LECTURE 20: PARTIAL FRACTIONS AND NUMERICAL INTEGRATION
MINGFENG ZHAO
February 27, 2015
Theorem 1.
Z
xp dx
Z
cos(ax) dx
Z
sin(ax) dx
Z
Z
Z
sec2 (ax) dx
csc2 (ax) dx
Z
eax dx
1
dx
a2 − x2
Z
1
dx
x2 + a2
Z
tan(x) dx
Z
sec(x) dx
√
=
 p+1

 x
+ C,
p+1

 ln |x| + C,
if p 6= −1
if p = −1.
1
sin(ax) + C
a
1
= − cos(ax) + C
a
1
=
tan(ax) + C
a
1
= − cot(ax) + C
a
1 ax
=
e +C
a
x
= sin−1
+C
a
x
1
=
tan−1
+C
a
a
=
=
− ln | cos(x)| + C
=
ln | sec(x) + tan(x)| + C
=
sec(x) + C.
Z
sec(x) tan(x) dx
Remark 1. One should remember the definitions of all trigonometric functions: sin(x), cos(x), tan(x), cot(x), sec(x),
csc(x), sin−1 (x), cos−1 (x), tan−1 (x), − cot−1 (x). And the following trigonometric identities should be remembered:
sin2 (θ) + cos2 (θ) = 1,
1 + tan2 (θ) = sec2 (θ)
sin(2θ) = 2 sin(θ) cos(θ),
cos(2θ) = 2 cos2 (θ) − 1 = 1 − 2 sin2 (θ) = cos2 (θ) − sin2 (θ)
1
2
MINGFENG ZHAO
Partial fraction decompositions
p(x)
be a proper rational function in reduced form. Assume the denominator q(x has been factored
q(x)
completed over the real numbers and m is a positive integer:
Let f (x) =
1. Repeated linear factor: A factor (x − r)m in the denominator requires the partial fractions
Am
A2
A1
+ ··· +
.
+
x − r (x − r)2
(x − r)m
2. Repeated irreducible quadratic factor: An irreducible factor (ax2 + bx + c)m in the denominator requires
the partial fractions
A1 x + B 1
A2 x + B 2
Am x + B m
+
+ ··· +
.
ax2 + bx + c (ax2 + bx + c)2
(ax2 + bx + c)m
Remark 2. Totally there are deg q(x) many coefficients to be determined.
Remark 3. A quadratic polynomial ax2 + bx + c is irreducible if and only if b2 − 4ac < 0.
Example 1. The polynomial 2x2 + x + 1 is irreducible, because 12 − 4 · 2 · 1 = 1 − 8 = −7 < 0.
x3 + 3x2 − 3x + 2
dx.
x3 − 2x2
x3 + 3x2 − 3x + 2
5x2 − 3x + 2
Using the long division, we know that
=1+
. Since x3 − 2x2 = x2 (x − 2), then the
3
2
x − 2x
x3 − 2x2
5x2 − 3x + 2
partial fraction decomposition of
has the form:
x3 − 2x2
Z
Example 2. Evaluate
A
B
C
5x2 − 3x + 2
= + 2+
.
3
2
x − 2x
x
x
x−2
Then
5x2 − 3x + 2
=
Ax(x − 2) + B(x − 2) + Cx2
= A(x2 − 2x) + B(x − 2) + Cx2
=
(A + C)x2 + (−2A + B)x − 2B.
So we have
(1)
A+C
=
5
(2)
−2A + B
=
−3
LECTURE 20: PARTIAL FRACTIONS AND NUMERICAL INTEGRATION
−2B
(3)
=
3
2
By (3), then B = −1. Plug B = −1 into (2), then −2A − 1 = −3, which implies that A = 1. Plug A = 1 into (1),
then C = 5 − A = 4. So we get
B = −1,
A = 1,
and C = 4.
2
That is, the partial fraction decomposition of
5x − 3x + 2
is:
x3 − 2x2
5x2 − 3x + 2
1
4
1
= − 2+
.
x3 − 2x2
x x
x−2
Therefore, we have
Z 3
x + 3x2 − 3x + 2
dx
x3 − 2x2
Z
Example 3. Evaluate
5x2 − 3x + 2
dx = x +
x3 − 2x2
1
= x + ln |x| + + 4 ln |x − 2| + C.
x
Z
=
Z
1 dx +
Z
1
dx −
x
Z
1
dx +
x2
Z
4
dx
x−2
7x2 − 13x + 13
dx.
(x − 2)(x2 − 2x + 3)
Since x2 − 2x + 3 = (x − 1)2 + 2 is irreducible, then the partial fraction decomposition for
7x2 − 13x + 13
is:
(x − 2)(x2 − 2x + 3)
7x2 − 13x + 13
A
Bx + C
=
+
.
(x − 2)(x2 − 2x + 3)
x − 2 x2 − 2x + 3
So we get
7x2 − 13x + 13
=
A(x2 − 2x + 3) + (Bx + C)(x − 2)
= A(x2 − 2x + 3) + Bx2 + (C − 2B)x − 2C
=
(A + B)x2 + (−2A + C − 2B)x + 3A − 2C.
Then
(4)
A+B
=
7
(5)
−2A + C − 2B
=
−13
(6)
3A − 2C
=
13
By (4), then B = 7 − A. Plug B = 7 − A into (5), then −2A + C − 2(7 − A) = −13, that is, C = 1. Plug C = 1 into
(6), then 3A − 2 = 13, that is, A = 5. Plug A = 5 into (4), then B = 7 − A = 2. So we get
A = 5,
B = 2,
and C = 1.
4
MINGFENG ZHAO
Then we have
7x2 − 13x + 13
5
2x + 1
=
+
.
(x − 2)(x2 − 2x + 3)
x − 2 x2 − 2x + 3
So
Z
Z
Z
7x2 − 13x + 13
5
2x + 1
2x + 1
dx
=
dx
+
dx
=
5
ln
|x
−
2|
+
dx.
(x − 2)(x2 − 2x + 3)
x−2
x2 − 2x + 3
x2 − 2x + 3
Z
2x + 1
For
dx, since x2 − 2x + 3 = (x − 1)2 − 12 + 3 = (x − 1)2 + 2, then
x2 − 2x + 3
Z
Z
2(u + 1) + 1
2x + 1
dx
=
du Let u = x − 1
2
x − 2x + 3
u2 + 2
Z
2u + 3
=
du
u2 + 2
Z
Z
du
2u
du
+
3
=
2
2
u +2
u +2
3
u
= ln |u2 + 2| + √ tan−1 √
+C
2
2
x−1
3
−1
2
√
+ C Since u = x − 1
= ln |(x − 1) + 2| + √ tan
2
2
3
x−1
√
= ln |x2 − 2x + 3| + √ tan−1
+ C.
2
2
Z
Therefore, we have
Z
3
7x2 − 13x + 13
dx = 5 ln |x − 2| + ln |x2 − 2x + 3| + √ tan−1
(x − 2)(x2 − 2x + 3)
2
x−1
√
2
+ C.
Numerical integration
Z
Let f be an integrable function on [a, b], usually it’s difficulty to compute
f (x) dx directly, instead, we can use
Z b
some numerical methods to approximate
f (x) dx. Just by the definition of definite integral, we have
a
Z
b
f (x) dx = lim
a
n→∞
n
X
f (x∗k )∆x,
k=1
where x∗k can be right endpoint, left endpoint, or middle point of subinterval [xk−1 , xk ].
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca