Section 6.3 Partial Fractions
2010 Kiryl Tsishchanka
Partial Fractions
EXAMPLE 1: Find
Z
dx
.
x(x + 1)
Solution: We have
therefore
Z
1
1
1
= −
x(x + 1)
x x+1
dx
=
x(x + 1)
EXAMPLE 2: Find
Z
Z Z
1
1
+
1−x 1+x
dx
.
1 + x2
Solution: We have
EXAMPLE 4: Find
x +C
dx = ln |x| − ln |x + 1| + C = ln x + 1
1
1
1
=
=
2
1−x
(1 − x)(1 + x)
2
dx
1
=
2
1−x
2
EXAMPLE 3: Find
1
1
−
x x+1
dx
.
1 − x2
Solution: We have
therefore
Z
Z Z
Z
1
1
+
1−x 1+x
1
1 1 + x +C
dx = (− ln |1 − x| + ln |1 + x|) + C = ln 2
2
1 − x
dx
= tan−1 x + C
1 + x2
dx
.
1 + x4
Solution: We have
1
1
= √
4
1+x
4 2
√
2
=
8
therefore
Z
√ !
√
2x − 2 2
2x + 2 2
√
√
−
x2 + 2x + 1 x2 − 2x + 1

!
√
√
2x + 2
1
2x − 2

√
√
+ 
−
2
2
4
x + 2x + 1 x − 2x + 1
√
dx
2
=
ln
4
1+x
8
1
1
1
x+ √
2
2
1
+
2
1



1
+
2
1
x− √
+
2
2
! √
√
√
x2 + 2x + 1
2 −1 √
√
+
tan ( 2x + 1) + tan−1 ( 2x − 1) + C
4
x2 − 2x + 1

Section 6.3 Partial Fractions
2010 Kiryl Tsishchanka
METHOD: Consider a rational function
P (x)
Q(x)
where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions provided that
the degree of P is less than the degree of Q. Such a rational function is called proper. If f is improper,
then we must take the preliminary step of dividing Q into P (by long division) until a remainder R(x) is
obtained such that deg(R) < deg(Q). The division statement is
f (x) =
f (x) =
R(x)
P (x)
= S(x) +
Q(x)
Q(x)
where S and R are also polynomials.
The next step is to factor the denominator Q(x) as far as possible. It can be shown that any polynomial
Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors (of
the form ax2 + bx + c, where b2 − 4ac < 0). For instance,
x2 − 1 = (x − 1)(x + 1)
10x2 − 11x − 6 = (2x − 3)(5x + 2)
x2 + 1, x2 + 2x + 5, etc. are irreducible quadratic factors
x3 + 1 = (x + 1)(x2 − x + 1)
x3 + 5 = (x +
√
√
√
3
5)(x2 − 3 5x + 3 25)
x3 + x2 + x + 1 = x2 (x + 1) + 1 · (x + 1) = (x + 1)(x2 + 1)
3x3 + 14x2 + 7x − 4 = 3x2 (x + 1) + 11x(x + 1) − 4(x + 1) = (x + 1)(3x − 1)(x + 4)
x4 − 16 = (x2 )2 − 42 = (x2 − 4)(x2 + 4) = (x − 2)(x + 2)(x2 + 4)
√
√
x4 + 16 = x4 + 8x2 + 16 − 8x2 = (x2 + 4)2 − 8x2 = (x2 + 2 2x + 4)(x2 − 2 2x + 4)
√
√
1
x5 + 1 = (x + 1)(x4 − x3 + x2 − x + 1) = (x + 1)(2x2 − x + 5x + 2)(2x2 − x − 5x + 2)
4
x6 − 3x5 + 10x3 − 15x2 + 9x − 2 = (x + 2)(x − 1)5
x9 + 6x8 + 21x7 + 51x6 + 81x5 + 87x4 + 32x3 − 63x2 − 108x − 108 = (x − 1)(x + 2)2 (x2 + x + 3)3
x11 − x + 1 ≈ (x + 1.068297)(x2 + 1.756597x + 1.131387)(x2 + 0.763439x + 1.100374)×
×(x2 − 0.456963x + 1.042811)(x2 − 1.418968x + 0.942936)(x2 − 1.712402x + 0.764660)
The third step is to express the proper rational function R(x)/Q(x) as a sum of partial fractions of the
form
Ax + B
A
or
i
2
(ax + b)
(ax + bx + c)j
A theorem in algebra guarantees that it is always possible to do this.
2
Section 6.3 Partial Fractions
2010 Kiryl Tsishchanka
There are four possible cases:
Case I: The denominator Q(x) is a product of distinct linear factors.
This means that we can write
Q(x) = (a1 x + b1 )(a2 x + b2 ) . . . (ak x + bk )
where no factor is repeated (and no factor is a constant multiple of another). For example,
(x − 3)(x + 3),
x(x − 1)(x − 2),
(x + 1)(x + 3)(x + 5), etc.
In this case the partial fraction theorem states that there exist constants A1 , A2 , . . . , Ak such that
R(x)
A1
A2
Ak
=
+
+ ...+
Q(x)
a1 x + b1 a2 x + b2
ak x + bk
(1)
For instance,
1
A
B
= +
,
x(x + 1)
x x+1
Here is a more complicated example:
where A = 1 and B = −1
A
B
C
1
=
+
+
(x − 2)(x + 1)(x + 4)
x−2 x+1 x+4
where
A=
1
,
18
1
B=− ,
9
and C =
1
18
Case II: The denominator Q(x) is a product of linear factors, some of which are repeated.
This means that we can write
Q(x) = (a1 x + b1 )r1 (a2 x + b2 )r2 . . . (ak x + bk )rk
where r1 r2 . . . rk > 1. For example,
(x + 1)2 ,
x3 (x − 2)(x + 4)2 ,
(x + 5)2 (x − 1)4 , etc.
In this case instead of the single term A1 /(a1 + b1 ) in (1), we would use
A1
A2
Ar1
+
+ ...+
2
a1 x + b1 (a1 x + b1 )
(a1 x + b1 )r1
(2)
For instance,
A
B
C
1
=
+
+
,
x(x + 1)2
x x + 1 (x + 1)2
where A = 1 and B = C = −1
Here is a more complicated example:
(x −
A
B
C
D
E
F
2x − 5
=
+
+
+
+
+
3
2
2
− 2)(x + 3)
x − 1 (x − 1)
x − 2 x + 3 (x + 3)
(x + 3)3
1)2 (x
where
A=−
5
,
256
B=
3
,
64
C=−
1
,
125
D=
3
881
,
32000
E=
57
,
800
and F =
11
80
Section 6.3 Partial Fractions
2010 Kiryl Tsishchanka
Case III: The denominator Q(x) contains irreducible quadratic factors, none of which is
repeated.
For example,
x2 + 1,
x(2x − 3)(x2 + x + 1),
x2 (x2 + 4)(x2 + 3x + 8), etc.
In this case if Q(x) has the factor ax2 + bx + c, where b2 − 4ac < 0, then, in addition to the partial fractions
in (1) and (2), the expression for R(x)/Q(x) will have a term of the form
Ax + B
ax2 + bx + c
(3)
For instance,
x(x2
1
A
Bx + C
= + 2
,
+ x + 1)
x x +x+1
where A = 1 and B = C = −1
Here is a more complicated example:
3x2 + x + 5
A
B
Cx + D
=
+
+
(x + 2)2 (x2 + 1)
x + 2 (x + 2)2
x2 + 1
where
1
A= ,
5
B = 3,
2
C= ,
5
and D = −
1
5
Case IV: The denominator Q(x) contains a repeated irreducible quadratic factor.
For example,
(x2 + x + 2)2 ,
(2x + 7)(x2 + 4)3 ,
x(x − 2)(x + 1)2 (2x2 + 3)(x2 + 1)3 (x2 + 4x + 5)4 , etc.
In this case if Q(x) has the factor (ax2 + bx + c)r , where b2 − 4ac < 0, then, instead of a single partial
fraction (3), the sum
A2 x + B2
Ar x + Br
A1 x + B1
+
+
.
.
.
+
ax2 + bx + c (ax2 + bx + c)2
(ax2 + bx + c)r
(4)
occurs in the partial fraction decomposition of R(x)/Q(x).
For instance,
x(x2
1
A
Bx + C
Dx + E
= + 2
+ 2
,
2
+ x + 1)
x
x + x + 1 (x + x + 1)2
where A = 1 and B = C = D = E = −1
Here is a more complicated example:
A
B
Cx + D
Ex + F
Gx + H
Ix + J
x3 + x2 + 1
= +
+ 2
+ 2
+ 2
+ 2
2
2
3
2
x(x − 1)(x + x + 1)(x + 1)
x
x−1 x +x+1
x +1
(x + 1)
(x + 1)3
where
A = −1,
1
B= ,
8
C = D = −1,
E=
15
,
8
1
F =− ,
8
4
3
G=H = ,
4
1
I =− ,
2
and J =
1
2
Section 6.3 Partial Fractions
EXAMPLES:
Z
1. Evaluate
2010 Kiryl Tsishchanka
x2 + 4x + 1
dx.
(x − 1)(x + 1)(x + 3)
Solution: This is a Type I integral. Therefore first we find constants A, B, and C such that
x2 + 4x + 1
A
B
C
=
+
+
(x − 1)(x + 1)(x + 3)
x−1 x+1 x+3
We have
B
C
A(x + 1)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x + 1)
A
+
+
=
x−1 x+1 x+3
(x − 1)(x + 1)(x + 3)
therefore
x2 + 4x + 1 = A(x + 1)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x + 1)
(5)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (5) and collect like terms, we get
x2 + 4x + 1 = (A + B + C)x2 + (4A + 2B)x + (3A − 3B − C)
hence

A+B+C =1



4A + 2B = 4



3A − 3B − C = 1
=⇒
3
A= ,
4
1
B= ,
2
and C = −
1
4
Method 2: If we put x = 1 in (5), we get
12 + 4 · 1 + 1 = A · 2 · 4
=⇒
A=
3
4
Similarly, if we put x = −1 in (5), we get
(−1)2 + 4 · (−1) + 1 = B · (−2) · 2
=⇒
B=
1
2
Finally, if we put x = −3 in (5), we get
(−3)2 + 4 · (−3) + 1 = C · (−4) · (−2)
So, we have
=⇒
C=−
1
4
x2 + 4x + 1
3
1
1
1
1
1
= ·
+ ·
− ·
(x − 1)(x + 1)(x + 3)
4 x−1 2 x+1 4 x+3
therefore
Z
Z
Z
Z
x2 + 4x + 1
dx
dx
dx
3
1
1
3
1
1
dx =
+
−
= ln |x−1|+ ln |x+1|− ln |x+3|+C
(x − 1)(x + 1)(x + 3)
4
x−1 2
x+1 4
x+3
4
2
4
2. Evaluate
Z
x2
dx
.
+x−2
5
Section 6.3 Partial Fractions
2. Evaluate
Z
Solution: Since
x2
2010 Kiryl Tsishchanka
dx
.
+x−2
x2 + x − 2 = (x − 1)(x + 2)
this is a Type I integral. Therefore first we find constants A and B such that
A
B
1
=
+
x2 + x − 2
x−1 x+2
We have
B
A(x + 2) + B(x − 1)
A
+
=
x−1 x+2
(x − 1)(x + 2)
therefore
1 = A(x + 2) + B(x − 1)
(6)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (6) and collect like terms, we get
1 = (A + B)x + (2A − B)
hence
(
A+B =0
2A − B = 1
=⇒
A=
1
3
and B = −
1
3
Method 2: If we put x = −2 in (6), we get
1 = B · (−3)
=⇒
B=−
1
3
Similarly, if we put x = 1 in (6), we get
1=A·3
So, we have
x2
therefore
Z
1
dx
=
2
x +x−2
3
3. Evaluate
Z
Z
dx
1
−
x−1 3
2x3 − 4x2 − x − 3
dx.
x2 − 2x − 3
=⇒
A=
1
3
1
1
1
1
1
= ·
− ·
+x−2
3 x−1 3 x+2
Z
dx
1
1
1 x − 1 +C
= ln |x − 1| − ln |x + 2| + C = ln x+2
3
3
3
x + 2
Solution: Since the degree of the numerator is greater than the degree of the denominator, we first perform
the long division. This enables us to write
Z Z Z
2
5x − 3
3
2x3 − 4x2 − x − 3
dx =
2x +
dx
dx =
2x + 2
+
x2 − 2x − 3
x − 2x − 3
x+1 x−3
= x2 + 2 ln |x + 1| + 3 ln |x − 3| + C
6
Section 6.3 Partial Fractions
4. Evaluate
Z
2010 Kiryl Tsishchanka
6x + 7
dx.
(x + 2)2
Solution: Since we have a repeated linear factor and no irreducible quadratic factors, this is a Type II
integral. Therefore first we find constants A and B such that
6x + 7
A
B
=
+
(x + 2)2
x + 2 (x + 2)2
We have
A
B
A(x + 2) + B
+
=
2
x + 2 (x + 2)
(x + 2)2
therefore
6x + 7 = A(x + 2) + B
(7)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (7) and collect like terms, we get
6x + 7 = Ax + (2A + B)
hence
(
A=6
2A + B = 7
=⇒
A = 6 and B = −5
Method 2: If we put x = −2 in (7), we get
6 · (−2) + 7 = B
=⇒
B = −5
Similarly, if we put x = 0 in (7), we get
6·0+7 =A·2+B
So, we have
7 = 2A + B
=⇒
A=6
6
5
6x + 7
=
−
2
(x + 2)
x + 2 (x + 2)2
therefore
5. Evaluate
=⇒
Z
Z
6x + 7
dx = 6
(x + 2)2
Z
dx
−5
x+2
Z
dx
= 6 ln |x + 2| + 5(x + 2)−1 + C
(x + 2)2
2x + 4
dx.
x3 − 2x2
7
Section 6.3 Partial Fractions
5. Evaluate
Z
2010 Kiryl Tsishchanka
2x + 4
dx.
x3 − 2x2
Solution: We have
x3 − 2x2 = x2 (x − 2)
Since we have a repeated linear factor and no irreducible quadratic factors, this is a Type II integral.
Therefore first we find constants A, B, and C such that
A B
C
2x + 4
= + 2+
− 2)
x x
x−2
x2 (x
We have
A B
C
Ax(x − 2) + B(x − 2) + Cx2
+ 2+
=
x
x
x−2
x2 (x − 2)
therefore
2x + 4 = Ax(x − 2) + B(x − 2) + Cx2
(8)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (8) and collect like terms, we get
2x + 4 = (A + C)x2 + (−2A + B)x − 2B
hence

A+C =0



−2A + B = 2



−2B = 4
=⇒
A = B = −2 and C = 2
Method 2: If we put x = 0 in (8), we get
2 · 0 + 4 = −2B
=⇒
B = −2
Similarly, if we put x = 2 in (8), we get
2 · 2 + 4 = C · 22
=⇒
C=2
Finally, if we put x = 1 in (8), we get
2 · 1 + 4 = A · (−1) + B · (−1) + C · 12
So, we have
therefore
=⇒
A = −2
2x + 4
1
1
1
=
−2
·
−
2
·
+
2
·
x2 (x − 2)
x
x2
x−2
Z
2x + 4
dx = −2
2
x (x − 2)
Z
dx
−2
x
Z
dx
+2
x2
Z
dx
x−2
x − 2 2
2
+ +C
= −2 ln |x| + + 2 ln |x − 2| + C = 2 ln x
x x
8
Section 6.3 Partial Fractions
6. Evaluate
Z
2010 Kiryl Tsishchanka
x2 + x − 2
dx.
3x3 − x2 + 3x − 1
Solution: We have
3x3 − x2 + 3x − 1 = x2 (3x − 1) + 1 · (3x − 1) = (3x − 1)(x2 + 1)
This is a Type III integral. Therefore first we find constants A, B, and C such that
x2 + x − 2
A
Bx + C
=
+ 2
3
2
3x − x + 3x − 1
3x − 1
x +1
We have
Bx + C
A(x2 + 1) + (Bx + C)(3x − 1)
A
+ 2
=
3x − 1
x +1
(3x − 1)(x2 + 1)
therefore
x2 + x − 2 = A(x2 + 1) + (Bx + C)(3x − 1)
(9)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (9) and collect like terms, we get
x2 + x − 2 = (A + 3B)x2 + (−B + 3C)x + (A − C)
hence

A + 3B = 1



−B + 3C = 1



A − C = −2
7
A=− ,
5
=⇒
4
B= ,
5
and C =
3
5
!
2
2
1
1
7
1
+ −2 = A
+ 1 , so A = − . Similarly, if we
Method 2: If we put x = 1/3 in (9), we get
3
3
3
5
3
put x = 0 in (9), we get 02 + 0 − 2 = A(02 + 1) + C(3 · 0 − 1), so C = . Finally, plugging in x = 1, we
5
4
get B = .
5
So, we have
hence
4
x + 35
x2 + x − 2
7
1
7
1
4
x
3
1
5
=− ·
+ 2
=− ·
+ · 2
+ · 2
3
2
3x − x + 3x − 1
5 3x − 1
x +1
5 3x − 1 5 x + 1 5 x + 1
Z
Note that
Z


dx

=
3x − 1 
therefore
7
x2 + x − 2
dx = −
3
2
3x − x + 3x − 1
5
3x − 1 = u
d(3x − 1) = du
3dx = du
1
dx = du
3
Z
Z
dx
4
+
3x − 1 5

Z

du
1

=
3
u

1
ln |u| + C
3
1
= ln |3x − 1| + C
3
and
=
Z
Z
x
3
dx +
2
x +1
5


x

dx
=

x2 + 1

Z
dx
+1
x2
x2 + 1 = u
d(x2 + 1) = du
2xdx = du
1
xdx = du
2

Z

du
1

=
2
u

1
ln |u| + C
2
1
= ln(x2 + 1) + C
2
=
7
2
3
x2 + x − 2
dx = − ln |3x − 1| + ln(x2 + 1) + tan−1 x + C
3
2
3x − x + 3x − 1
15
5
5
9
Section 6.3 Partial Fractions
7. Evaluate
Z
2010 Kiryl Tsishchanka
2x2 + 2x + 1
dx.
x3 + x2 + x
Solution: We have
x3 + x2 + x = x(x2 + x + 1)
This is a Type III integral. Therefore first we find constants A, B, and C such that
A
Bx + C
2x2 + 2x + 1
= + 2
3
2
x +x +x
x
x +x+1
We have
A
Bx + C
A(x2 + x + 1) + (Bx + C)x
+ 2
=
x x +x+1
x(x2 + x + 1)
therefore
2x2 + 2x + 1 = A(x2 + x + 1) + (Bx + C)x
(10)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (10) and collect like terms, we get
2x2 + 2x + 1 = (A + B)x2 + (A + C)x + A
hence

A+B =2



A+C =2



A=1
Method 2: If we put x = 0 in (10), we get
=⇒
2 · 02 + 2 · 0 + 1 = A
A=B=C =1
=⇒
A=1
Similarly, if we put x = 1 in (10), we get
2 · 12 + 2 · 1 + 1 = (1 + B) · 12 + (1 + C) · 1 + 1
=⇒
5=B+C +3
Finally, plugging in x = −1, we get
2 · (−1)2 + 2 · (−1) + 1 = (1 + B) · (−1)2 + (1 + C) · (−1) + 1
=⇒
From B + C = 2 and B − C = 0 it follows that B = C = 1.
So, we have
=⇒
1 = B−C +1
B+C =2
=⇒
B−C = 0
1
x+1
2x2 + 2x + 1
= + 2
3
2
x +x +x
x x +x+1
therefore (see Appendix B)
Z
2x2 + 2x + 1
dx =
x3 + x2 + x
= ln |x| +
Z
Z
dx
+
x
Z
x+1
dx = ln |x| +
2
x +x+1
u − 21 + 1
du = ln |x| +
u2 + 43
Z

Z
u + 21
du = ln |x| +
u2 + 43
Z

x+1

dx = 
3
1 2

x+ 2 + 4
u
1
3 du +
2
2
u +4
Z
1
x+ =u
2
1
d x+
= du
2
dx = du
u2 +
1
√ 2 du
3
2
1
1
2u
1
2x + 1
3
1
2
= ln |x| + ln u +
+ √ tan−1 √ + C = ln |x| + ln(x2 + x + 1) + √ tan−1 √
+C
2
4
2
3
3
3
3
10





Section 6.3 Partial Fractions
8. Evaluate
Z
2010 Kiryl Tsishchanka
3x4 + 4x3 + 16x2 + 20x + 9
dx.
(x + 2)(x2 + 3)2
Solution: This is a Type IV integral. Therefore first we find constants A, B, C, D, and E such that
A
Bx + C
Dx + E
3x4 + 4x3 + 16x2 + 20x + 9
=
+ 2
+ 2
2
2
(x + 2)(x + 3)
x+2
x +3
(x + 3)2
We have
Bx + C
Dx + E
A(x2 + 3)2 + (Bx + C)(x + 2)(x2 + 3) + (Dx + E)(x + 2)
A
+ 2
+ 2
=
x+2
x +3
(x + 3)2
(x + 2)(x2 + 3)2
therefore
3x4 + 4x3 + 16x2 + 20x + 9 = A(x2 + 3)2 + (Bx + C)(x + 2)(x2 + 3) + (Dx + E)(x + 2)
(11)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (11) and collect like terms, we get
3x4 + 4x3 + 16x2 + 20x + 9
= (A + B)x4 + (2B + C)x3 + (6A + 3B + 2C + D)x2 + (6B + 3C + 2D + E)x + (9A + 6C + 2E)
hence

A+B =3





 2B + C = 4
6A + 3B + 2C + D = 16



6B + 3C + 2D + E = 20



9A + 6C + 2E = 9
=⇒
A = 1,
B = 2,
C = 0,
D = 4,
and E = 0
Method 2: If we put x = −2 in (11), we get A = 1. Plugging in 0 and three other different values of x into
(11), we get C = E = 0, B = 2, and D = 4.
So, we have
hence
Note that
Z
3x4 + 4x3 + 16x2 + 20x + 9
1
2x
4x
=
+ 2
+ 2
2
2
(x + 2)(x + 3)
x + 2 x + 3 (x + 3)2
3x4 + 4x3 + 16x2 + 20x + 9
dx =
(x + 2)(x2 + 3)2
Z


Z
x2 + 3 = u
du
2x
 d(x2 + 3) = du  =
dx
=
x2 + 3
u
2xdx = du
= ln |u| + C
1
dx +
x+2
Z
and
= ln(x2 + 3) + C
therefore
Z
Z
Z
2x
dx +
2
x +3
Z
(x2
4x
dx
+ 3)2


x2 + 3 = u
Z
 d(x2 + 3) = du 
4x
du


dx = 
=2
2xdx = du 
(x2 + 3)2
u2
4xdx = 2du
=2
Z
u−2 du = 2
u−2+1
2
+ C = −2u−1 + C = − 2
+C
−2 + 1
x +3
2
3x4 + 4x3 + 16x2 + 20x + 9
dx = ln |x + 2| + ln(x2 + 3) − 2
+C
2
2
(x + 2)(x + 3)
x +3
11
Section 6.3 Partial Fractions
9. Evaluate
Z
x5
2010 Kiryl Tsishchanka
dx
.
+ 2x3 + x
Solution: Since
x5 + 2x3 + x = x(x4 + 2x2 + 1) = x(x2 + 1)2
this is a Type IV integral. Therefore first we find constants A, B, C, D, and E such that
x(x2
We have
A Bx + C
Dx + E
1
= + 2
+ 2
2
+ 1)
x
x +1
(x + 1)2
A Bx + C
Dx + E
A(x2 + 1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x
+ 2
+ 2
=
x
x +1
(x + 1)2
x(x2 + 1)2
therefore
1 = A(x2 + 1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x
(12)
We now can proceed in two different ways:
Method 1: If we expand the parentheses on the right-hand side of (12) and collect like terms, we get
1 = (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A
hence


A+B =0






C=0



2A + B + D = 0





C +E =0




 A=1
=⇒
A = 1,
B = −1,
C = 0,
D = −1,
and E = 0
Method 2: If we put x = 0 in (12), we get A = 1. Plugging in four other different values of x into (12), we
get B = D = −1 and C = E = 0.
So, we have
x(x2
hence
Z
Note that
Z


x

dx = 
2
x +1

therefore
1
dx =
2
x(x + 1)2
x2 + 1 = u
d(x2 + 1) = du
2xdx = du
1
xdx = du
2
Z
1
1
x
x
= − 2
− 2
2
+ 1)
x x + 1 (x + 1)2

Z
1
dx −
x
Z
x
dx −
2
x +1
Z

(x2
x
dx
+ 1)2
x2 + 1 = u
d(x2 + 1) = du
2xdx = du
1
xdx = du
2

Z

 1 Z du
Z

x
1


du
1

dx = 
=
u−2 du
=
=
(x2 + 1)2
u2
2

 2
2
u

and
1
= ln |u| + C
2
1
1
1
1 u−2+1
= ln(x2 + 1) + C
+ C = − u−1 + C = −
+C
= ·
2
2
2 −2 + 1
2
2(x + 1)
Z
1
1
1
2
dx
=
ln
|x|
−
ln(x
+
1)
+
+C
x(x2 + 1)2
2
2(x2 + 1)
12
Section 6.3 Partial Fractions
2010 Kiryl Tsishchanka
Appendix A
1. If f (x) =
1
, then
1+x
f ′ (x) = ((1 + x)−1 )′ = (−1)(1 + x)−2 · (1 + x)′ = (−1)(1 + x)−2 · 2 = −
1
(1 + x)2
and F (x) = ln |1 + x| .
2. If f (x) =
1
, then
1 + x2
f ′ (x) = ((1 + x2 )−1 )′ = (−1)(1 + x2 )−2 · (1 + x2 )′ = (−1)(1 + x2 )−2 · 2x = −
2x
(1 + x2 )2
and F (x) = tan−1 x .
3. If f (x) =
1
, then
1 + x3
f ′ (x) = ((1 + x3 )−1 )′ = (−1)(1 + x3 )−2 · (1 + x3 )′ = (−1)(1 + x3 )−2 · 3x2 = −
3x2
(1 + x3 )2
and
1
1
1
F (x) = − ln x2 − x + 1 + ln(x + 1) + √ tan−1
6
3
3
4. If f (x) =
2x − 1
√
3
1
, then
1 + x4
f ′ (x) = ((1 + x4 )−1 )′ = (−1)(1 + x4 )−2 · (1 + x4 )′ = (−1)(1 + x4 )−2 · 4x3 = −
and
4x3
(1 + x4 )2
√
√
√ 1 2
2
F (x) = √ − ln x − 2x + 1 + ln x + 2x + 1
−2 tan−1 1 − 2x
4 2
√
2x + 1
+2 tan−1
5. If f (x) =
1
, then
1 + x5
f ′ (x) = ((1 + x5 )−1 )′ = (−1)(1 + x5 )−2 · (1 + x5 )′ = (−1)(1 + x5 )−2 · 5x4 = −
5x4
(1 + x5 )2
and
√ √ 1 √
1
2
5 − 1 ln x +
5 − 1 x + 1 − 1 + 5 ln x −
1+ 5 x+1
2
2


!
√
√
q
p
√
√ 4x + 5 − 1 
−4x + 5 + 1
p
+2 2 5 + 5 tan−1  q
+4 ln(x + 1) − 2 10 − 2 5 tan−1
√
√ 10 − 2 5
2 5+ 5
1
F (x) =
20
√
2
13
Section 6.3 Partial Fractions
2010 Kiryl Tsishchanka
Appendix B
x+1
dx, we first rewrite the denominator as
x2 + x + 1
2 2
1
1
1
1
2
2
2
−
+1= x+
x + x + 1 = x + 2x · + 1 = x + 2x · +
2
2
2
2
= x+
To find
Z
We have
2
1
−
+1
2
2
2
1
1
3
1
− +1= x+
+
2
4
2
4
1
2
2

1
=
u
x
+
Z
Z
 Z u− 1 +1
 2
x+1
x+1


2
1
du
dx =
=
2 3 dx =  d x +
2
2+ 3
=
du
1

x +x+1

u
x+ 2 + 4
4
2
dx = du
Z
Z
Z
Z
Z
u + 12
u
1
u
1
1
1
=
√ 2 du
3 du =
3 du +
3 du =
3 du +
2
2
2
2
2
2
u +4
u +4
u +4
u +4
u2 + 23

Note that
Z
and
Z
u2




u

3 du = 
2

u +4


1
du =
+ a2
Z
3
u + =v
4
3
2
= dv
d u +
4
2udu = dv
1
udu = dv
2
1
a2
u2
a2
+1
du =
Z


 1Z 1
1
1
3

2
+C
dv = ln |v| + C = ln u +
=
 2
v
2
2
4


1
a2
1
=
a
hence

2
 u
=v
 a u

 d
= dv
1

a
du
=
 1
u 2
+1
 du = dv
a
 a
du = adv
Z
Z
v2



Z

1
= 1
adv
 a2
2
v +1


1
1
1
u
dv = tan−1 v + C = tan−1 + C
+1
a
a
a
1
2
−1
√ 2 du = √ tan
3
3
u2 +
2
2u
√
3
+C
Therefore
Z
Z
Z
2u
1
1 2
x+1
u
3
1
1
−1
2
√
+C
+ · √ tan
dx =
du +
√ 2 du = ln u +
x2 + x + 1
2
2
4
2
u2 + 43
3
3
u2 + 23
=
14
1
2x + 1
1
+C
ln(x2 + x + 1) + √ tan−1 √
2
3
3