −3𝑥𝑥 2 −3𝑥𝑥
(𝑥𝑥 2 +𝑥𝑥−1)3
𝑑𝑑𝑑𝑑
6. ∫
𝑥𝑥 5 −2𝑥𝑥 4 +4𝑥𝑥 3 −2𝑥𝑥 2
(𝑥𝑥 2 −𝑥𝑥 +1)3
7. ∫
𝑥𝑥 5 +2𝑥𝑥 3 −𝑥𝑥+3
(𝑥𝑥 2 +1)3
8. ∫
𝑥𝑥 3 +4𝑥𝑥 −5
(𝑥𝑥 2 +2)3
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑥𝑥 4 −𝑥𝑥 3 +1
9. ∫ 𝑥𝑥 2 (𝑥𝑥 2 2 𝑑𝑑𝑑𝑑
+1)
10. ∫
2𝑥𝑥 4 +13𝑥𝑥 3 +13𝑥𝑥 2 −12𝑥𝑥 +2
𝑥𝑥 2 (𝑥𝑥 2 +3𝑥𝑥 −1)2
𝑑𝑑𝑥𝑥
𝑥𝑥 5 +𝑥𝑥 4 −4𝑥𝑥+4
11. ∫ 𝑥𝑥 2 (𝑥𝑥 2 −𝑥𝑥 +2)2 𝑑𝑑𝑑𝑑
12. ∫
−2𝑥𝑥 5 −4𝑥𝑥 4 +12𝑥𝑥 3 +9𝑥𝑥 2 −6𝑥𝑥 +1
𝑥𝑥 2 (𝑥𝑥 2 +3𝑥𝑥−1)2
13. ∫
4𝑥𝑥 4 +𝑥𝑥 3 −25𝑥𝑥 2 −9𝑥𝑥 +30
(𝑥𝑥+2)(𝑥𝑥 2 −3)2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
14. ∫
4𝑥𝑥 4 −7𝑥𝑥 3 +5𝑥𝑥 2 −𝑥𝑥 +1
(2𝑥𝑥−1)(𝑥𝑥 2 −𝑥𝑥 +1)2
15. ∫
3𝑥𝑥 4 +2𝑥𝑥 3 +8𝑥𝑥 2 +7𝑥𝑥 +2
(𝑥𝑥−1)(𝑥𝑥 2 +2)2
16. ∫
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑥𝑥 4 −5𝑥𝑥 2 −10𝑥𝑥 2 +26𝑥𝑥 +33
(𝑥𝑥+3)(𝑥𝑥 2 −𝑥𝑥−3)2
𝑑𝑑𝑑𝑑
THE END