Integrating Rational Functions
Solutions
Practice Problems: Evaluate the following Integrals
Z
(1)
Z
(2)
Solutions: Z
(1) Evaluate
4x
dx
(x − 1)2 (x + 1)
Z
x2
dx
x2 − 1
Z
x3 + 10x2 + 3x + 36
dx
(x − 1)(x2 + 4)2
(3)
x2 + 4
dx
3x3 + 4x2 − 4x
(4)
4x
dx
(x − 1)2 (x + 1)
Observation: The function
4x
is a proper rational function. Find the PFD and then integrate.
(x − 1)2 (x + 1)
4x
A
B
C
=
+
+
(x − 1)2 (x + 1)
x − 1 (x − 1)2
x+1
4x = A(x − 1)(x + 1) + B(x + 1) + C(x − 1)2
2
2
4x = A(x − 1) + Bx + B + C(x − 2x + 1)
2
x = 1 : 4 = 2B → B = 2
x = −1 : −4 = 4C → C = −1
2
0x + 4x + 0 = (A + C)x + (B − 2C)x + (−A + B + C).
Equating the coefficients of the x2 term gives
0 = A + C −→ A = −C = 1.
Therefore
Z
2
−1
1
+
+
dx
2
x − 1 (x − 1)
x+1
2
= ln |x − 1| −
− ln |x + 1| + C.
x−1
4x
dx =
(x − 1)2 (x + 1)
Z
Observation: Use the substitution u = x − 1 to get
Z
2
2
dx = −
+ C.
(x − 1)2
x−1
See solution video
Z
x2 + 4
(2) Evaluate
dx
3
3x + 4x2 − 4x
x2 + 4
x2 + 4
Observation: The function
=
is a proper rational function. Find the
3x3 + 4x2 − 4x
x(x + 2)(3x − 2)
PFD and then integrate.
x2 + 4
A
B
C
= +
+
x(x + 2)(3x − 2)
x
x + 2 3x − 2
x2 + 4 = A(x + 2)(3x − 2) + Bx(3x − 2) + Cx(x + 2)
1
Calculus II Resources
Integration Techniques
Corresponding to each linear term we choose an ‘intelligent’ choice of x:
x = 0 : 4 = A(−4) −→ A = −1
1
x = −2 : 8 = B(16) −→ B =
2
4
2
8
5
2
x= :
+4=C
−→ C = .
3
9
3
3
2
Therefore
Z
x2 + 4
dx =
3x3 + 4x2 − 4x
1/2
5/2
−1
+
+
dx
x
x + 2 3x − 2
1
51
= − ln |x| + ln |x + 2| +
ln |3x − 2| + C.
2
23
Z
Observation: Use the substitution u = 3x − 2 ( 31 du = dx) to get
Z
51
5/2
dx =
ln |3x − 2| + C.
3x − 2
23
See solution video
Z
x2
dx
(3) Evaluate
x2 − 1
x2
is NOT a proper rational function. We need to use a polynomial
x2 − 1
long division to rewrite f as the sum of a polynomial and a proper rational function.
1
2
2
x −1
x
− x2 + 1
Observation: The function f (x) =
1
This gives x2 = (x2 − 1) + 1 so that
x2
1
=1+ 2
.
2
x −1
x −1
Note: Alternatively, one can add zero to the numerator to find
x2 − 1 + 1
1
x2
=
=1+ 2
.
2
−1
x −1
x −1
x2
We now find the PFD of the proper rational function
1
x2 −1
=
1
(x+1)(x−1) .
A
B
1
=
+
(x + 1)(x − 1)
x+1 x−1
1 = A(x − 1) + B(x + 1)
Corresponding to each linear term we choose an ‘intelligent’ choice for x:
x = 1 : 1 = B(2) −→ B =
1
2
1
x = −1 : 1 = A(−2) −→ A = − .
2
Therefore
2
Calculus II Resources
Integration Techniques
Z
x2
dx =
2
x −1
−1/2
1/2
+
dx
x+1 x−1
1
1
= x − ln |x + 1| + ln |x − 1| + C.
2
2
Z
1+
See solution video
Z 3
x + 10x2 + 3x + 36
dx
(4) Evaluate
(x − 1)(x2 + 4)2
x3 + 10x2 + 3x + 36
is a proper rational function. Find the PFD and then
Observation: The function
(x − 1)(x2 + 4)2
integrate.
x3 + 10x2 + 3x + 36
A
Bx + C
Dx + E
=
+ 2
+ 2
(x − 1)(x2 + 4)2
x−1
x +4
(x + 4)2
x3 + 10x2 + 3x + 36 = A(x2 + 4)2 + (Bx + C)(x − 1)(x2 + 4) + (Dx + E)(x − 1)
3
2
4
3
x + 10x + 3x + 36 = (A + B)x + (C − B)x + (8A + 4B − C + D)x
x = 1 : 50 = A(25) → A = 2
2
+ (−4B + 4C − D + E)x + (16A − 4C − E)
Equating the corresponding coefficients leads to a system of equations:
x4 : A + B = 0 −→ B = −A = −2
x3 : C − B = 1 −→ C = 1 + B = −1
x2 : 8A + 4B − C + D = 10 −→ 8(2) + 4(−2) − (−1) + D = 10 −→ D = 1
x1 :
− 4B + 4C − D + E = 3
0
x : 16A − 4C − E = 36 −→ 16(2) − 4(−1) − E = 36 −→ E = 0
Therefore
Z
x3 + 10x2 + 3x + 36
dx =
(x − 1)(x2 + 4)2
−2x − 1
x
2
+ 2
+ 2
dx
x−1
x +4
(x + 4)2
Z
Z
Z
Z
2
2x
1
x
=
dx −
dx
−
dx
+
dx
x−1
x2 + 4
x2 + 4
(x2 + 4)2
x 1 1
1
= 2 ln |x − 1| − ln |x2 + 4| − arctan
−
+ C.
2
2
2 x2 + 4
Z
Observation: Use the substitution u = x2 + 4 to evaluate the second and fourth integrals. In the third
integral we have used the common antiderivative
Z
x
1
1
+ C.
dx = arctan
2
2
x +a
a
a
See solution video
3