Lecture Notes Partial Fractions page 1
Sample Problems
Compute each of the following integrals.
1.
Z
1
x
2
4
dx
2.
Z
2x
(x + 3) (3x + 1)
dx
3.
Z
x + 5
x
2
2x 3
dx
4.
Z
x
4
+ x
3
5x
2
+ 26x 21
x
2
+ 3x 4
dx
5.
Z
x
2
+ x 3
(x + 1) (x 2) (x 5)
dx
6.
Z
2x 1
(x 5)
2
dx
7.
Z
x + 3
(x 1)
3
dx
8.
Z
x
4
x
4
1
dx
9.
Z
sec x dx
10.
Z
csch x dx
Practice Problems
1.
Z
1
x
2
+ 3x
dx
2.
Z
x 5
x
2
2x 8
dx
3.
Z
1
x
2
a
2
dx
4.
Z
x 1
x
2
4
dx
5.
Z
x 1
x
2
+ 4
dx
6.
Z
x
2
x
2
+ 2x 3
dx
7.
Z
2x
3
x
2
10x 4
x
2
4
dx
8.
Z
5x 17
x
2
6x + 9
dx
9.
Z
2x
2
+ 7x + 3
x
2
+ 1
dx
10.
Z
2x
2
x + 20
(x 2) (x
2
+ 9)
dx
11.
Z
x
4
x
4
16
dx
12.
Z
2x + 1
x
2
+ 1
dx
13.
Z
x
2
+ 2
x (x
2
+ 6)
dx
14.
Z
x + 6
(x + 3)
2
dx
15.
Z
2x 3
x
2
+ 9
dx
16.
Z
x
2
+ 2x 1
x
3
x
dx
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 2
Sample Problems - Answers
1.)
1
4
ln jx 2j
1
4
ln jx + 2j + C 2.)
3
4
ln jx + 3j
1
12
ln j3x + 1j + C 3.) 2 ln jx 3j ln jx + 1j + C
4.)
x
3
3
x
2
+ 5x +
13
5
ln jx + 4j +
2
5
ln jx 1j + C 5.)
1
6
ln jx + 1j
1
3
ln jx 2j +
2
3
ln jx 5j + C
6.) 2 ln jx 5j
9
x 5
+ C 7.)
1
x 1
2
(x 1)
2
+ C =
x 1
(x 1)
2
+ C
8.) x
1
2
arctan x
1
4
ln jx + 1j +
1
4
ln jx 1j + C 9.) ln jsec x + tan xj + C = ln jsec x tan xj + C
10.) ln je
x
1j ln (e
x
+ 1) + C
Practice Problems - Answers
1.)
1
3
ln jxj
1
3
ln jx + 3j + C 2.)
7
6
ln jx + 2j
1
6
ln jx 4j + C 3.)
1
2a
ln jx aj
1
2a
ln jx + aj + C
4.)
1
4
ln jx 2j +
3
4
ln jx + 2j + C 5.)
1
2
ln
x
2
+ 4
1
2
tan
1
1
2
x + C
6.) x +
1
4
ln jx 1j
9
4
ln jx + 3j + C 7.) x
2
x 3 ln jx 2j + ln jx + 2j + C
8.) 5 ln jx 3j +
2
x 3
+ C 9.) 2x +
7
2
ln
x
2
+ 1
+ tan
1
x + C
10.) 2 ln jx 2j
1
3
tan
1
x
3
+ C 11.) x +
1
2
ln jx 2j
1
2
ln jx + 2j tan
1
x
2
+ C
12.) tan
1
x + ln
x
2
+ 1
+ C 13.)
1
3
ln
x
3
+ 6x
+ C 14.) ln jx + 3j
9
x + 3
+ C
15.) ln
x
2
+ 9
tan
1
x
3
+ C 16.) ln jxj + ln jx 1j ln jx + 1j + C
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 3
Sample Problems - Solutions
Compute each of the following integrals.
1.
Z
1
x
2
4
dx
Solution: We factor the denominator: x
2
4 = (x + 2) (x 2). Next, we re-write the fraction
1
x
2
4
as a
sum (or di¤erence) of fractions with denominators x + 2 and x 2. This means that we n eed to solve for A
and B in the equation
A
x + 2
+
B
x 2
=
1
x
2
4
To simplify the left-hand side, we bring the fractions to the common denominator:
A (x 2)
(x + 2) (x 2)
+
B (x + 2)
(x 2) (x + 2)
=
Ax 2A + Bx + 2B
x
2
4
=
(A + B) x 2A + 2B
x
2
4
Thus we have
(A + B) x 2A + 2B
x
2
4
=
1
x
2
4
We clear the denominators by multiplication
(A + B) x 2A + 2B = 1
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. In other words,
(A + B) x 2A + 2B = 0x + 1
This gives us an equation for each coe¢ cient, forming a system of linear equations:
A + B = 0
2A + 2B = 1
We solve this system and obtain A =
1
4
and B =
1
4
.
So our fraction,
1
x
2
4
can be re-written as
1
4
x + 2
+
1
4
x 2
. We check:
1
4
x + 2
+
1
4
x 2
=
1
4
(x 2)
(x + 2) (x 2)
+
1
4
(x + 2)
(x 2) (x + 2)
=
1
4
(x 2) +
1
4
(x + 2)
(x + 2) (x 2)
=
1
4
x +
1
2
+
1
4
x +
1
2
(x + 2) (x 2)
=
1
x
2
4
Now we can easily integrate:
Z
1
x
2
4
dx =
Z
1
4
x + 2
+
1
4
x 2
dx =
1
4
Z
1
x + 2
dx +
1
4
Z
1
x 2
dx =
1
4
ln jx + 2j +
1
4
ln jx 2j + C
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 4
Method 2: The values of A and B can be found using a slightly di¤erent method as follows. Consider …rst
the equation
A
x + 2
+
B
x 2
=
1
x
2
4
We bring the fractions to the common denominator:
A (x 2)
(x + 2) (x 2)
+
B (x + 2)
(x 2) (x + 2)
=
1
x
2
4
and then multiply both sides by the denominator:
A (x 2) + B (x + 2) = 1
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x = 2 into both sides:
A (0) + B (4) = 1
B =
1
4
Let us substitute x = 2 into both sides:
A (4) + B (0) = 1
A =
1
4
and so A =
1
4
and B =
1
4
.
2.
Z
2x
(x + 3) (3x + 1)
dx
Solution: We re-write the fraction
2x
(x + 3) (3x + 1)
as a sum (or di¤erence) of fractions with denominators
x + 3 and 3x + 1. This means that we need to solve for A and B in the equation
A
x + 3
+
B
3x + 1
=
2x
(x + 3) (3x + 1)
To simplify the left-hand side, we bring the fractions to the common denominator:
A
x + 3
+
B
3x + 1
=
A (3x + 1)
(x + 3) (3x + 1)
+
B (x + 3)
(x + 3) (3x + 1)
=
A (3x + 1) + B (x + 3)
(x + 3) (3x + 1)
=
3Ax + A + Bx + 3B
(x + 3) (3x + 1)
=
(3A + B) x + A + 3B
(x + 3) (3x + 1)
Thus we have
(3A + B) x + A + 3B
(x + 3) (3x + 1)
=
2x
(x + 3) (3x + 1)
We clear the denominators by multiplication
(3A + B) x + A + 3B = 2x
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. In other words,
(3A + B) x + A + 3B = 2x + 0
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 5
This gives us an equation for each coe¢ cient, forming a system of linear equations:
3A + B = 2
A + 3B = 0
We solve this system and obtain A =
3
4
and B =
1
4
.
So our fraction,
2x
(x + 3) (3x + 1)
can be re-written as
3
4
x + 3
+
1
4
3x + 1
. We check:
3
4
x + 3
+
1
4
3x + 1
=
3
4
(3x + 1)
(x + 3) (3x + 1)
+
1
4
(x + 3)
(x + 3) (3x + 1)
=
3
4
(3x + 1)
1
4
(x + 3)
(x + 3) (3x + 1)
=
9
4
x +
3
4
1
4
x
3
4
(x + 3) (3x + 1)
=
2x
(x + 3) (3x + 1)
Now we can easily integrate:
Z
2x
(x + 3) (3x + 1)
dx =
Z
3
4
x + 3
+
1
4
3x + 1
dx =
3
4
Z
1
x + 3
dx
1
4
Z
1
3x + 1
dx =
3
4
ln jx + 3j
1
12
ln j3x + 1j + C
The second integral can be computed using the substitution u = 3x + 1.
Method 2: The values of A and B can be found using a slightly di¤erent me thod as follows. Consider …rst
the equation
A
x + 3
+
B
3x + 1
=
2x
(x + 3) (3x + 1)
We bring the fractions to the common denominator:
A (3x + 1)
(x + 3) (3x + 1)
+
B (x + 3)
(x + 3) (3x + 1)
=
2x
(x + 3) (3x + 1)
and then multiply both sides by the denominator:
A (3x + 1) + B (x + 3) = 2x
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x =
1
3
into both sides:
A (0) + B
1
3
+ 3
= 2
1
3
8
3
B =
2
3
B =
1
4
Let us substitute x = 3 into both sides:
A (3 (3) + 1) + B (3 + 3) = 2 (3)
8A + B (0) = 6
8A = 6
A =
3
4
and so A =
3
4
and B =
1
4
.
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 6
3.
Z
x + 5
x
2
2x 3
dx
Solution: We factor the denominator: x
2
2x3 = (x + 1) (x 3). Next, we re-write the fraction
x + 5
x
2
2x 3
as a sum (or di¤erence) of fractions with denominators x + 1 and x 3. This means that we need to solve
for A and B in th e equation
A
x + 1
+
B
x 3
=
x + 5
x
2
2x 3
To simplify the left-hand side, we bring the fractions to the common denominator:
A (x 3)
(x + 1) (x 3)
+
B (x + 1)
(x 3) (x + 1)
=
Ax 3A + Bx + B
x
2
2x 3
=
(A + B) x 3A + B
x
2
2x 3
Thus
(A + B) x 3A + B
x
2
2x 3
=
x + 5
x
2
2x 3
We clear the denominators by multiplication
(A + B) x 3A + B = x + 5
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. This gives us an equation for each coe¢ cient, forming a system of linear
equations:
A + B = 1
3A + B = 5
We solve the system and obtain A = 1 and B = 2.
So we have that our fraction,
x + 5
x
2
2x 3
can be re-written as
1
x + 1
+
2
x 3
. We check:
1
x + 1
+
2
x 3
=
1 (x 3)
(x + 1) (x 3)
+
2 (x + 1)
(x 3) (x + 1)
=
(x 3) + 2 (x + 1)
(x + 1) (x 3)
=
x + 3 + 2x + 2
x
2
2x 3
=
x + 5
x
2
2x 3
Now we can easily integrate:
Z
x + 5
x
2
2x 3
dx =
Z
1
x + 1
+
2
x 3
dx =
Z
1
x + 1
dx + 2
Z
1
x 3
dx = ln jx + 1j + 2 ln jx 3j + C
Method 2: The values of A and B can be found using a slightly di¤erent me thod as follows. Consider …rst
the equation
A
x + 1
+
B
x 3
=
x + 5
x
2
2x 3
We bring the fractions to the common denominator:
A (x 3)
(x 3) (x + 1)
+
B (x + 1)
(x 3) (x + 1)
=
x + 5
x
2
2x 3
and then multiply both sides by the denominator:
A (x 3) + B (x + 1) = x + 5
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x = 3 into both sides:
A (0) + B (4) = 3 + 5
4B = 8
B = 2
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 7
Let us substitute x = 1 into both sides:
A (4) + B (0) = 1 + 5
4A = 4
A = 1
and so A = 1 and B = 2.
4.
Z
x
4
+ x
3
5x
2
+ 26x 21
x
2
+ 3x 4
dx
Solution: This rational function is an improper fraction since the numerator has a higher degree than the
denominator. We …rst perform long division. This process is similar to long division among numbers. For
example, to simplify
38
7
; we perform the long division 38 7 = 5 R 3 which is the same thing as to say that
38
7
= 5
3
7
. The division:
x
2
2x + 5
x
2
+ 3x 4 ) x
4
+ x
3
5x
2
+ 26x 21
x
4
3x
3
+ 4x
2
2x
3
x
2
+ 26x 21
2x
3
+ 6x
2
8x
5x
2
+ 18x 21
5x
2
15x + 20
3x 1
Step 1:
x
4
x
2
= x
2
x
2
x
2
+ 3x 4
= x
4
+ 3x
3
4x
2
x
4
+ 3x
3
4x
2
= x
4
3x
3
+ 4x
2
We add that to the original polynomial shown above.
Step 2:
2x
3
x
2
= 2x
2x
x
2
+ 3x 4
= 2x
3
6x
2
+ 8x
1
2x
3
6x
2
+ 8x
= 2x
3
+ 6x
2
8x
We add that to the original polynomial shown above.
Step 3:
5x
2
x
2
= 5
5
x
2
+ 3x 4
= 5x
2
+ 15x 20
1
5x
2
+ 15x 20
= 5x
2
15x + 20
We add that to the original polynomial shown above.
The result of this computation is that
x
4
+ x
3
5x
2
+ 26x 21
x
2
+ 3x 4
= x
2
2x + 5 +
3x 1
x
2
+ 3x 4
very much like
38
7
= 5 +
3
7
. Thus
Z
x
4
+ x
3
5x
2
+ 26x 21
x
2
+ 3x 4
dx =
Z
x
2
2x + 5 +
3x 1
x
2
+ 3x 4
dx =
Z
x
2
2x + 5 dx +
Z
3x 1
x
2
+ 3x 4
dx
=
x
3
3
x
2
+ 5x + C
1
+
Z
3x 1
x
2
+ 3x 4
dx
We apply the method of partial fractions to compute
Z
3x 1
x
2
+ 3x 4
dx.
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 8
We factor the denominator: x
2
+ 3x 4 = (x + 4) (x 1). Next, we re-write the fraction
3x 1
x
2
+ 3x 4
as a
sum (or di¤erence) of fractions with denominators x + 4 and x 1. This means that we n eed to solve for A
and B in the equation
A
x + 4
+
B
x 1
=
3x 1
x
2
+ 3x 4
To simplify the left-hand side, we bring the fractions to the common denominator:
A (x 1)
(x + 4) (x 1)
+
B (x + 4)
(x + 4) (x 1)
=
Ax A + Bx + 4B
x
2
+ 3x 4
=
(A + B) x A + 4B
x
2
+ 3x 4
Thus
(A + B) x A + 4B
x
2
+ 3x 4
=
3x 1
x
2
+ 3x 4
We clear the denominators by multiplication
(A + B) x A + 4B = 3x 1
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. This gives us an equation for each coe¢ cient that forms a system of linear
equations:
A + B = 3
A + 4B = 1
We solve the system and obtain A =
13
5
and B =
2
5
.
So our fraction,
3x 1
x
2
+ 3x 4
can be re-written as
13
5
x + 4
+
2
5
x 1
. We check:
13
5
x + 4
+
2
5
x 1
=
13
5
(x 1)
(x + 1) (x 4)
+
2
5
(x + 4)
(x 4) (x + 1)
=
13
5
(x 1) +
2
5
(x + 4)
(x + 1) (x 4)
=
13
5
x
13
5
+
2
5
x +
8
5
(x + 1) (x 4)
=
15
5
x
5
5
x
2
+ 3x 4
=
3x 1
x
2
+ 3x 4
Now we can easily integrate:
Z
3x 1
x
2
+ 3x 4
dx =
Z
13
5
x + 4
+
2
5
x 1
dx =
Z
13
5
x + 4
dx +
Z
2
5
x 1
dx
=
13
5
Z
1
x + 4
dx +
2
5
Z
1
x 1
dx =
13
5
ln jx + 4j +
2
5
ln jx 1j + C
Thus the …nal answer is
Z
x
4
+ x
3
5x
2
+ 26x 21
x
2
+ 3x 4
dx =
=
x
3
3
x
2
+ 5x + C
1
+
Z
3x 1
x
2
+ 3x 4
dx =
x
3
3
x
2
+ 5x + C
1
+
13
5
ln jx + 4j +
2
5
ln jx 1j + C
2
=
x
3
3
x
2
+ 5x +
13
5
ln jx + 4j +
2
5
ln jx 1j + C
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 9
Method 2: The values of A and B can be found using a slightly di¤erent me thod as follows. Consider …rst
the equation
A
x + 4
+
B
x 1
=
3x 1
x
2
+ 3x 4
We bring the fractions to the common denominator:
A (x 1)
(x + 4) (x 1)
+
B (x + 4)
(x + 4) (x 1)
=
3x 1
(x + 1) (x 4)
and then multiply both sides by the denominator:
A (x 1) + B (x + 4) = 3x 1
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x = 1 into both sides:
A (1 1) + B (1 + 4) = 3x 1
A 0 + B 5 = 3 1 1
5B = 2
B =
2
5
Let us substitute x = 4 into both sides:
A (4 1) + B (4 + 4) = 3 (4) 1
5A = 13
A =
13
5
and so A =
4
5
and B =
11
5
.
5.
Z
x
2
+ x 3
(x + 1) (x 2) (x 5)
dx
Solution: We re-write the fraction
x
2
+ x 3
(x + 1) (x 2) (x 5)
as a sum (or di¤erence) of fractions with denomi-
nators x + 1, x 2 and x 5. This means that we need to solve for A , B, and C in the equation
A
x + 1
+
B
x 2
+
C
x 5
=
x
2
+ x 3
(x + 1) (x 2) (x 5)
To simplify the left-hand side, we bring the fractions to the common denominator:
A
x + 1
+
B
x 2
+
C
x 5
=
A (x 2) (x 5)
(x + 1) (x 2) (x 5)
+
B (x + 1) (x 5)
(x + 1) (x 2) (x 5)
+
C (x + 1) (x 2)
(x + 1) (x 2) (x 5)
=
A
x
2
7x + 10
(x + 1) (x 2) (x 5)
+
B
x
2
4x 5
(x + 1) (x 2) (x 5)
+
C
x
2
x 2
(x + 1) (x 2) (x 5)
=
A
x
2
7x + 10
+ B
x
2
4x 5
+ C
x
2
x 2
(x + 1) (x 2) (x 5)
=
Ax
2
7Ax + 10A + Bx
2
4Bx 5B + Cx
2
Cx 2C
(x + 1) (x 2) (x 5)
=
(A + B + C) x
2
+ (7A 4B C) x + 10A 5B 2C
(x + 1) (x 2) (x 5)
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 10
Thus
(A + B + C) x
2
+ (7A 4B C) x + 10A 5B 2C
(x + 1) (x 2) (x 5)
=
x
2
+ x 3
(x + 1) (x 2) (x 5)
We clear the denominators by multiplication
(A + B + C) x
2
+ (7A 4B C) x + 10A 5B 2C = x
2
+ x 3
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. We have an equation for each coe¢ cient that gives us a system of linear
equations:
A + B + C = 1
7A 4B C = 1
10A 5B 2C = 3
We solve the system by elimination: …rst we will eliminate C from the second and third equations. To
eliminate C from the second equation, we simply add the …rst and sec ond equations.
A + B + C = 1
7A 4B C = 1
6A 3B = 2
To eliminate C from the third equation, we multiply the …rst equation by 2 and add that to the third equation.
2A + 2B + 2C = 2
10A 5B 2C = 3
12A 3B = 1
We now have a system of linear equations in two variables:
6A 3B = 2
12A 3B = 1
We will eliminate B by adding the opposite of the …rst equation to the second equation.
6A + 3B = 2
12A 3B = 1
18A = 3
A =
1
6
Using the equation 6A + 3B = 2 we can now solve for B.
6
1
6
+ 3B = 2
1 + 3B = 2
3B = 1
B =
1
3
Using the …rst equation, we can now solve for C.
A + B + C = 1
1
6
+
1
3
+ C = 1
1
2
+ C = 1
C =
3
2
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 11
Thus A =
1
6
, B =
1
3
, and C =
3
2
So we have that our fraction,
x
2
+ x 3
(x + 1) (x 2) (x 5)
can be re-written as
1
6
x + 1
+
1
3
x 2
+
3
2
x 5
. We check:
1
6
x + 1
+
1
3
x 2
+
3
2
x 5
=
1
6
(x 2) (x 5)
(x + 1) (x 2) (x 5)
+
1
3
(x + 1) (x 5)
(x + 1) (x 2) (x 5)
+
3
2
(x + 1) (x 2)
(x + 1) (x 2) (x 5)
=
1
6
(x 2) (x 5)
1
3
(x + 1) (x 5) +
3
2
(x + 1) (x 2)
(x + 1) (x 2) (x 5)
=
1
6
x
2
7x + 10
1
3
x
2
4x 5
+
3
2
x
2
x 2
(x + 1) (x 2) (x 5)
=
1
6
x
2
+
7
6
x
5
3
1
3
x
2
+
4
3
x +
5
3
+
3
2
x
2
3
2
x 3
(x + 1) (x 2) (x 5)
=
1
6
1
3
+
3
2
x
2
+
7
6
+
4
3
3
2
x
5
3
+
5
3
3
(x + 1) (x 2) (x 5)
=
x
2
+ x 3
(x + 1) (x 2) (x 5)
Now we can easily integrate:
Z
x
2
+ x 3
(x + 1) (x 2) (x 5)
dx =
Z
1
6
x + 1
+
1
3
x 2
+
3
2
x 5
dx
=
1
6
Z
1
x + 1
dx
1
3
Z
1
x 2
dx +
3
2
Z
1
x 5
dx
=
1
6
ln jx + 1j
1
3
ln jx 2j +
2
3
ln jx 5j + C
Method 2: The values of A; B, and C can be found using a slightly di¤erent method as follows. Consider
…rst the equation
A
x + 1
+
B
x 2
+
C
x 5
=
x
2
+ x 3
(x + 1) (x 2) (x 5)
We bring the fractions to the common denominator:
A (x 2) (x 5)
(x + 1) (x 2) (x 5)
+
B (x + 1) (x 5)
(x + 1) (x 2) (x 5)
+
C (x + 1) (x 2)
(x + 1) (x 2) (x 5)
=
x
2
+ x 3
(x + 1) (x 2) (x 5)
and then multiply both sides by the denominator:
A (x 2) (x 5) + B (x + 1) (x 5) + C (x + 1) (x 2) = x
2
+ x 3
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x = 2 into both sides:
A (2 2) (2 5) + B (2 + 1) (2 5) + C (2 + 1) (2 2) = 2
2
+ 2 3
0A 9B + 0C = 3
9B = 3
B =
1
3
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 12
Let us substitute x = 1 into both sides:
A (1 2) (1 5) + B (1 + 1) (1 5) + C (1 + 1) (1 2) = (1)
2
+ (1) 3
A (3) (6) + 0B + 0C = 3
18A = 3
A =
1
6
Let us substitute x = 5 into both sides:
A (5 2) (5 5) + B (5 + 1) (5 5) + C (5 + 1) (5 2) = 5
2
+ 5 3
A (0) + B (0) + C (6) (3) = 27
18C = 27
C =
3
2
and so A =
1
6
, B =
1
3
, and C =
2
3
.
6.
Z
2x 1
(x 5)
2
dx
Solution: We will re-w rite the fraction
2x 1
(x 5)
2
as a sum (or di¤erence) of fractions with denominators x 5
and (x 5)
2
. This means that we need to solve for A and B in th e equation
A
x 5
+
B
(x 5)
2
=
2x 1
(x 5)
2
To simplify the left-hand side, we bring the fractions to the common denominator:
A
x 5
+
B
(x 5)
2
=
A (x 5)
(x 5)
2
+
B
(x 5)
2
=
A (x 5) + B
(x 5)
2
=
Ax 5A + B
(x 5)
2
Thus we have
Ax 5A + B
(x 5)
2
=
2x 1
(x 5)
2
We clear the denominators by multiplication
Ax 5A + B = 2x 1
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. This gives us an equation for each coe¢ cient, forming a system of linear
equations:
A = 2
5A + B = 1
We solve this system and obtain A = 2 and B = 9.
So our fraction,
2x 1
(x 5)
2
can be re-written as
2
x 5
+
9
(x 5)
2
. We check:
2
x 5
+
9
(x 5)
2
=
2 (x 5)
(x 5)
2
+
9
(x 5)
2
=
2x 10 + 9
(x 5)
2
=
2x 1
(x 5)
2
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 13
Now we can easily integrate:
Z
2x 1
(x 5)
2
dx =
Z
2
x 5
+
9
(x 5)
2
dx = 2
Z
1
x 5
dx + 9
Z
1
(x 5)
2
dx = 2 ln jx 5j
9
x 5
+ C
Method 2: The values of A and B can be found using a slightly di¤erent me thod as follows. Consider …rst
the equation
A
x 5
+
B
(x 5)
2
=
2x 1
(x 5)
2
We bring the fractions to the common denominator:
A (x 5)
(x 5)
2
+
B
(x 5)
2
=
2x 1
(x 5)
2
and then multiply both sides by the denominator:
A (x 5) + B = 2x 1
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x = 5 into both sides:
A (0) + B = 9
B = 9
The other value of x can be arbitrarily chosen. (There is no value that would eliminate B from the equation.)
For easy substitution, let us substitute x = 0 into both sides and also substitute B = 9:
A (5) + 9 = 1
5A = 10
A = 2
and so A = 2 and B = 9.
7.
Z
x + 3
(x 1)
3
dx
Solution: We re-write the fraction
x + 3
(x 1)
3
as a sum (or di¤erence) of fractions with denominators x 1,
(x 1)
2
and (x 1)
3
. This means that we need to solve for A, B, and C in the equation
A
x 1
+
B
(x 1)
2
+
C
(x 1)
3
=
x + 3
(x 1)
3
To simplify the left-hand side, we bring the fractions to the common denominator:
A
x 1
+
B
(x 1)
2
+
C
(x 1)
3
=
A (x 1)
2
(x 1)
3
+
B (x 1)
(x 1)
3
+
C
(x 1)
3
=
A (x 1)
2
+ B (x 1) + C
(x 1)
3
=
A
x
2
2x + 1
+ B (x 1) + C
(x 1)
3
=
Ax
2
2Ax + A + Bx B + C
(x 1)
3
=
Ax
2
+ (2A + B) x + A B + C
(x 1)
3
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 14
Thus
Ax
2
+ (2A + B) x + A B + C
(x 1)
3
=
x + 3
(x 1)
3
We clear the denominators by multiplication
Ax
2
+ (2A + B) x + A B + C = x + 3
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. We have an equation for each coe¢ cient that gives us a system of linear
equations:
A = 0
2A + B = 1
A B + C = 3
Since A = 0; this is really a system in two variables:
B = 1
B + C = 3
We solve this system and obtain B = 1 and C = 4.
So our fraction,
x + 3
(x 1)
3
can be re-written as
1
(x 1)
2
+
4
(x 1)
3
. We check:
1
(x 1)
2
+
4
(x 1)
3
=
1 (x 1)
(x 1)
3
+
4
(x 1)
3
=
x 1 + 4
(x 1)
3
=
x + 3
(x 1)
3
Now we can easily integrate:
Z
x + 3
(x 1)
3
dx =
Z
1
(x 1)
2
+
4
(x 1)
3
dx =
Z
1
(x 1)
2
dx + 4
Z
1
(x 1)
3
dx
=
1
x 1
4
2
1
(x 1)
2
+ C =
1
x 1
2
(x 1)
2
+ C
=
1 (x 1)
(x 1)
2
2
(x 1)
2
+ C =
x + 1 2
(x 1)
2
+ C =
x 1
(x 1)
2
+ C
Both …nal answers are acceptable.
Method 2: The values of A; B, and C can be found using a slightly di¤erent method as follows. Consider
…rst the equation
A
x 1
+
B
(x 1)
2
+
C
(x 1)
3
=
x + 3
(x 1)
3
We bring the fractions to the common denominator:
A (x 1)
2
(x 1)
3
+
B (x 1)
(x 1)
3
+
C
(x 1)
3
=
x + 3
(x 1)
3
and then multiply both sides by the denominator:
A (x 1)
2
+ B (x 1) + C = x + 3
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x = 1 into both sides:
A (0) + B (0) + C = 1 + 3
C = 4
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 15
There is no value other than 1 that would eliminate A or B from the equation. Our method will still work.
For easy substitution, let us substitute x = 0 into both sides and also substitute C = 4:
A (x 1)
2
+ B (x 1) + C = x + 3
A (0 1)
2
+ B (0 1) + 4 = 0 + 3
A B + 4 = 3
A B = 1
Let us substitute x = 2 into both sides:
A (x 1)
2
+ B (x 1) + C = x + 3
A (2 1)
2
+ B (2 1) + 4 = 2 + 3
A + B + 4 = 5
A + B = 1
We now solve the system of equations
A B = 1
A + B = 1
and obtain A = 0 and B = 1. Recall that we already have C = 4.
8.
Z
x
4
x
4
1
dx
Solution: This rational function is an improper fraction since the numerator has the same degree as the
denominator. We …rst perform long division. This one is an easy one; the method featured below is called
smuggling.
x
4
x
4
1
=
x
4
1 + 1
x
4
1
=
x
4
1
x
4
1
+
1
x
4
1
= 1 +
1
x
4
1
Thus
Z
x
4
x
4
1
dx =
Z
1 +
1
x
4
1
dx =
Z
1 dx +
Z
1
x
4
1
dx = x + C
1
+
Z
1
x
4
1
dx
We apply the method of partial fractions to compute
Z
1
x
4
1
dx.
We factor the d enomin ator: x
4
1 =
x
2
+ 1
(x + 1) (x 1). Next, we re-write the fraction
1
x
4
1
as a sum
(or di¤erence) of fractions with denominators x
2
+ 1 and x + 1, and x 1. In the fraction with quadratic
denominator, the numerator is linear. This means that we need to solve for A and B in the equation
Ax + B
x
2
+ 1
+
C
x + 1
+
D
x 1
=
1
x
4
1
To simplify the left-hand side, we bring the fractions to the common denominator:
Ax + B
x
2
+ 1
+
C
x + 1
+
D
x 1
=
(Ax + B) (x + 1) (x 1)
(x
2
+ 1) (x + 1) (x 1)
+
C
x
2
+ 1
(x 1)
(x
2
+ 1) (x + 1) (x 1)
+
D
x
2
+ 1
(x + 1)
(x
2
+ 1) (x + 1) (x 1)
=
(Ax + B) (x + 1) (x 1) + C
x
2
+ 1
(x 1) + D
x
2
+ 1
(x + 1)
x
4
1
=
(Ax + B)
x
2
1
+ C
x
3
x
2
+ x 1
+ D
x
3
+ x
2
+ x + 1
x
4
1
=
Ax
3
+ Bx
2
Ax B + Cx
3
Cx
2
+ Cx C + Dx
3
+ Dx
2
+ Dx + D
x
4
1
=
(A + C + D) x
3
+ (B C + D) x
2
+ (A + C + D) x B C + D
x
4
1
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 16
Thus
(A + C + D) x
3
+ (B C + D) x
2
+ (A + C + D) x B C + D
x
4
1
=
1
x
4
1
We clear the denominators by multiplication
(A + C + D) x
3
+ (B C + D) x
2
+ (A + C + D) x B C + D = 1
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. This gives us an equation for each coe¢ cient that forms a system of linear
equations:
A + C + D = 0
B C + D = 0
A + C + D = 0
B C + D = 1
We will solve this system by elimination. First, we will eliminate A using the …rst equation. The second
and fourth equations do not have A in them, so there is nothing to do there. To eliminate A from the third
equation, we add the …rst one to it.
B C + D = 0
2C + 2D = 0
B C + D = 1
We now have three equations with three unknowns. We will use the …rst equation to eliminate B. In case
of the second equation, again, there is nothing to do. We add the …rst equation to the third one to eliminate
B.
2C + 2D = 0
2C + 2D = 1
Adding the two equations eliminates C and gives us 4D = 1 and so D =
1
4
. Next, we compute C using the
equation
2C + 2D = 0
2C + 2
1
4
= 0
2C =
1
2
C =
1
4
We can compute A using the …rst equation, A + C + D = 0
A + C + D = 0
A
1
4
+
1
4
= 0
A = 0
and we can compute B using the second equation,
B C + D = 0
B
1
4
+
1
4
= 0
B =
1
2
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 17
Thus A = 0; B =
1
2
; C =
1
4
; and D =
1
4
:
So our fraction,
1
x
4
1
can be re-written as
1
2
x
2
+ 1
1
4
x + 1
+
1
4
x 1
. We check:
1
2
x
2
+ 1
1
4
x + 1
+
1
4
x 1
=
1
2
(x + 1) (x 1)
(x
2
+ 1) (x + 1) (x 1)
1
4
x
2
+ 1
(x 1)
(x
2
+ 1) (x + 1) (x 1)
+
1
4
x
2
+ 1
(x + 1)
(x
2
+ 1) (x + 1) (x 1)
=
1
2
(x + 1) (x 1)
1
4
x
2
+ 1
(x 1) +
1
4
x
2
+ 1
(x + 1)
(x
2
+ 1) (x + 1) (x 1)
=
1
2
x
2
1
1
4
x
3
x
2
+ x 1
+
1
4
x
3
+ x
2
+ x + 1
x
4
1
1
2
x
2
+
1
2
1
4
x
3
+
1
4
x
2
1
4
x +
1
4
+
1
4
x
3
+
1
4
x
2
+
1
4
x +
1
4
x
4
1
=
1
4
+
1
4
x
3
+
1
2
+
1
4
+
1
4
x
2
+
1
4
+
1
4
x +
1
2
+
1
4
+
1
4
x
4
1
=
1
x
4
1
Now we re-write the integral:
Z
1
x
4
1
dx =
Z
1
2
x
2
+ 1
1
4
x + 1
+
1
4
x 1
dx =
1
2
Z
1
x
2
+ 1
dx
1
4
Z
1
x + 1
dx +
1
4
Z
1
x 1
dx
=
1
2
arctan x
1
4
ln jx + 1j +
1
4
ln jx 1j + C
Thus the …nal answer is
Z
x
4
x
4
1
dx =
Z
1 +
x
4
x
4
1
dx =
Z
1 dx +
Z
1
x
4
1
dx
= x + C
1
1
2
arctan x
1
4
ln jx + 1j +
1
4
ln jx 1j + C
2
= x
1
2
arctan x
1
4
ln jx + 1j +
1
4
ln jx 1j + C
Method 2: The values of A; B; C; and D can be found using a slightly di¤erent method as follows. Consider
…rst the equation
Ax + B
x
2
+ 1
+
C
x + 1
+
D
x 1
=
1
x
4
1
We bring the fractions to the common denominator:
(Ax + B) (x + 1) (x 1)
(x
2
+ 1) (x + 1) (x 1)
+
C
x
2
+ 1
(x 1)
(x
2
+ 1) (x + 1) (x 1)
+
D
x
2
+ 1
(x + 1)
(x
2
+ 1) (x + 1) (x 1)
=
1
(x
2
+ 1) (x + 1) (x 1)
and then multiply both sides by the denominator:
(Ax + B) (x + 1) (x 1) + C
x
2
+ 1
(x 1) + D
x
2
+ 1
(x + 1) = 1
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 18
The equation above is about two functions; the two sides must be e qual for all values of x. Let u s substitute
x = 1 into both sides:
(Ax + B) (x + 1) (x 1) + C
x
2
+ 1
(x 1) + D
x
2
+ 1
(x + 1) = 1
(A (1) + B) ((1) + 1) ((1) 1) + C
(1)
2
+ 1
((1) 1) + D
(1)
2
+ 1
((1) + 1) = 1
(A + B) (0) (2) + C (2) (2) + D (2) (0) = 1
4C = 1
C =
1
4
Let us substitute x = 1 into both sides:
(Ax + B) (x + 1) (x 1) + C
x
2
+ 1
(x 1) + D
x
2
+ 1
(x + 1) = 1
(A (1) + B) ((1) + 1) (1 1) + C
1
2
+ 1
(1 1) + D
1
2
+ 1
(1 + 1) = 1
(A + B) (2) (0) + C (2) (0) + D (2) (2) = 1
4D = 1
D =
1
4
Let us substitute x = 0 into both sides and also C =
1
4
and D =
1
4
:
(Ax + B) (x + 1) (x 1) + C
x
2
+ 1
(x 1) + D
x
2
+ 1
(x + 1) = 1
(A (0) + B) ((0) + 1) (0 1) + C
0
2
+ 1
(0 1) + D
0
2
+ 1
(0 + 1) = 1
(B) (1) (1) + C (1) (1) + D (1) (1) = 1
B C + D = 1
B
1
4
+
1
4
= 1
B +
1
2
= 1
B =
1
2
B =
1
2
Let us substitute x = 2 into both sides and also B =
1
2
, C =
1
4
and D =
1
4
:
(Ax + B) (x + 1) (x 1) + C
x
2
+ 1
(x 1) + D
x
2
+ 1
(x + 1) = 1
(A (2) + B) ((2) + 1) (2 1) + C
2
2
+ 1
(2 1) + D
2
2
+ 1
(2 + 1) = 1
(2A + B) (3) (1) + C (5) (1) + D (5) (3) = 1
3 (2A + B) + 5C + 15D = 1
6A + 3B + 5C + 15D = 1
6A + 3
1
2
+ 5
1
4
+ 15
1
4
= 1
6A
3
2
5
4
+
15
4
= 1
6A +
6 5 + 15
4
= 1
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 19
6A +
4
4
= 1
6A + 1 = 1
6A = 0
A = 0
and so A = 0, B =
1
2
, C =
1
4
, and D =
1
4
.
9.
Z
sec x dx =
1
2
ln
1 + sin x
1 sin x
+ C = ln jsec x + tan xj + C = ln jsec x tan xj + C
Solution:
Z
sec x dx =
Z
1
cos x
dx =
Z
1
cos x
cos x
cos x
dx =
Z
cos x
cos
2
x
dx =
Z
cos x
1 sin
2
x
dx
Now let u = sin x. Then du = cos xdx.
Z
cos x
1 sin
2
x
dx =
Z
1
1 sin
2
x
cos xdx =
Z
1
1 u
2
du =
Z
1
(1 u) (1 + u)
du
This integral can be computed by partial fractions:
A
1 u
+
B
1 + u
=
1
1 u
2
The left-hand side can be re-written
A
1 u
+
B
1 + u
=
A (1 + u)
(1 u) (1 + u)
+
B (1 u)
(1 + u) (1 u)
=
Au + A Bu + B
1 u
2
=
(A B) u + A + B
1 u
2
So we have
(A B) u + A + B
1 u
2
=
1
1 u
2
We clear the denominators by multiplication
(A B) u + A + B = 1
The equation above is about two polynomials: they are equal to each other as fun ctions and so they must be
identical, coe¢ cint by coe¢ cient. This gives us an equation for each coe¢ cient that forms a system of linear
equations:
A B = 0
A + B = 1
we solve this system and obtain A = B =
1
2
. Indeed,
1
2
1
1 + u
+
1
1 u
=
1
2
1 u + 1 + u
(1 u) (1 + u)
=
1
2
2
1 u
2
=
1
1 u
2
Now for the integral:
Z
1
1 u
2
du =
Z
1
(1 u) (1 + u)
du =
Z
1
2
1
1 u
+
1
1 + u
du =
1
2
Z
1
1 u
du +
1
2
Z
1
1 + u
du
=
1
2
ln j1 uj +
1
2
ln j1 + uj + C =
1
2
ln j1 + uj
1
2
ln j1 uj + C
=
1
2
ln j1 + sin xj
1
2
ln j1 sin xj + C
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 20
Note the sign in
Z
1
1 u
du = ln j1 uj + C is caused by the chain rule. Aslo note that the …nal answer
can be re-written in several forms:
1
2
ln j1 + sin xj
1
2
ln j1 sin xj =
=
1
2
ln
1 + sin x
1 sin x
=
1
2
ln
1 + sin x
1 sin x
1 + sin x
1 + sin x
=
1
2
ln
(1 + sin x)
2
1 sin
2
x
=
1
2
ln
(1 + sin x)
2
cos
2
x
= ln
(1 + sin x)
2
cos
2
x
!
1=2
= ln
s
(1 + sin x)
2
cos
2
x
= ln
1 + sin x
cos x
= ln
1
cos x
+
sin x
cos x
= ln jsec x + tan xj
So, our result can also be presented as ln jsec x + tan xj + C
Another form can be obtained as shown below.
1
2
ln j1 + sin xj
1
2
ln j1 sin xj =
=
1
2
ln
1 + sin x
1 sin x
=
1
2
ln
1 + sin x
1 sin x
1 sin x
1 sin x
=
1
2
ln
1 sin
2
x
(1 sin x)
2
=
1
2
ln
cos
2
x
(1 sin x)
2
= ln
cos
2
x
(1 sin x)
2
1=2
= ln
s
cos
2
x
(1 sin x)
2
= ln
cos x
1 sin x
= ln
1
1 sin x
cos x
= ln
1 sin x
cos x
1
= ln
1
cos x
sin x
cos x
1
= ln jsec x tan xj
So, our resu lt can also be presented as ln jsec x tan xj + C : Actually, there are more forms possible, but
we will stop here.
10.
Z
csch x dx
Solution: This is an interesting application of partial fractions.
Z
csch x dx =
Z
2
e
x
e
x
dx =
Z
2
e
x
1
e
x
dx
we now multiply both numerator and denominator by e
x
.
Z
2
e
x
1
e
x
dx =
Z
2e
x
(e
x
)
2
1
dx
We proc eed with a substitution: let u = e
x
. Then du = e
x
dx and so
Z
2
(e
x
)
2
1
(e
x
dx) =
Z
2
u
2
1
du
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013
Lecture Notes Partial Fractions page 21
This is now an integral we can easily compute via partial f ractions. We easily decompose
2
u
2
1
as
1
u 1
1
u + 1
Z
csch x dx =
Z
2
u
2
1
du =
Z
1
u 1
1
u + 1
du =
Z
1
u 1
du
Z
1
u + 1
du = ln ju 1j ln ju + 1j + C
= ln je
x
1j ln (e
x
+ 1) + C
For more d oc ume nts like this, visit our page at https://teaching.martahidegkuti.com and click on Lecture Notes.
E-mail questions or comments to [email protected].
c
copyright Hidegkuti, Powell, 2012 Last revised: February 24, 2013