CHAPTER 7. TECHNIQUES OF INTEGRATION
Example 4. Use polynomial division to find
�
44
x4 + x3 − 2x2 + 17x + 2
dx.
x2 + 1
Solution. We will first put a x2 on top, because multiplying this by x2 + 1 on
the side will allow us to kill the x4 underneath (note, we need to keep track
of the x3 column, so either leave that column black, or write “0x3 ” in it):
x2
x2 + 1 ) x4 + x3 − 2x2 + 17x + 2
−(x4
+ x2 )
x3 − 3x2
Next, we put x on top because when we multiply this by x2 + 1 we can kill
off the x3 . Then we put −3 on top (but, you can’t see that −3 is the correct
thing until you’ve finished the step involving the x on top though)
x2 + x − 3
x2 + 1 ) x4 + x3 − 2x2 + 17x + 2
−(x4 +
x2 )
x3 − 3x2 + 17x
−(x3
+ x)
− 3x2 + 16x + 2
− (− 3x2
− 3)
16x + 5
Continuing in this way, we find that the quotient is x2 +x−3 and the remainder
is 16x + 5. In other words
x4 + x3 − 2x2 + 17x + 2
x2 + 1
=
x2 + x − 3 +
16x + 5
.
x2 + 1
Now we can find the integral
� 4
�
x + x3 − 2x2 + 17x + 2
16x + 5
2
dx
=
x
+
x
−
3
+
dx
x2 + 1
x2 + 1
�
�
�
x
1
2
= x + x − 3 dx + 16
dx + 5
dx
2
2
x +1
x +1
x3 x2
=
+
− 3x + 8 ln |x2 + 1| + 5 tan−1 (x)
3
2
Example 5.
(a) Use partial fractions to split the fraction
x2
−3x − 7
into two
+ 7x + 12
fractions.
(b) Verify that your partial fraction solution works, by combining your answers
with a� common denominator.
3x + 7
(c) Find
− 2
dx.
x + 7x + 12
Solution. (a) We factor x2 + 7x + 12 as (x + 3)(x + 4) and so our set up is
−3x − 7
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
CHAPTER 7. TECHNIQUES OF INTEGRATION
45
We multiply both sides of this equation by (x + 3)(x + 4), this cancels all the
denominators
−3x − 7
(x + 3)(x + 4)
A
(x + 3)(x + 4)
B
(x + 3)(x + 4)
·
=
·
+
·
(x + 3)(x + 4)
1
x+3
1
x+4
1
−3x − 7 = A(x + 4) + B(x + 3)
Now we plug in values of x that make one of the terms equal to 0
x = −4
⇒ 12 − 7 = A(0) + B(−1) ⇒ 5 = −B ⇒ B = −5
x = −3
⇒ 9 − 7 = A(1) + B(0) ⇒ A = 2
Thus, we know
−3x − 7
2
5
=
−
(x + 3)(x + 4)
x+3 x+4
(b) Just this once, we verify that the partial fractions we obtained in the
previous part were correct.
5
2
(x + 4)
5 (x + 3)
2
−
=
·
−
x+3 x+4
x + 3 (x + 4) x + 4 (x + 3)
2x + 8 − (5x + 15)
=
(x + 3)(x + 4)
−3x − 7
= 2
x + 7x + 12
(c) We use the partial fractions to finish the integral
�
�
3x + 7
−3x − 7
− 2
dx =
dx
2
x + 7x + 12
x + 7x + 12
�
2
5
=
−
dx
x+3 x+4
= 2 ln |x + 3| − 5 ln |x + 4|
Example 6. Find
�
x2 + 3x + 1
dx.
x3 + x
Solution. We start by factoring the bottom, and then setting up the partial
fractions using the factors of the bottom
x2 + 3x + 1
A Bx + C
=
+ 2
.
x(x2 + 1)
x
x +1
Multiplying both sides by x(x2 + 1) we get:
x2 + 3x + 1 = A(x2 + 1) + (Bx + C)x
It is true that we could plug in x = 0 and solve for A, but we can’t do this
again and solve for B and C, so we’ll combine both approaches.
CHAPTER 7. TECHNIQUES OF INTEGRATION
46
If we plug in x = 0 we gen
x = 0 ⇒ 1 = A(1) + 0 =⇒ A = 1
Now we combine this with the equation involving B and C
x2 + 3x + 1 = 1(x2 + 1) + (Bx + C)x
We start by multiplying everything out, and then gathering terms on the right
hand side.
x2 + 3x + 1 = 1(x2 + 1) + (Bx + C)x
= x2 + 1 + Bx2 + Cx
x2 + 3x + 1 = (1 + B)x2 + Cx + 1
Now we set up new equations by making the coefficients from one side equal
to the coefficients from the other side.
x2 coeffs : 1 = 1 + B
x coeffs : 3 = C
.
constants : 1 = 1
This gives us A = 1, B = 0 and C = 3. Now we can integrate
� 2
�
x + 3x + 1
1
3
dx =
+ 2
dx
2
x(x + 1)
x x +1
= ln |x| + 3 tan−1 (x)
Example 7. Use completing the square to find
Solution.
x2 + 6x +7
÷2 ↓
∧2
3−−−→9
→
�
x2
1
dx.
+ 6x + 7
x2 + 6x +9 − 9+7
÷2 ↓
��
∧2
3−−−→ 9
Note that x2 + 6x + 9 = (x + 3)2 , and simplify −9 + 7 to −2 to get
x2 + 6x + 7
=
(x + 3)2 − 2
Now we can finish the integral
�
�
1
1
dx =
dx
2
x + 6x + 7
(x + 3)2 − 2
�
1
√
=
du (u = x + 3)
2
u − ( 2)2
�
�
� u − √2 �
1
�
�
√ �
= √ ln �
�
2 2
u + 2�
�
�
� x + 3 − √2 �
1
�
�
√ �
= √ ln �
�
2 2
x − 3 + 2�
CHAPTER 7. TECHNIQUES OF INTEGRATION
47
Extra examples
Example 8. [Stewart, 6e, #4b] Set up the partial fraction expansion for
2x + 1
(x + 1)3 (x2 + 4)2
Solution.
2x + 1
A
B
C
Dx + E
Ex + F
=
+
+
+ 2
+ 2
3
2
2
2
3
(x + 1) (x + 4)
x + 1 (x + 1)
(x + 1)
x +4
(x + 4)2
Example 9. [Stewart, 6e, #17] Find
� 2
4y 2 − 7y − 12
dy
1 y(y + 2)(y − 3)
Solution.
4y 2 − 7y − 12
A
B
C
= +
+
y(y + 2)(y − 3)
y
y+2 y−3
2
4y − 7y − 12 = A(y + 2)(y − 3) + By(y − 3) + Cy(y + 2)
y = 0 =⇒ −12 = A(2)(−3) + B(0) + C(0) =⇒ −12 = −6A =⇒ A = 2
y = −2 =⇒ 4(4) + 14 − 12 = A(0) + B(−2)(−5) + C(−2)0 =⇒ 18 = 10B =⇒ B = 9/5
y = 3 =⇒ 4(9) − 21 − 12 = A(0) + B(0) + C(3)(5) =⇒ 3 = 15C =⇒ C = 1/5
� 2
� 2
4y 2 − 7y − 12
2
9/5
1/5
dy =
+
+
dy
y+2 y−3
1 y(y + 2)(y − 3)
1 y
�2
�
9
1
= 2 ln |y| + ln |y + 2| + ln |y − 3|��
5
5
1
9
= 2 ln |2| + ln |4| +
5
9
= 2 ln |2| + ln |4| −
5
Example 10. Find
�
1
9
1
ln | − 1| − (2 ln |1| + ln |3| + ln | − 2|)
5
5
5
9
1
ln |3| − ln |2|)
5
5
2x3 − 3x + 7
dx
x+1
Solution. We start with polynomial division. Note how we write “0x2 ” in
one of the columns to help keep things lined up.
2
�2x3 −
x + 1 2x +
−(2x3 +
2x − 1
0x2 −3x+7
2x2 )
−2x2 −3x+7
−( − 2x2 −2x)
− x+7
−(− x−1)
8
CHAPTER 7. TECHNIQUES OF INTEGRATION
48
Now we can finish the integral
�
�
2x3 − 3x + 7
8
dx = 2x2 − 2x − 1 +
dx
x+1
x+1
2
= x3 − x2 − x + 8 ln |x + 1|
3
Example 11. Find
�
x
dx
x2 + 4x + 10
Solution. We start by completing the square
x2 + 4x + 10 = x2 + 4x + 4 − 4 + 10 = (x + 2)2 + 6
Now our integral becomes
�
x
dx =
2
x + 4x + 10
�
x
dx
(x + 2)2 + 6
To finish we let u = x + 2, but there’s a problem here
�
x
du
2
u +6
We still need to get rid of that last x on top. We solve u = x + 2 for x to get
x = u − 2 and so our integral becomes
�
�
u−2
u
2
√
du =
−
du
2
2
2
u +6
u + 6 u + ( 6)2
�
�
1
1
u
2
−1
√
= ln |u + 6| − 2 · √ tan
2
2 6
6
�
�
1
1
x
+2
2
−1
√
= ln |(x + 2) + 6| − √ tan
2
6
6
Example 12. Apply partial fractions to split up
(x2
x
+ 2x + 2)(x2 − x + 3)
Solution. We set up the partial fraction and clear denominators
x
Ax + B
Cx + D
=
+
(x2 + 2x + 2)(x2 − x + 3)
x2 + 2x + 2 x2 − x + 3
x = (Ax + B)(x2 − x + 3) + (Cx + D)(x2 + 2x + 2)
In the homework we would stop here, but I’ll show you how to finish it.
x = Ax3 − Ax2 + 3Ax + Bx2 − Bx + 3B + Cx3 + 2Cx2 + 2Cx + Dx2 + 2Dx + 2D
x = (A + C)x3 + (−A + B + 2C + D)x2 + (3A − B + 2C + 2D)x + (3B + 2D)
Now we equate coefficients
Left hand side
Right hand side:
CHAPTER 7. TECHNIQUES OF INTEGRATION
constants terms
x-coefficients
x2 -coefficients
x3 -coefficients
49
0 = 3B + 2D
1 = 3A − B + 2C + 2D
0 = −A + B + 2C + D
0=A+C
From the first equation we see that 2D = −3B and so D = − 32 B. From the
last equation we see that C = −A. Plugging these into the other equations
(i.e. the second and third equations) we get
1 = 3A − B + 2(−A) + 2(− 32 B)
0 = −A + B + 2(−A) + (− 32 B)
Simplifying we get
1 = A − 4B
0 = −3A − 12 B
From the second equation get B = −6A. Plugging this into the first equation
we get 1 = A − 4(−6A), 1 = 25A, A = 1/25.
Now that we know A = 1/25, we get B = −6/25, C = −1/25 and D =
3 −6
9
− 2 25 = 25
.
Thus, our partial fraction solution is
1
6
1
9
x − 25
− 25
x + 25
x
25
= 2
+
(x2 + 2x + 2)(x2 − x + 3)
x + 2x + 2 x2 − x + 3
Example 13. Set up (but do not solve) the following as a partial fraction:
x4 + 3x2 − 17x + 11
x(x + 1)2 (x2 + 2x − 7)
Solution. Applying the above rules we see that we will have fractions with
denominators of x, x + 1, (x + 1)2 , and x2 + 2x − 7. This gives
x4 + 3x2 − 17x + 11
A
B
C
Dx + E
= +
+
+ 2
2
2
2
x(x + 1) (x + 2x − 7)
x x + 1 (x + 1)
x + 2x − 7
Example 14. Set up (but do not solve) the following as a partial fraction:
x7 − 1324x2 + x − 10
(x2 + 1)(x2 + 10)4
Solution. Applying the above rules we see that we will have fractions with
denominators of x2 + 1, x2 + 10, (x2 + 10)2 , (x2 + 10)3 , (x2 + 10)4 . This gives
x7 − 1324x2 + x − 10
Ax + B Cx + D Dx + E
Fx + G
Hx + I
=
+
+
+
+
(x2 + 1)(x2 + 10)4
x2 + 1 x2 + 10 (x2 + 10)2 (x2 + 10)3 (x2 + 10)4
CHAPTER 7. TECHNIQUES OF INTEGRATION
50
Example 15. Work Example 13 further until we get the equations for the constant
terms, the x-coefficients, the x2 -coefficients, etc.
Solution. We start by clearing fractions. In other words, we multiply
A
B
C
Dx + E
x4 + 3x2 − 17x + 11
= +
+
+ 2
2
2
2
x(x + 1) (x + 2x − 7)
x x + 1 (x + 1)
x + 2x − 7
by
x(x + 1)2 (x2 + 2x − 7)
to get
1
x4 + 3x2 − 17x + 11 = A(x + 1)2 (x2 + 2x − 7) + Bx(x + 1)(x2 + 2x − 7)
+ Cx(x2 + 2x − 7) + (Dx + E)x(x + 1)2
We start the easy way, by finding some numbers that we can plug in for x
which will kill all but one of the unknown coefficients. We can plug in x = 0
and x = −1 since each of these will make most of the terms equal to 0.
Plugging in x = 0 gives
x=0⇒
11 = −7A + 0B + 0C + (0D + E)0
A = −11/7 .
Plugging in x = −1 gives
x = −1 ⇒
32 = 0A + 0B + 8C + (−D + E)0
C=4.
Now we use these values of A and C to set up the remaining equations.
Plugging these into our big polynomial equation above gives
x4 + 3x2 − 17x + 11 = − 11
(x + 1)2 (x2 + 2x − 7) + Bx(x + 1)(x2 + 2x − 7)
7
+ 4x(x2 + 2x − 7) + (Dx + E)x(x + 1)2
Now we multiply this out
x4 + 3x2 − 17x + 11 = Bx4 + 3Bx3 − 5Bx2 − 7Bx −
16 3
x
7
2
+ Dx4 + 2Dx3 + Dx2 + Ex3 + 2Ex +
78 2
x − 64
x
7
7
11 4
Ex − 7 x + 11
+
and then collect the powers of x together:
x4 + 3x2 − 17x + 11 = (B + D −
11
)x4
7
+ (3B + 2D + E −
+ (−5B + D + 2E +
78
)x2
7
16
)x3
7
+ (−7B + E −
64
)x
7
+ 11
From this we equate the coefficients from the right and left hand sides:
Left hand side
Right hand side:
constants terms
11 = 11
64
x-coefficients
− 17 = −7B + E −
7
x2 -coefficients
3 = −5B + D + 2E +
78
7
CHAPTER 7. TECHNIQUES OF INTEGRATION
51
x3 -coefficients
0 = 3B + 2D + E −
x4 -coefficients
1=B+D−
11
7
16
7
Well, we’ll (mostly) stop here, but note that you could easily solve the second
equation for E, in terms of B. Plugging this into the other equations eliminates
E. You could do the same thing with the last equation to eliminate D. This
makes all the equations only involve B. So, you solve for B, and then you know
D and E too. You could also solve this whole set of equations using linear
algebra. In any case, this example just asked for us to set these equations up,
so we’re done now.
For curiosity sake, I’ll show you the final answer, but again, this isn’t
something we were supposed to find this time:
109
x − 195
11 1
5/8
4
56
56
−
+
+
+
.
7 x
x + 1 (x + 1)2 x2 + 2x − 7