INTERGRATION OF RATIONAL
FRACTIONS
Overview
A rational function has the form
𝑝(𝑥)
𝑟 𝑥 =
𝑞(𝑥)
where 𝑝 and 𝑞 are polynomials. For example,
𝑥2 − 3
𝑓 𝑥 = 4
𝑥 +3
and
𝑡 6 + 4𝑡 2 − 3
𝑔 𝑡 =
7𝑡 5 + 3𝑡
5
ℎ 𝑠 = 3
𝑠 − 4𝑠 2 + 3𝑠 − 8
are all rational functions. A rational function is called proper if the
degree of the numerator is less than the degree of the denominator,
and improper otherwise. Thus,𝑓 and ℎ are proper rational functions,
while 𝑔 is an improper rational function.
Indefinite integrals (antiderivatives) of rational functions can always
be found by the following steps:
1. Polynomial Division: Divide the denominator into the numerator
(if needed) to write the integrand as a polynomial plus a proper
rational function.
2. Partial Fraction Expansion: Expand the proper rational function
using partial fractions.
3. Completing the Square: If any terms involve quadratics, eliminate
the linear term if needed by completing the square.
4. Term by Term Integration: Use elementary integral formulas and
substitution.
Partial Fractions
It is shown in algebra that every rational fraction whose numerator
is of lower degree than the denominator can be broken up into socalled partial fractions of the exact form below. It follows that every
rational fraction can be integrated in elementary terms.
𝐴
,
𝑎𝑥+𝑏 𝑛
𝐴 2𝑎𝑥+𝑏
𝐴
,
𝑎𝑥 2 +𝑏𝑥+𝑐 𝑛 𝑎𝑥 2 +𝑏𝑥+𝑐
The first two leads to powers, if 𝑛 > 1, to logarithms, if 𝑛 = 1; the
third leads to an arctangent. We can also integrate
𝐴
, 𝑛 > 1,
2
𝑛
𝑎𝑥 +𝑏𝑥+𝑐
by trigonometric substitution.
Case I: Distinct Linear Factors
The simplest case is that in which the denominator can be broken up
into real linear factors, none of which is repeated. In this case we may
always rewrite the given fraction (provided the numerator is of lower
degree than the denominator) as a sum of fractions whose numerator
are constants and whose respective denominators are the factors of
the original denominator.
𝑑𝑥
𝑥 2 −4
Example 1: Evaluate
Solution:
- Transform/decompose the integrand into partial fractions
1
1
=
2
𝑥 −4
𝑥−2 𝑥+2
𝐴
𝐵
=
+
𝑥−2
𝑥+2
Multiplying both sides of the equation by 𝑥 − 2 𝑥 + 2
1 =𝐴 𝑥+2 +𝐵 𝑥−2
Solve for the constants A and B by elimination,
When 𝑥 = 2,
1 = 𝐴 2 + 2 + 𝐵(2 − 2),
Solving for 𝐴 gives us, 𝐴 = 14
When 𝑥 = −2,
1 = 𝐴 −2 + 2 + 𝐵 −2 − 2
Solving for 𝐵 gives us, B = −14
Substituting the values of 𝐴 and 𝐵, gives us
1
1
1
−
4
4
=
+
2
𝑥 −4
𝑥−2
𝑥+2
𝑑𝑥
=
𝑥2 − 4
1
=
4
1
4
−14
+
𝑑𝑥
𝑥−2
𝑥+2
𝑑𝑥
1
−
𝑥−2
4
𝑑𝑥
𝑥+2
1
1
= ln 𝑥 − 2 − ln 𝑥 + 2 + 𝐶
4
4
Example 2. Evaluate
3𝑥+7 𝑑𝑥
𝑥 2 −2𝑥−3
Solution:
- Transform/decompose the integrand into partial fractions
𝐴
𝐵
3𝑥 + 7
3𝑥 + 7
=
+
=
2
𝑥−3
𝑥+1
𝑥 − 2𝑥 − 3
𝑥−3 𝑥+1
Multiplying both sides of the equation by 𝑥 − 3 𝑥 + 1
3𝑥 + 7
=𝐴 𝑥+1 +𝐵 𝑥−3
Solve for the constants A and B by elimination,
When 𝑥 = 3,
3 3 + 7 = 𝐴 3 + 1 + 𝐵(3 − 3)
Solving for 𝐴 gives us, 𝐴 = 4
When 𝑥 = −1,
3 −1 + 7 = 𝐴 −1 + 1 + 𝐵(−1 − 3)
Solving for 𝐵 gives us, 𝐵 = −1
Substituting the values of 𝐴 and 𝐵, gives us
4
1
3𝑥 + 7
=
−
2
𝑥−3
𝑥+1
𝑥 − 2𝑥 − 3
We are now ready to integrate.
3𝑥 + 7 𝑑𝑥
=
2
𝑥 − 2𝑥 − 3
=4
4
1
−
𝑑𝑥
𝑥−3
𝑥+1
𝑑𝑥
−
𝑥−3
𝑑𝑥
𝑥+1
= 4 ln 𝑥 − 3 − ln 𝑥 + 1 + 𝐶
Example 3. Evaluate
𝑥+1
𝑥 3 +𝑥 2 −6𝑥
Solution:
- Transform/decompose the integrand into partial fractions
𝑥+1
𝑥+1
=
3
2
𝑥 + 𝑥 − 6𝑥
𝑥 𝑥2 + 𝑥 − 6
𝑥+1
=
𝑥 𝑥+3 𝑥−2
𝐴
𝐵
𝐶
= +
+
𝑥 𝑥+3 𝑥−2
Multiplying both sides of the equation by 𝑥 𝑥 + 3 𝑥 − 2
𝑥 + 1 = 𝐴 𝑥 + 3 𝑥 − 2 + 𝐵𝑥 𝑥 − 2 + 𝐶𝑥 𝑥 + 3
Solve for the constants A, B and C by elimination,
1 = 𝐴(3)(−2)
𝐴 = −16
𝑥 = −3;
−3 + 1 = 𝐵(−3)(−3 − 2)
2
𝐵 = −15
𝑥 = 2;
2 + 1 = 𝐶(2)(2 + 3)
3
𝐶 = 10
When 𝑥 = 0;
Substituting the values of 𝐴, B and 𝐶, gives us
2
3
𝑥+1
−16
−15
10
=
+
+
3
2
𝑥 + 𝑥 − 6𝑥
𝑥
𝑥+3 𝑥−2
We are now ready to integrate.
𝑥+1
𝑑𝑥=
3
2
𝑥 + 𝑥 − 6𝑥
2
3
−16
−15
+
+ 10 𝑑𝑥
𝑥
𝑥+3 𝑥−2
1
=−
6
𝑑𝑥 2
−
𝑥 15
𝑑𝑥
3
+
𝑥+3
10
𝑑𝑥
𝑥−2
1
2
3
= − ln 𝑥 − ln 𝑥 + 3 + ln 𝑥 − 2 + 𝐶
6
15
10
Example 4. Evaluate
Solution:
let:
𝑧=𝑥
𝑧 = 𝑥2
𝑑𝑧
(4−𝑧) 𝑧
𝑑𝑧 = 2𝑥 𝑑𝑥
substitute
𝑑𝑧
=
(4 − 𝑧) 𝑧
=
2𝑥 𝑑𝑥
4 − 𝑥2 𝑥
2 𝑑𝑥
2−𝑥 2+𝑥
Transform/decompose the integrand
2
𝐴
𝐵
=
+
2−𝑥 2+𝑥
2−𝑥
2+𝑥
2 = 𝐴 2+𝑥 +𝐵 2−𝑥
Solve for the constants A and B by elimination,
When:
𝑥=2
2= 𝐴 2+2
𝐴 = 12
𝑥 = −2
2 = 𝐵 2 − (−2)
𝐵 = 12
Substituting the values of 𝐴 and 𝐵, gives us
1
1
1
2
2
=
+
2−𝑥 2+𝑥
2−𝑥
2+𝑥
We are now ready to integrate.
2 𝑑𝑥
2−𝑥 2+𝑥
=
1
=
2
1
2
2−𝑥
+
1
2
2+𝑥
𝑑𝑥
1
+
2−𝑥
2
𝑑𝑥
𝑑𝑥
2+𝑥
1
1
= − ln 2 − 𝑥 + ln 2 + 𝑥 + 𝐶
2
2
Substituting back the original variable where 𝑥 = 𝑧
𝑑𝑧
1
1
(4 − 𝑧) 𝑧 = − 2 ln 2 − 𝑧 + 2 ln 2 + 𝑧 + 𝐶
Example 5. Evaluate
𝑥 2 +1
𝑥−2 𝑥−1 2𝑥+1
𝑑𝑥
Solution:
Transform/decompose the integrand
𝑥2 + 1
𝑥 − 2 𝑥 − 1 2𝑥 + 1
𝐴
𝐵
𝐶
=
+
+
𝑥−2
𝑥−1
2𝑥 + 1
𝑥 2 + 1 = 𝐴 𝑥 − 1 2𝑥 + 1 + 𝐵 𝑥 − 2 2𝑥 + 1 + 𝐶 𝑥 − 2 𝑥 − 1
Solve for the constants A, B and C
When:
𝑥=2
22 + 1 = 𝐴(2 − 1) 2 2 + 1
𝐴=1
𝑥=1
12 + 1 = 𝐵 1 − 2 2 1 + 1
𝐵 = −23
𝑥 = −12
−12
2
+ 1 = 𝐶 −12 −2 −12 −1
Substituting the values of 𝐴, 𝐵 and 𝐶, gives us
2
1
𝑥2 + 1
−
1
3 +
3
=
+
𝑥 − 2 𝑥 − 1 2𝑥 + 1
𝑥−2
𝑥−1
2𝑥 + 1
𝐶 = 13
We are now ready to integrate.
𝑥2 + 1
𝑑𝑥
=
𝑥 − 2 𝑥 − 1 2𝑥 + 1
=
2
−3
1
1
3
+
+
𝑑𝑥
𝑥−2
𝑥−1
2𝑥 + 1
𝑑𝑥
2
−
𝑥−2
3
𝑑𝑥
1
+
𝑥−1
3
𝑑𝑥
2𝑥 + 1
2
1 1
= ln 𝑥 − 2 − ln 𝑥 − 1 + ∙ ln 2𝑥 + 1 + 𝐶
3
3 2
2
1
= ln 𝑥 − 2 − ln 𝑥 − 1 + ln 2𝑥 + 1 + 𝐶
3
6
Example 6. Evaluate
𝑥 4 −𝑥 3 −3𝑥 2 −2𝑥+2
𝑑𝑥.
𝑥 3 +𝑥 2 −2𝑥
Solution:
Since the numerator is of higher degree than the denominator, then
divide the denominator into the numerator
𝑥−2
𝑥 3 + 𝑥 2 − 2𝑥 𝑥 4 − 𝑥 3 − 3𝑥 2 − 2𝑥 + 2
𝑥 4 + 𝑥 3 − 2𝑥 2
−2𝑥 3 − 𝑥 2 − 2𝑥 + 2
−2𝑥 3 − 2𝑥 2 + 4𝑥
𝑥 2 − 6𝑥 + 2
𝑥 4 − 𝑥 3 − 3𝑥 2 − 2𝑥 + 2
𝑑𝑥 =
3
2
𝑥 + 𝑥 − 2𝑥
=
𝑥 2 − 6𝑥 + 2
𝑥−2+ 3
𝑑𝑥
2
𝑥 + 𝑥 − 2𝑥
𝑥 𝑑𝑥 − 2
𝑑𝑥 +
𝑥 2 − 6𝑥 + 2
𝑑𝑥
𝑥 3 + 𝑥 2 − 2𝑥
Decompose the third integrand.
𝑥 2 − 6𝑥 + 2
𝑥 2 − 6𝑥 + 2
=
𝑥 3 + 𝑥 2 − 2𝑥
𝑥 𝑥2 + 𝑥 − 2
𝑥 2 − 6𝑥 + 2
=
𝑥 𝑥−1 𝑥+2
𝐴
𝐵
𝐶
= +
+
𝑥
𝑥−1
𝑥+2
(1)
𝑥 2 − 6𝑥 + 2 = 𝐴 𝑥 − 1 𝑥 + 2 + 𝐵𝑥 𝑥 + 2 + 𝐶𝑥 𝑥 − 1
(2)
𝑥 2 − 6𝑥 + 2 = 𝐴 𝑥 2 + 𝑥 − 2 + 𝐵 𝑥 2 + 2𝑥 + 𝐶 𝑥 2 − 𝑥
(3)
Evaluate 𝐴, 𝐵 and 𝐶 using Eq. (2)
From (2):
when 𝑥 = 0
2 = 𝐴(−1)(2)
𝐴 = −1
𝑥=1
12 − 6(1) + 2 = 𝐵(1)(1 + 2)
𝐵 = −1
𝑥 = −2
−2 2 − 6 −2 + 2 = 𝐶(−2)(−2 − 1)
𝐶=3
Substitute values of A, B and C then integrate
𝑥 4 − 𝑥 3 − 3𝑥 2 − 2𝑥 + 2
𝑑𝑥 =
𝑥 3 + 𝑥 2 − 2𝑥
=
𝑥 𝑑𝑥 − 2
𝑥𝑑𝑥 − 2
𝑑𝑥 +
−1
−1
3
+
+
𝑑𝑥
𝑥
𝑥−1
𝑥+2
𝑑𝑥 −
𝑑𝑥
−
𝑥
𝑑𝑥
+3
𝑥−1
𝑑𝑥
𝑥+2
𝑥2
=
− 2𝑥 − ln 𝑥 − ln 𝑥 − 1 + 3 ln 𝑥 + 2 + 𝐶
2
Now try these
Practice Problem
1.
2𝑥 + 11
𝑑𝑥
2
𝑥 +𝑥−6
Ans.
2.
𝑑𝑥
𝑥 2 + 4𝑥
Ans.
3.
3 ln 𝑥 − 2 − ln 𝑥 + 3 + 𝐶
1
𝑥
ln 𝑥+4 + 𝐶
4
3𝑥 2 + 8𝑥 − 12
𝑑𝑥
3
2
𝑥 + 7𝑥 + 12𝑥
Ans.
3 ln 𝑥 + 3 − ln 𝑥 + ln 𝑥 + 4 + 𝐶
Case II: Repeated linear Factors
If the denominator contains a factor 𝑥 − 𝑎 𝑟 , the above method
fails, since there would be 𝑟 partial fractions with denominator
𝑥 − 𝑎 , and these could be combined into a single fraction with
denominator 𝑥 − 𝑎 . In this case, corresponding to the factor
𝑥 − 𝑎 𝑟 , we assume 𝑟 partial fractions of the form
𝐴
𝐵
𝐶
𝑍
+
+
+∙∙∙∙∙∙ +
2
3
𝑥−𝑎
𝑥−𝑎
𝑥−𝑎
𝑥−𝑎 𝑟
Example 1. Evaluate
𝑥 3 −1
𝑑𝑥.
𝑥 𝑥+1 3
Solution:
Decompose the integrand.
𝐴
𝐵
𝐶
𝐷
𝑥3 − 1
= +
+
+
(1)
2
3
𝑥
𝑥
+
1
𝑥
+
1
𝑥+1 3
𝑥 𝑥+1
𝑥 3 − 1 = 𝐴 𝑥 + 1 3 + 𝐵𝑥 𝑥 + 1 2 + 𝐶𝑥 𝑥 + 1 + 𝐷𝑥
(2)
𝑥 3 − 1 = 𝐴 𝑥 3 + 3𝑥 2 + 3𝑥 + 1 + 𝐵𝑥 𝑥 2 + 2𝑥 + 1 + 𝐶𝑥 𝑥 + 1 + 𝐷𝑥
(3)
𝑥 3 − 1 = 𝐴 𝑥 3 + 3𝑥 2 + 3𝑥 + 1 + 𝐵 𝑥 3 + 2𝑥 2 + 𝑥 + 𝐶 𝑥 2 + 𝑥 + 𝐷𝑥
To get the necessary four equations for the determination of
𝐴, 𝐵, 𝐶, 𝐷, two methods are once available. Specific values of 𝑥 can
be used in the identity (2), or the coefficients of like powers of 𝑥 in
the two members of (3) can be equated.
We naturally employ whatever combination of these methods yields
simple equations to be solved for the unknowns 𝐴, 𝐵, etc.
From (2)
When:
𝑥 = 0:
−1 = 𝐴 0 + 1 3
𝐴 = −1
𝑥 = −1:
−1 3 − 1 = 𝐷(−1)
𝐷=2
From (3), equate coefficients of like powers
coefficients of 𝑥 3 :
1=𝐴+𝐵
𝐵=2
but 𝐴 = −1
coefficients of 𝑥 2 :
0 = 3𝐴 + 2𝐵 + 𝐶
but 𝐴 = −1 and 𝐵 = 2
𝐶 = −1
whence
𝑥3 − 1
𝑑𝑥 =
𝑥 𝑥+1 3
=−
−1
2
−1
2
+
+
+
𝑑𝑥
𝑥
𝑥+1
𝑥+1 2
𝑥+1 3
𝑑𝑥
+2
𝑥
𝑑𝑥
−
𝑥+1
𝑥 + 1 −2 𝑑𝑥 + 2
𝑥 + 1 −3 𝑑𝑥
𝑥 + 1 −1
𝑥 + 1 −2
= − ln 𝑥 + 2 ln 𝑥 + 1 −
+2
+𝐶
−1
−2
1
1
= − ln 𝑥 + 2 ln 𝑥 + 1 +
−
+𝐶
𝑥+1
𝑥+1 2
Example 2. Evaluate
3𝑥+5
𝑑𝑥.
𝑥 3 −𝑥 2 −𝑥+1
Solution:
Decompose the integrand.
3𝑥 + 5
3𝑥 + 5
=
𝑥3 − 𝑥2 − 𝑥 + 1
𝑥+1 𝑥−1 2
𝐴
𝐵
𝐶
=
+
+
𝑥+1
𝑥−1
𝑥−1 2
3𝑥 + 5 = 𝐴 𝑥 − 1 2 + 𝐵 𝑥 + 1 𝑥 − 1 + 𝐶 𝑥 + 1
(1)
3𝑥 + 5 = 𝐴 𝑥 2 − 2𝑥 + 1 + 𝐵 𝑥 2 − 1 + 𝐶 𝑥 + 1
(3)
(2)
Evaluate A, B and C
From (2):
when 𝑥 = −1
𝑥=1
3 −1 + 5 = 𝐴 −1 − 1 2
𝐴 = 12
3 1 + 5 = 𝐶(1 + 1)
𝐶=4
From (3), equate coefficients of like powers
coefficients of 𝑥 2
0=𝐴+𝐵
but 𝐴 = 12
Note: Synthetic division may be used to factor the denominator.
𝐵 = −12
We are ready to integrate
1
2
−12
4
+
+
𝑑𝑥
2
𝑥+1
𝑥−1
𝑥−1
3𝑥 + 5
𝑑𝑥 =
𝑥3 − 𝑥2 − 𝑥 + 1
1
=
2
𝑑𝑥
1
−
𝑥+1
2
𝑑𝑥
+4
𝑥−1
𝑑𝑥
𝑥−1 2
1
=
2
𝑑𝑥
1
−
𝑥+1
2
𝑑𝑥
+4
𝑥−1
𝑥 − 1 −2 𝑑𝑥
1
1
𝑥 − 1 −1
= ln 𝑥 + 1 − ln 𝑥 − 1 + 4
+𝐶
2
2
−1
=
1
1
4
ln 𝑥 + 1 − ln 𝑥 − 1 −
+𝐶
2
2
𝑥−1
Example 3. Evaluate
𝑥+1
𝑥 2 𝑥−1
𝑑𝑥
Solution:
Decompose the integrand.
𝐴 𝐵
𝐶
+ 2+
𝑥 𝑥
𝑥−1
(1)
𝑥 + 1 = 𝐴𝑥 𝑥 − 1 + 𝐵 𝑥 − 1 + 𝐶𝑥 2
(2)
𝑥 + 1 = 𝐴 𝑥 2 − 2𝑥 + 𝐵 𝑥 − 1 + 𝐶𝑥 2
(3)
𝑥+1
𝑥2 𝑥 − 1
=
Evaluate A, B and C
From (2):
when 𝑥 = 1
1 + 1 = 𝐶(12 )
𝐶=2
𝑥=0
1 = 𝐵(0 − 1)
𝐵 = −1
From (3), equate coefficients of like powers
coefficients of 𝑥 2
0=𝐴+𝐶
but C = 2
𝐴 = −2
We are ready to integrate
𝑥+1
𝑑𝑥 =
𝑥2 𝑥 − 1
−2 −1
2
+ 2+
𝑑𝑥
𝑥
𝑥
𝑥−1
= −2
𝑑𝑥
−
𝑥
𝑑𝑥
+2
𝑥2
= −2
𝑑𝑥
−
𝑥
−2
𝑥
𝑑𝑥
𝑥−1
𝑑𝑥 + 2
𝑑𝑥
𝑥−1
𝑥 −1
= −2 ln 𝑥 −
+ 2 ln 𝑥 − 1 + 𝐶
−1
1
= −2 ln 𝑥 + + 2 ln 𝑥 − 1 + 𝐶
𝑥
Example 4. Evaluate
𝑥
𝑑𝑥
𝑥+2 3
Solution:
Decompose the integrand.
𝑥
𝐴
𝐵
𝐶
=
+
+
𝑥+2 3
𝑥+2
𝑥+2 2
𝑥+2 3
(1)
𝑥 =𝐴 𝑥+2 2+𝐵 𝑥+2 +𝐶
(2)
𝑥 + 1 = 𝐴 𝑥 2 + 4𝑥 + 4 + 𝐵 𝑥 + 2 + 𝐶
(3)
Evaluate A, B and C
From (2):
when 𝑥 = −2
−2 = 𝐶
𝐶 = −2
From (3), equate coefficients of like powers
coefficients of 𝑥 2
0=𝐴
𝐴=0
coefficients of 𝑥
1 = 4𝐴 + 𝐵, but 𝐴 = 0
𝐵=1
Substitute values of A, B and C then integrate
𝑥
𝑑𝑥 =
𝑥+2 3
0
1
−2
+
+
𝑑𝑥
𝑥+2
𝑥+2 2
𝑥+2 3
=
𝑑𝑥
−2
𝑥+2 2
𝑑𝑥
𝑥+2 3
=
𝑥 + 2 −2 𝑑𝑥 − 2
𝑥 + 2 −3 𝑑𝑥
𝑥 + 2 −1
𝑥 + 2 −2
=
−2
+𝐶
−1
−2
1
1
=−
+
+𝐶
𝑥+2
𝑥+2 2
Now try these
Practice Problem
1.
1
𝑑𝑥
2
𝑥 𝑥+2
Ans.
2.
3.
1
1
ln
𝑥
−
ln 𝑥 + 2
4
4
1
+ 2 𝑥 + 2 −1 + 𝐶
5𝑥 − 4 𝑑𝑥
𝑥 3 + 4𝑥 2
3
3
−1
Ans. 𝑥 + ln 𝑥 − ln 𝑥 + 4
2
2
𝑥4 + 1
+𝐶
𝑑𝑥
2
2
𝑥 𝑥+1
Ans.
x − 2ln x − 𝑥 −1 − 2 𝑥 + 1 −1 + 𝐶
Case III: Quadratic Factors
Corresponding to a factor in the denominator of one form
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 with 𝑏 2 − 4𝑎𝑐 < 0, we assume the partial fraction
𝐴 2𝑎𝑥+𝑏 +𝐵
, where A and B are to be determined.
𝑎𝑥 2 +𝑏𝑥+𝑐
Note: the term 2𝑎𝑥 + 𝑏 is the derivative of the denominator.
Example 1: Evaluate
𝑥 2 +4𝑥+10
𝑑𝑥
𝑥 3 +2𝑥 2 +5𝑥
Solution:
Decompose the integrand
𝑥 2 + 4𝑥 + 10
𝑥 2 + 4𝑥 + 10
=
3
2
𝑥 + 2𝑥 + 5𝑥
𝑥(𝑥 2 + 2𝑥 + 5)
𝐴
𝐵 2𝑥 + 2
𝐶
= + 2
+
𝑥 (𝑥 +2𝑥 + 5) (𝑥 2 +2𝑥 + 5)
(1)
𝑥 2 + 4𝑥 + 10 = 𝐴(𝑥 2 +2𝑥 + 5) + 𝐵𝑥 2𝑥 + 2 + 𝐶𝑥
(2)
𝑥 2 + 4𝑥 + 10 = 𝐴(𝑥 2 +2𝑥 + 5) + 𝐵 2𝑥 2 + 2𝑥 + 𝐶𝑥
(3)
Evaluate A, B and C
From (2):
when 𝑥 = 0
10 = 𝐴 5
𝐴=2
From (3), equate coefficients of like powers
coefficients of 𝑥 2
1 = 𝐴 + 2𝐵, but 𝐴 = 2
𝐵 = −12
coefficients of 𝑥
4 = 2𝐴 + 2𝐵 + 𝐶, but 𝐴 = 2 &𝐵 = −12
𝐶=1
Substitute values of A, B and C then integrate
𝑥 2 + 4𝑥 + 10
𝑑𝑥 =
𝑥 3 + 2𝑥 2 + 5𝑥
−12 2𝑥 + 2
2
1
+
+
𝑑𝑥
𝑥 (𝑥 2 +2𝑥 + 5) (𝑥 2 +2𝑥 + 5)
=2
𝑑𝑥 1
−
𝑥
2
2𝑥 + 2 𝑑𝑥
+
(𝑥 2 +2𝑥 + 5)
𝑑𝑥
(𝑥 2 +2𝑥 + 5)
=2
𝑑𝑥 1
−
𝑥
2
2𝑥 + 2 𝑑𝑥
+
(𝑥 2 +2𝑥 + 5)
𝑑𝑥
(𝑥 2 +2𝑥 + 1) + 4
1
= 2 ln 𝑥 − ln(𝑥 2 +2𝑥 + 5) +
2
𝑑𝑥
𝑥+1 2+4
We evaluate
let
𝑑𝑥
separately
𝑥+1 2 +4
𝑎=2
𝑎2 = 4
𝑢 =𝑥+1
𝑑𝑢 = 𝑑𝑥
and
𝑑𝑥
=
𝑥+1 2+1
𝑑𝑢
1
𝑢
= 𝐴𝑟𝑐𝑡𝑎𝑛
𝑢2 + 𝑎2 𝑎
𝑎
1
𝑥+1
= 𝐴𝑟𝑐𝑡𝑎𝑛
+𝐶
2
2
We continue with the integration of the original example.
𝑥 2 + 4𝑥 + 10
1
2
𝑑𝑥
=
2
ln
𝑥
−
ln(𝑥
+2𝑥 + 5) +
3
2
𝑥 + 2𝑥 + 5𝑥
2
𝑑𝑥
𝑥+1 2+4
1
1
𝑥+1
2
= 2 ln 𝑥 − ln(𝑥 +2𝑥 + 5) + 𝐴𝑟𝑐𝑡𝑎𝑛
+𝐶
2
2
2
Example 2: Evaluate
𝑥 2 +2
𝑑𝑥.
𝑥 3 −1
Solution:
Decompose the integrand
𝑥2 + 2
𝑥2 + 2
=
3
𝑥 −1
𝑥 − 1 𝑥2 + 𝑥 + 1
𝐴
𝐵 2𝑥 + 1
𝐶
=
+ 2
+ 2
𝑥−1
𝑥 +𝑥+1
𝑥 +𝑥+1
𝑥 2 + 2 = 𝐴 𝑥 2 + 𝑥 + 1 + 𝐵(𝑥 − 1) 2𝑥 + 1 + 𝐶(𝑥 − 1)
𝑥 2 + 2 = 𝐴 𝑥 2 + 𝑥 + 1 + 𝐵 2𝑥 2 − 𝑥 − 1 + 𝐶(𝑥 − 1)
Evaluate A, B and C
From (2):
when 𝑥 = 1
12 + 2 = 𝐴 12 + 1 + 1
𝐴=1
From (3), equate coefficients of like powers
coefficients of 𝑥 2
1 = 𝐴 + 2𝐵, but 𝐴 = 1
𝐵=0
coefficients of 𝑥
0 = 𝐴 − 𝐵 + 𝐶, but 𝐴 = 1 & 𝐵 = 0
𝐶 = −1
(1)
(2)
(3)
Substitute values of A, B and C then integrate
𝑥2 + 2
𝑑𝑥 =
3
𝑥 −1
1
0 2𝑥 + 1
−1
+ 2
+ 2
𝑑𝑥
𝑥−1
𝑥 +𝑥+1
𝑥 +𝑥+1
=
𝑑𝑥
−
𝑥−1
=
𝑑𝑥
−
𝑥−1
=
𝑑𝑥
−
𝑥−1
= ln 𝑥 − 1 −
= ln 𝑥 − 1 −
= ln 𝑥 − 1 −
𝑑𝑥
𝑥2 + 𝑥 + 1
𝑑𝑥
𝑥 2 + 𝑥 + 14 + 34
𝑑𝑥
2
𝑥 + 12 +
1
3
2
2
3
2
3
𝐴𝑟𝑐𝑡𝑎𝑛
𝐴𝑟𝑐𝑡𝑎𝑛
𝐴𝑟𝑐𝑡𝑎𝑛
3
2
2
𝑥 + 12
3
2
2 𝑥 + 12
3
2𝑥 + 1
3
+𝐶
+𝐶
+𝐶
Example 3: Evaluate
𝑥 3 +𝑥 2 +𝑥+2
𝑑𝑥.
𝑥 4 +3𝑥 2 +2
Solution:
Decompose the integrand
𝑥3 + 𝑥2 + 𝑥 + 2
𝑥3 + 𝑥2 + 𝑥 + 2
= 2
4
2
𝑥 + 3𝑥 + 2
𝑥 + 1 𝑥2 + 2
𝐴 2𝑥
𝐵
𝐶 2𝑥
𝐷
= 2
+ 2
+ 2
+ 2
𝑥 +1
𝑥 +1
𝑥 +2
𝑥 +2
𝑥 3 + 𝑥 2 + 𝑥 + 2 = 𝐴 2𝑥 𝑥 2 + 2 + 𝐵 𝑥 2 + 2 + 𝐶 2𝑥 𝑥 2 + 1 + 𝐷 𝑥 2 + 1
𝑥 3 + 𝑥 2 + 𝑥 + 2 = 𝐴 2𝑥 3 + 4𝑥 + 𝐵 𝑥 2 + 2 + 𝐶 2𝑥 3 + 2𝑥 + 𝐷 𝑥 2 + 1
From (3), equate coefficients of like powers
coefficients of 𝑥 3
1 = 2𝐴 + 2𝐶
(4)
coefficients of 𝑥 2
1=𝐵+𝐷
(5)
coefficients of 𝑥
1 = 4𝐴 + 2𝐶
(6)
coefficients of 𝑥 0
2 = 2𝐵 + 𝐷
(7)
(1)
(2)
(3)
Solve for A, B, C and D.
- from (4) and (6)
𝐴=0
and
𝐶 = 12
- from (5) and (7)
𝐵=1
and
𝐷=0
Substitute values of A, B and C then integrate
𝑥3 + 𝑥2 + 𝑥 + 2
𝑑𝑥
4
2
=
𝑥 + 3𝑥 + 2
=
1
2𝑥
0 2𝑥
1
0
2
+ 2
+ 2
+ 2
𝑑𝑥
2
𝑥 +1
𝑥 +1
𝑥 +2
𝑥 +2
𝑑𝑥
1
+
2
𝑥 +1
2
2𝑥 𝑑𝑥
𝑥2 + 2
1
= 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + ln 𝑥 2 + 2 + 𝐶
2
Example 4: Evaluate
𝑥
𝑑𝑥.
𝑥 2 +6𝑥+13
Solution:
Decompose the integrand
𝑥
𝐴 2𝑥 + 6
𝐵
𝑥 2 + 6𝑥 + 13 = 𝑥 2 + 6𝑥 + 13 + 𝑥 2 + 6𝑥 + 13
𝑥 = 𝐴 2𝑥 + 6 + 𝐵
(1)
(2)
Solve for A and B
From (2) when 𝑥 = −3
−3 = 𝐴 2 −3 + 6 + 𝐵
𝐵 = −3
when 𝑥 = 0
0 = 6𝐴 + 𝐵,
but 𝐵 = −3
𝐴 = 12
Substitute values of A and B then integrate
1
2𝑥 + 6
2
𝑥 2 + 6𝑥 + 13
𝑥
𝑑𝑥 =
2
𝑥 + 6𝑥 + 13
=
1
2
−3
+ 2
𝑑𝑥
𝑥 + 6𝑥 + 13
2𝑥 + 6 𝑑𝑥
−3
2
𝑥 + 6𝑥 + 13
𝑑𝑥
𝑥 2 + 6𝑥 + 13
1
= ln 𝑥 2 + 6𝑥 + 13 − 3
2
𝑑𝑥
𝑥 2 + 6𝑥 + 9 + 4
1
= ln 𝑥 2 + 6𝑥 + 13 − 3
2
𝑑𝑥
𝑥 + 3 2 + 22
1
1
𝑥+3
2
= ln 𝑥 + 6𝑥 + 13 − 3 𝐴𝑟𝑐𝑡𝑎𝑛
2
2
2
+𝐶
1
3
𝑥+3
2
= ln 𝑥 + 6𝑥 + 13 − 𝐴𝑟𝑐𝑡𝑎𝑛
+𝐶
2
2
2
Now try these
Practice Problem
1.
4𝑥 + 5
𝑑𝑥
2
𝑥 + 4𝑥 + 20
Ans.
2.
2 ln 𝑥 2 + 4𝑥 + 20
3
𝑥+2
− 2 Arctan 4
4𝑑𝑥
𝑥 3 − 4𝑥 2 − 8𝑥
1
1
Ans. 2 ln 𝑥 − 4 ln 𝑥 2 − 4𝑥 + 8
3.
+𝐶
1
𝑥−2
+ 2 𝐴𝑟𝑐𝑡𝑎𝑛 2
+𝐶
9𝑦 + 14
𝑑𝑥
2
𝑦−2 𝑦 +4
Ans.
4ln 𝑦 − 2
− 2 ln 𝑦 2 + 4
1
𝑦
+ 2 𝐴𝑟𝑐𝑡𝑎𝑛 2 + 𝐶
Case IV: Repeated Quadratic Factors
The case of repeated quadratic factors occurs less often.
Corresponding to a factor 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑟 , we write 𝑟 partial
fractions with linear numerators of the form
𝐴 2𝑥 + 𝑏 + 𝐵
𝐶 2𝑥 + 𝑏 + 𝐷
𝐻 2𝑥 + 𝑏 + 𝐼
+
+∙∙∙ +
2
2
2
𝑎𝑥 + 𝑏𝑥 + 𝑐
𝑎𝑥 + 𝑏𝑥 + 𝑐
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑟
where A, B, C, etc. are constants to be determined
Example 1: Evaluate
𝑥2
𝑑𝑥
𝑥 2 +4𝑥+5 2
Solution:
Decompose the integrand
𝐴 2𝑥 + 4 + 𝐵 𝐶 2𝑥 + 4 + 𝐷
𝑥2
= 2
+ 2
2
2
𝑥
+
4𝑥
+
5
𝑥 + 4𝑥 + 5 2
𝑥 + 4𝑥 + 5
𝑥 2 = 𝐴 2𝑥 + 4 𝑥 2 + 4𝑥 + 5 + 𝐵 𝑥 2 + 4𝑥 + 5 + 𝐶 2𝑥 + 4 + 𝐷
𝑥 2 = 𝐴 2𝑥 3 + 12𝑥 2 + 26𝑥 + 20 + 𝐵 𝑥 2 + 4𝑥 + 5 + 𝐶 2𝑥 + 4 + 𝐷
Solve for A, B, C and D
equate coefficients of like powers
coefficients of 𝑥 3
0 = 2𝐴
(1)
coefficients of 𝑥 2
1 = 12𝐴 + 𝐵
(2)
coefficients of 𝑥
0 = 26𝐴 + 4𝐵 + 2𝐶
(3)
coefficients of 𝑥 0
0 = 20𝐴 + 5𝐵 + 4𝐶 + 𝐷
(4)
From the 4 equations above, we have
𝐴 = 0,
𝐵 = 1,
𝐶 = −2,
and
𝐷=3
Substitute values of A, B, C and D then integrate
𝑥2
𝑑𝑥 =
𝑥 2 + 4𝑥 + 5 2
0 2𝑥 + 4 + 1 −2 2𝑥 + 4 + 3
+ 2
𝑑𝑥
𝑥 2 + 4𝑥 + 5
𝑥 + 4𝑥 + 5 2
=
𝑑𝑥
−2
𝑥 2 + 4𝑥 + 5
=
𝑑𝑥
−2
𝑥 2 + 4𝑥 + 4 + 1
+3
=
𝑑𝑥
𝑥 2 + 4𝑥 + 5 2
𝑥 2 + 4𝑥 + 5 −2 2𝑥 + 4 𝑑𝑥
𝑑𝑥
𝑥 2 + 4𝑥 + 4 + 1 2
𝑑𝑥
−2
𝑥+2 2+1
+3
2𝑥 + 4 𝑑𝑥
+3
𝑥 2 + 4𝑥 + 5 2
𝑥 2 + 4𝑥 + 5 −2 2𝑥 + 4 𝑑𝑥
𝑑𝑥
𝑥+2 2+1 2
𝑥 2 + 4𝑥 + 5 −1
𝑥2
+3
𝑑𝑥 = 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2 − 2
2
2
−1
𝑥 + 4𝑥 + 5
We integrate
𝑑𝑥
𝑥+2 2+1 2
𝑑𝑥
separately
𝑥+2 2 +1 2
let 𝑢 = 𝑥 + 2
𝑑𝑥
=
𝑥+2 2+1 2
𝑑𝑢 = 𝑑𝑥
𝑑𝑢
𝑢2 + 1 2
𝑑𝑢 = 𝑠𝑒𝑐 2 𝑧 𝑑𝑧
Let: 𝑢 = tan 𝑧
𝑠𝑒𝑐 2 𝑧 𝑑𝑧
=
𝑡𝑎𝑛2 𝑧 + 1 2
=
𝑠𝑒𝑐 2 𝑧 𝑑𝑧
=
𝑠𝑒𝑐 2 𝑧 2
𝑑𝑧
=
𝑠𝑒𝑐 2 𝑧
𝑐𝑜𝑠 2 𝑧 𝑑𝑧
But 𝑐𝑜𝑠 2 𝑧 = 12 1 + cos 2𝑧
1
=
2
=
=
1
2
1
1 + cos 2𝑧 𝑑𝑧 =
2
𝑧+
1 1
2 2
sin 2𝑧 =
1
2
1
𝑑𝑧 +
2
cos 2𝑧 𝑑𝑧
1
𝑧 + sin 2𝑧 + 𝐶, but 𝑧 = 𝐴𝑟𝑐𝑡𝑎𝑛 𝑢
4
1
1
𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2 + sin 2 𝐴𝑟𝑐𝑡𝑎𝑛 (𝑥 + 2) + 𝐶
2
4
The integral therefore is,
𝑥 2 + 4𝑥 + 5 −1
𝑥2
𝑑𝑥 = 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2 − 2
−1
𝑥 2 + 4𝑥 + 5 2
1
1
+ 3 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2 + sin 2 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2 + 𝐶
2
4
2
3
= 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2 + 2
+ 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2
𝑥 + 4𝑥 + 5
2
3
+ sin 2 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 + 2 + 𝐶
4
Example 2: Evaluate
𝑥 2 +𝑥+2
𝑑𝑥 .
𝑥 2 +2𝑥+3 2
Solution:
Decompose the integrand
𝐴 2𝑥 + 2 + 𝐵 𝐶 2𝑥 + 2 + 𝐷
𝑥2 + 𝑥 + 2
= 2
+ 2
2
2
𝑥
+
2𝑥
+
3
𝑥 + 2𝑥 + 3 2
𝑥 + 2𝑥 + 3
𝑥 2 + 𝑥 + 2 = 𝐴 2𝑥 + 2 𝑥 2 + 2𝑥 + 3 + 𝐵 𝑥 2 + 2𝑥 + 3 + 𝐶 2𝑥 + 2 + 𝐷
𝑥 2 + 𝑥 + 2 = 𝐴 2𝑥 3 + 6𝑥 2 + 10𝑥 + 6 + 𝐵 𝑥 2 + 4𝑥 + 5 + 𝐶 2𝑥 + 4 + 𝐷
Solve for A, B, C and D
equate coefficients of like powers
coefficients of 𝑥 3
0 = 2𝐴
(1)
coefficients of 𝑥 2
1 = 6𝐴 + 𝐵
(2)
coefficients of 𝑥
1 = 10𝐴 + 4𝐵 + 2𝐶
(3)
coefficients of 𝑥 0
2 = 6𝐴 + 5𝐵 + 4𝐶 + 𝐷
(4)
From the 4 equations above, we have
3
𝐴 = 0,
𝐵 = 1,
𝐶 = − 2, and
𝐷=3
Substitute values of A, B, C and D then integrate
3
− 2 2𝑥 + 2 + 3
𝑥2 + 𝑥 + 2
𝑑𝑥 =
𝑥 2 + 2𝑥 + 3 2
0 2𝑥 + 2 + 1
+
𝑑𝑥
2
2
2
𝑥 + 2𝑥 + 3
𝑥 + 2𝑥 + 3
=
𝑑𝑥
3
−
𝑥 2 + 2𝑥 + 3
2
=
𝑑𝑥
3
−
𝑥 2 + 2𝑥 + 1 + 2 2
+3
2𝑥 + 2 𝑑𝑥
+3
2
2
𝑥 + 2𝑥 + 3
𝑑𝑥
𝑥 2 + 2𝑥 + 3 2
𝑥 2 + 2𝑥 + 3 −2 2𝑥 + 2 𝑑𝑥
𝑑𝑥
𝑥 2 + 2𝑥 + 1 + 2 2
=
𝑑𝑥
3
−
𝑥+1 2+2 2
+3
𝑑𝑥
𝑥+1 2+2 2
𝑥 2 + 2𝑥 + 3 −2 2𝑥 + 2 𝑑𝑥
Following the solution as example 1 above we have,
1
𝑥+1
3 𝑥 2 + 2𝑥 + 3 −1
𝑥2 + 𝑥 + 2
𝐴𝑟𝑐𝑡𝑎𝑛
−
𝑑𝑥 =
2
−1
𝑥 2 + 2𝑥 + 3 2
2
2
3
𝑥+1
1
𝑥+1
+
𝐴𝑟𝑐𝑡𝑎𝑛
+ sin 2 𝐴𝑟𝑐𝑡𝑎𝑛
2
4 2
2
2
+𝐶
Now try these
Practice Problem
1.
𝑑𝑥
𝑥 2 + 2𝑥 + 10 2
Ans.
𝑥+1
1
𝑥+1
+ 54 Arctan 3
18 𝑥 2 +2𝑥+10
+𝐶