Section 7.4 Integration of Rational Functions By Partial
Functions
Ruipeng Shen
November 21
Introduction
Let us first introduce a basic formula
Z
1
dx
= ln |ax + b| + C
ax + b
a
and consider an example.
Z
1
Example 1. Find
dx.
x2 − a2
Solution
We can first rewrite the integrand into
1
1
1
1
=
−
x2 − a2
2a x − a x + a
Thus we have
Z
x − a
1
1
1
+C
dx
=
[ln
|x
−
a|
−
ln
|x
+
a|]
+
C
=
ln
x2 − a2
2a
2a x + a General Method
guidelines
In order to integrate a rational function f (x) = P (x)/Q(x), we follow the
(I) If f (x) is improper, that is, deg(P ) ≥ deg(Q), then we must take the preliminary step of
dividing Q into P until a remainder R(x) is obtained such that deg(R) < deg(Q). The
division statement is
P (x)
R(x)
f (x) =
= S(x) +
Q(x)
Q(x)
where S and R are also polynomials.
(II) Next we 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 (in the form of ax + b) and irreducible
quadratic factors (in the form of ax2 + bx + c, where b2 − 4ac < 0)
(III) The third step is to express the proper rational function R(x)/Q(x) as a sum of partial
fractions in the form
A
(ax + b)i
Ax + B
(ax2 + bx + c)j
Please refer to the table below in order to determine the partial fractions that should be
included.
1
(IV) Integrate by basic formulas and techniques.
If Q(x) contains the factor
Then the following partial fractions are included in the decomposition
(ax + b)m
Am
A1
A2
+ ··· +
+
2
ax + b (ax + b)
(ax + b)m
(ax2 + bx + c)n
An x + B n
A1 x + B 1
A2 x + B 2
+ ··· +
+
ax2 + bx + c (ax2 + bx + c)2
(ax2 + bx + c)n
Z
Example 2. Find
Solution
x3 + x
dx
x−1
We can rewrite the integrand into
x3 + x
2
= x2 + x + 2 +
x−1
x−1
Thus we have
Z
x3 + x
dx =
x−1
Z x2 + x + 2 +
2
x−1
dx
x3
x2
+
+ 2x + 2 ln |x − 1| + C.
3
2
Z
x2 + 2x − 1
dx
Example 3. Evaluate
2x3 + 3x2 − 2x
=
Solution
We can factor the denominator as
2x3 + 3x2 − 2x = x(2x2 + 3x − 2) = x(2x − 1)(x + 2)
Thus we know that the partial fraction decomposition has the form
A
B
C
x2 + 2x − 1
= +
+
.
2x3 + 3x2 − 2x
x
2x − 1 x + 2
Now let us determine the unknowns A, B and C. Multiplying 2x3 + 3x2 − 2x = x(2x − 1)(x + 2)
on both sides, we have
x2 + 2x − 1 =A(2x − 1)(x + 2) + Bx(x + 2) + Cx(2x − 1)
=(2A + B + 2C)x2 + (3A + 2B − C)x − 2A
Thus we have
2A + B + 2C
3A + 2B − C
−2A
= 1
= 2
= −1
Finally we can solve A = 1/2, B = 1/5 and C = −1/10. As a result we have
Z
Z 1 1 1
1
1
1
x2 + 2x − 1
dx =
· + ·
−
·
dx
2x3 + 3x2 − 2x
2 x 5 2x − 1 10 x + 2
1
1
1
= ln |x| +
ln |2x − 1| −
ln |x + 2| + C.
2
10
10
Z 4
x − 2x2 + 4x + 1
Example 4. Find
dx
x3 − x2 − x + 1
2
Solution
By long division we have
x4 − 2x2 + 4x + 1
4x
=x+1+ 3
.
x3 − x2 − x + 1
x − x2 − x + 1
We can factor the denominator as x3 − x2 − x + 1 = (x − 1)2 (x + 1) thus we have a decomposition
of partial fractions in the form of
x3
C
4x
A
B
+
=
+
2
− −x+1
x − 1 (x − 1)
x+1
x2
This is equivalent to
4x = A(x − 1)(x + 1) + B(x + 1) + C(x − 1)2
Plugging x = 1 in, we have B = 2. Plugging x = −1 in, we have C = −1. Considering the
coefficient of x2 , we have A = 1. Therefore we have
Z 4
Z x − 2x2 + 4x + 1
1
2
1
dx
=
x
+
1
+
+
−
dx
x3 − x2 − x + 1
x − 1 (x − 1)2
x+1
x2
2
= + x + ln |x − 1| −
− ln |x + 1| + C
2
x−1
x − 1
2
x2
+ C.
+ ln = +x−
2
x−1
x + 1
More Basic Formulas
In order to deal with the integral of
formulas
Z
x
dx
1
=
arctan
+C
x2 + a2
a
a
Z
2x2 − x + 4
dx.
Example 5. Evaluate
x3 + 4x
Solution
Z
Ax + B
, we have the basic
ax2 + bx + c
x
1
dx = ln x2 + a2 + C
x2 + a 2
2
We can factor the denominator as x3 +4x = x(x2 +4), thus we have the decomposition
A Bx + C
2x2 − x + 4
= + 2
x3 + 4x
x
x +4
Multiplying by x(x2 + 4), we have
2x2 − x + 4 = A(x2 + 4) + x(Bx + C)
Plugging in x = 0, we have A = 1. This gives B = 1 and C = −1. Thus we have
Z
Z 2x2 − x + 4
1
x−1
dx =
+
dx
x3 + 4x
x x2 + 4
Z 1
x
1
=
+
−
dx
x x2 + 4 x2 + 4
1
1
= ln |x| + ln(x2 + 4) − arctan(x/2) + C.
2
2
Z
2
4x − 3x + 2
Example 6. Evaluate
dx
4x2 − 4x + 3
3
Solution
We can rewrite the integrand into
4x2 − 3x + 2
x−1
x−1
=1+ 2
=1+
4x2 − 4x + 3
4x − 4x + 3
(2x − 1)2 + 2
u+1
, dx = (1/2)du, we have
2
Z
Z
Z
4x2 − 3x + 2
x−1
dx = 1 · dx +
dx
4x2 − 4x + 3
(2x − 1)2 + 2
Z
(u + 1)/2 − 1
· (1/2)du
=x +
u2 + 2
Z
1
u−1
=x +
du
4
u2 + 2
Z
Z
1
u
1
1
=x +
du −
du
4
u2 + 2
4
u2 + 2
√
1
1
=x + ln(u2 + 2) − √ arctan(u/ 2) + C
8
4 2
2x − 1
1
1
2
√
+ C.
=x + ln(4x − 4x + 3) − √ arctan
8
4 2
2
Apply the substitution u = 2x − 1 =⇒ x =
Example 7. Write out the form of the partial fraction decomposition of the function
x3 + x2 + 1
.
x(x − 1)(x2 + x + 1)(x2 + 1)3
Solution
According to the rule given in the table, we have
x3 + x2 + 1
x(x − 1)(x2 + x + 1)(x2 + 1)3
A
B
Cx + D
Ex + F
Gx + H
Ix + J
= +
+ 2
+ 2
+ 2
+ 2
.
2
x
x−1 x +x+1
x +1
(x + 1)
(x + 1)3
Z
1 − x + 2x2 − x3
dx.
Example 8. Evaluate
x(x2 + 1)2
Solution
The partial fraction decomposition is in the form of
1 − x + 2x2 − x3
Dx + E
A Bx + C
= + 2
+ 2
2
2
x(x + 1)
x
x +1
(x + 1)2
Multiplying by x(x2 + 1)2 , we have
1 − x + 2x2 − x3 =A(x2 + 1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x
=(A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A.
Thus we have the equation system
A+B =0
C = −1
C + E = −1
A=1
2A + B + D = 2
4
We have the solution A = 1, B = −1, C = −1, D = 1 and E = 0. Thus we have
Z
Z 1 − x + 2x2 − x3
1 −x − 1
x
dx
=
dx
+
+
x(x2 + 1)2
x
x2 + 1
(x2 + 1)2
Z 1
x
1
x
=
dx
− 2
− 2
+ 2
x x + 1 x + 1 (x + 1)2
1
1
1
= ln |x| − ln(x2 + 1) − arctan x − · 2
+ C.
2
2 x +1
How to integrate
Ax + B
(x2 + a2 )m
Z
First of all, by substitution we have
−1
x
1
dx =
.
·
(x2 + a2 )m
2(m − 1) (x2 + a2 )m−1
In addition, Integration by parts gives us
Z
1
x
dx = 2
(x2 + a2 )n
(x + a2 )n
x
= 2
(x + a2 )n
x
= 2
(x + a2 )n
x
= 2
(x + a2 )n
Z
−
x
1
(x2 + a2 )n
0
dx
Z
2x
x · (−n) · 2
dx
(x + a2 )n+1
Z
(x2 + a2 ) − a2
+ 2n
dx
(x2 + a2 )n+1
Z
Z
1
1
2
+ 2n
dx − 2na
dx
(x2 + a2 )n
(x2 + a2 )n+1
−
Thus we have the induction formula
Z
Z
1
2n − 1
x
1
dx =
dx +
.
(x2 + a2 )n+1
2na2
(x2 + a2 )n
2na2 (x2 + a2 )n
5