Math 166
October 1, 2015
Integration of rational functions
In this section we learn how to integrate rational functions. That is, we learn
to do integrals of the form
Z
P (x)
dx
Q(x)
where P, Q are polynomials.
1
1.1
The technique
Check degrees first
The technique for integrating rational functions depends first on deg P , the degree of the polynomial in the numerator, and deg Q, the degree of the polynomial
in the denominator.
• If deg P < deg Q, then do partial fraction decomposition.
• If deg P ≥ deg Q, then first do long division, then do partial
fraction decomposition if necessary.
1.2
Partial fraction decomposition
The idea of partial fraction decomposition is to express
fractions. There are three main steps:
1
P (x)
as a sum of simpler
Q(x)
• Factor Q(x) completely into irreducible linear and quadratic terms.
• Set up the decomposition with unknown coefficients.
• Solve for the unknown coefficients using tricks like the cover up method,
plugging in your favorite number, or clearing the denominator and grouping like terms.
1.2.1
Set up
• If (x − a) is a linear factor of Q(x) and (x − a)m is the highest power
that divides Q(x), then assign to this factor the sum
A1
A2
Am
+
+ ··· +
.
x − a (x − a)2
(x − a)m
Do this for each distinct linear factor.
• If (x2 + bx + c) is an irreducible quadratic factor of Q(x) and (x2 +
bx + c)n is the highest power that divides Q(x), then assign to this factor
the sum
B1 x + C1
B2 x + C2
Bn x + Cn
+
+ ··· + 2
.
x2 + bx + c (x2 + bx + c)2
(x + bx + c)n
Do this for each distinct irreducible quadratic factor.
1.2.2
Solve
• For distinct linear factors use the cover up method.
• For repeated linear factors first use cover up method, then pick your
favorite number.
• For irreducible quadratic factors clear the denominator, group terms,
and compare coefficients on both sides of the equation.
2
1.3
Integrate
Once you’ve finished the algebraic work of partial fraction decomposition and/or
long division, all that’s left is to integrate the sum of simpler terms. You may
need to do u-substitution or trigonometric substitution.
2
Problems
Problem 1
Z
x+3
dx
2x3 − 8x
We do partial fraction decomposition:
(1) Factor
x+3
x+3
x+3
=
=
2x3 − 8x
2x(x2 − 4)
2x(x − 2)(x + 2)
(2) Set up
A
B
C
x+3
=
+
+
2x(x − 2)(x + 2)
2x x − 2 x + 2
(3) Solve: Look at the left hand side of the equation above and
(a) cover up 2x and plug in x = 0: this gives A = −3/4
(b) cover up x − 2 and plug in x = 2: this gives B = 5/16
(c) cover up x + 2 and plug in x = −2: this gives C = 1/16.
Therefore
Z
−3/4
5/16
1/16
+
+
dx
2x
x−2 x+2
3
5
1
= − ln |x| +
ln |x − 2| +
ln |x + 2| + C
8
16
16
x+3
dx =
2x3 − 8x
Problem 2
Z Z
x2
x
dx
+ 2x + 1
We do partial fraction decomposition:
3
(1) Factor
x2
x
x
=
+ 2x + 1
(x + 1)2
(2) Set up
x
A
B
=
+
2
(x + 1)
x + 1 (x + 1)2
(3) Solve: multiply both sides of the above equation by the denominator of the
right hand side and observe that
x = A(x + 1) + B.
Next we consider this new equation and plug in
(a) x = −1 and get B = −1.
(b) x = 0 and get A = 1.
Therefore
Z
Problem 3
1
−1
+
dx
x + 1 (x + 1)2
1
+C
= ln |x + 1| +
x+1
x
dx =
2
x + 2x + 1
Z
Z x2 + 2x + 1
dx
(x2 + 1)2
We do partial fraction decomposition:
(1) Factoring is already done for us.
(2) Set up
Ax + B
Cx + D
x2 + 2x + 1
= 2
+ 2
2
2
(x + 1)
x +1
(x + 1)2
(3) Solve: multiply both sides of the above equation by the denominator of the
right hand side, multiply out, and group terms:
x2 + 2x + 1 = (Ax + B)(x2 + 1) + Cx + D
= Ax3 + Bx2 + Ax + B + Cx + D
= Ax3 + Bx2 + (A + C)x + (B + D).
Next we compare coefficients on both sides of the above equation.
4
(a) Coefficient on x3 on the right hand side is 0 which means A = 0
(b) Coefficient on x2 on the right hand side is 1 which means B = 1
(c) Coefficient on x on the right hand side is 2 which means C = 2
(d) Constant term on the right hand side is 1 which means D = 0
Therefore
Z
x2 + 2x + 1
dx =
(x2 + 1)2
Z 1
2x
+
x2 + 1 (x2 + 1)2
1
= tan−1 x − 2
+ C.
x +1
Problem 4
Z
dx
1
dx
(x2 − 1)2
We do partial fraction decomposition:
(1) Factor
1
1
=
(x2 − 1)2
(x − 1)2 (x + 1)2
(2) Set up
1
(x −
1)2 (x
+
1)2
=
A
B
C
D
+
+
+
2
x − 1 (x − 1)
x + 1 (x + 1)2
(3) Solve: multiply both sides of the above equation by the denominator of the
right hand side:
1 = A(x − 1)(x + 1)2 + B(x + 1)2 + C(x + 1)(x − 1)2 + D(x − 1)2 .
Next we consider this new equation and plug in
(a) x = 1 ⇒ B = 1/4
(b) x = −1 ⇒ D = 1/4
(c) x = 0 ⇒ 1 = A + 1/4 + C + 1/4 ⇒ A + C = 1/2
(d) x = 2 ⇒ 1 = 9A + 9/4 + 3C + 1/4 ⇒ 9A + 3C = −3/2
The equation in (c) gives A = 1/2 − C which, plugging into (d), gives
9(1/2 − C) + 3C = −3/2
⇒
which implies A = −1/2.
5
−6C = −6
⇒C=1
Therefore
Z
Z 1
1
−1/2
1/4
1/4
dx =
+
dx
+
+
(x2 − 1)2
x − 1 (x − 1)2
x + 1 (x + 1)2
1
1
1
= − ln |x − 1| −
+ ln |x + 1| −
+C
2
4(x − 1)
4(x + 1)
Problem 5
Z
x4 + x2 − 1
dx
x3 + x
We start with long division:
x
x3 + x))x4 + x2 + 1
x4 + x2
5
1
which gives that
1
x4 + x2 − 1
=x+ 3
x3 + x
x +x
Now we do partial fraction decomposition for the term
1
:
x3 + x
(1) Factor
1
1
=
x3 + x
x(x2 + 1)
(2) Set up
1
A Bx + C
= + 2
x(x2 + 1)
x
x +1
(3) Solve: multiply both sides of the above equation by the denominator of the
right hand side:
1 = A(x2 + 1) + (Bx + C)x
(a) Plug in x = 0 and get A = 1.
(b) Rewrite the equation as
1 = x2 + 1 + Bx2 + Cx = (B + 1)x2 + Cx + 1.
Therefore B = −1 and C = 0.
6
Therefore
Z
x4 + x2 − 1
dx =
x3 + x
Z −x
1
dx
x+ + 2
x x +1
1
1
= x2 + ln |x| − ln |x2 + 1| + C.
2
2
7