TataKelola dan KamSiber Kecerdasan Buatan v022.pdf
17 integrals of rational functions x
1. In this section we will show how to integrate
rational functions, that is, functions of the form
P(x)
Q(x)
where P and Q are polynomials.
Integrals of Rational Functions
2. In this section we will show how to integrate
rational functions, that is, functions of the form
P(x)
Q(x)
where P and Q are polynomials.
Rational Decomposition Theorem
Given reduced P/Q where deg P < deg Q,
then P/Q = F1 + F2 + .. + Fn where
Fi = orA
(ax + b)k
Ax + B
(ax2 + bx + c)k
and that (ax + b)k or (ax2 + bx + c)k are
factors of Q(x).
Integrals of Rational Functions
3. Therefore finding the integrals of rational
functions is reduced to finding integrals of
orA
(ax + b)k
Ax + B
(ax2 + bx + c)k
Integrals of Rational Functions
4. Therefore finding the integrals of rational
functions is reduced to finding integrals of
orA
(ax + b)k
Ax + B
(ax2 + bx + c)k
Integrals of Rational Functions
The integrals ∫ are straight forward
with the substitution method by setting
u = ax + b.
dx
(ax + b)k
5. Therefore finding the integrals of rational
functions is reduced to finding integrals of
orA
(ax + b)k
Ax + B
(ax2 + bx + c)k
Integrals of Rational Functions
The integrals ∫ are straight forward
with the substitution method by setting
u = ax + b.
dx
(ax + b)k
To integrate ∫ dx, we need to
complete the square of the denominator.
Ax + B
(ax2 + bx + c)k
7. Complete the square of x2 + 6x + 10 – it’s irreducible.
Integrals of Rational Functions
x
x2 + 6x + 10Example A. Find ∫ dx
8. x2 + 6x + 10 = (x2 + 6x ) + 10
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
9. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
10. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
11. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
x
x2 + 6x + 10∫ dx
= ∫
x
(x + 3)2 + 1
dx
Hence
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
12. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
x
x2 + 6x + 10∫ dx
Set u = x + 3
= ∫
x
(x + 3)2 + 1
dx
substitution
Hence
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
13. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
x
x2 + 6x + 10∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x
(x + 3)2 + 1
dx
substitution
Hence
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
14. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
x
x2 + 6x + 10∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x
(x + 3)2 + 1
dx
substitution
= ∫
u – 3
u2 + 1
du
Hence
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
15. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
x
x2 + 6x + 10∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x
(x + 3)2 + 1
dx
substitution
= ∫
u – 3
u2 + 1
du
= ∫
u
u2 + 1 du – 3 ∫
1
u2 + 1 du
Hence
Integrals of Rational Functions
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
16. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
x
x2 + 6x + 10∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x
(x + 3)2 + 1
dx
substitution
= ∫
u – 3
u2 + 1
du
= ∫
u
u2 + 1 du – 3 ∫
1
u2 + 1 du
Hence
Integrals of Rational Functions
Set w = u2 + 1
substitution
Complete the square of x2 + 6x + 10 – it’s irreducible.
x
x2 + 6x + 10Example A. Find ∫ dx
17. Set w = u2 + 1
= 2u
substitution
dw
du
Integrals of Rational Functions
= ∫
u
u2 + 1 du – 3 ∫
1
u2 + 1 du Set w = u2 + 1
18. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
Integrals of Rational Functions
= ∫
u
u2 + 1 du – 3 ∫
1
u2 + 1 du Set w = u2 + 1
19. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
= ∫
u
w
– 3 ∫
1
u2 + 1 du
dw
2u
Integrals of Rational Functions
= ∫
u
u2 + 1 du – 3 ∫
1
u2 + 1 du Set w = u2 + 1
20. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
= ∫
u
w
– 3 ∫
1
u2 + 1 du
dw
2u
= ½ ∫ 1
w – 3 tan-1(u) + cdw
Integrals of Rational Functions
21. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
= ∫
u
w
– 3 ∫
1
u2 + 1 du
dw
2u
= ½ ∫ 1
w – 3 tan-1(u) + cdw
= ½ Ln(w) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
22. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
= ∫
u
w
– 3 ∫
1
u2 + 1 du
dw
2u
= ½ ∫ 1
w – 3 tan-1(u) + cdw
= ½ Ln(lwl) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
23. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
= ∫
u
w
– 3 ∫
1
u2 + 1 du
dw
2u
= ½ ∫ 1
w – 3 tan-1(u) + cdw
= ½ Ln(lwl) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
3x
x2 + 7x + 10Example: Find ∫ dx
24. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
= ∫
u
w
– 3 ∫
1
u2 + 1 du
dw
2u
= ½ ∫ 1
w – 3 tan-1(u) + cdw
= ½ Ln(lwl) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
Since x2 + 7x + 10 = (x + 2)(x + 5), we can
decompose the rational expression.
3x
x2 + 7x + 10Example: Find ∫ dx
25. Set w = u2 + 1
= 2u
substitution
dw
du
du =
dw
2u
= ∫
u
w
– 3 ∫
1
u2 + 1 du
dw
2u
= ½ ∫ 1
w – 3 tan-1(u) + cdw
= ½ Ln(lwl) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
3x
x2 + 7x + 10Example: Find ∫ dx
Since x2 + 7x + 10 = (x + 2)(x + 5), we can
decompose the rational expression.
Specifically 3x
(x + 2)(x + 5) =
A
(x + 2)
+ B
(x + 5)
26. Integrals of Rational Functions
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
28. Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
29. Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
30. Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
31. Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
32. Integrals of Rational Functions
Hence 3x
(x + 2)(x + 5) =
-2
(x + 2)
+ 5
(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
33. Integrals of Rational Functions
Hence x
(x + 2)(x + 5) =
-2
(x + 2)
+ 5
(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
3x
x2 + 7x + 10So ∫ dx
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
34. Integrals of Rational Functions
Hence x
(x + 2)(x + 5) =
-2
(x + 2)
+ 5
(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
3x
x2 + 7x + 10So ∫ dx
= ∫ dx
-2
(x + 2)
+ 5
(x + 5)
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
35. Integrals of Rational Functions
Hence x
(x + 2)(x + 5) =
-2
(x + 2)
+ 5
(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
3x
x2 + 7x + 10So ∫ dx
= -2Ln(lx + 2l) + 5Ln(lx + 5l) + c
= ∫ dx
-2
(x + 2)
+ 5
(x + 5)
Clear denominators: 3x
(x + 2)(x + 5) =
A
(x + 2)
+
B
(x + 5)
37. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
1
(x2 + 1)2xExample: Find ∫ dx
38. Integrals of Rational Functions
1
(x2 + 1)2xExample: Find ∫ dx
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
39. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
1
(x2 + 1)2xExample: Find ∫ dx
40. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
1
(x2 + 1)2xExample: Find ∫ dx
41. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
1
(x2 + 1)2xExample: Find ∫ dx
42. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
There is no x4-term,
1
(x2 + 1)2xExample: Find ∫ dx
43. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0,
1
(x2 + 1)2xExample: Find ∫ dx
44. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
1
(x2 + 1)2xExample: Find ∫ dx
45. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term,
1
(x2 + 1)2xExample: Find ∫ dx
46. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term, hence Bx3 = 0,
1
(x2 + 1)2xExample: Find ∫ dx
47. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term, hence Bx3 = 0, so B = 0
1
(x2 + 1)2xExample: Find ∫ dx
48. Integrals of Rational Functions
1
(x2 + 1)2x =
Ax + B
(x2 + 1)
+ Cx + D
(x2 + 1)2 + E
x
Clear denominators
1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term, hence Bx3 = 0, so B = 0
So we've1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
1
(x2 + 1)2xExample: Find ∫ dx
49. Integrals of Rational Functions
There is no x-term in
1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
50. Integrals of Rational Functions
There is no x-term in
1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0.
51. Integrals of Rational Functions
There is no x-term in
1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2
(x2 + 1)+Cx2 +1(x2 + 1)2
52. Integrals of Rational Functions
There is no x-term in
1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2
(x2 + 1)+Cx2 +1(x2 + 1)2
Expand this
53. Integrals of Rational Functions
There is no x-term in
1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2
(x2 + 1)+Cx2 +1(x2 + 1)2
Expand this
1 = -x4 – x2
Cx2 + x4 + 2x2 + 1+
54. Integrals of Rational Functions
There is no x-term in
1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2
(x2 + 1)+Cx2 +1(x2 + 1)2
Expand this
1 = -x4 – x2
Cx2 + x4 + 2x2 + 1+
Hence Cx2 + x2 = 0 or C = -1
55. Integrals of Rational Functions
There is no x-term in
1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2
(x2 + 1)+Cx2 +1(x2 + 1)2
Expand this
1 = -x4 – x2
Cx2 + x4 + 2x2 + 1+
Hence Cx2 + x2 = 0 or C = -1
Put it all together
1
(x2 + 1)2x =
-x
(x2 + 1)
+ -x
(x2 + 1)2 + 1
x
61. Integrals of Rational Functions
1
(x2 + 1)2xTherefore ∫ dx
= ∫
-x dx
(x2 + 1)
+ -x dx
(x2 + 1)2 + dx
x∫ ∫
substitution
set u = x2 + 1
du
dx = 2x
du
2xdx == ∫
-x
u
du
2x
+ ∫
-x
u2
du
2x
+ Ln(lxl)
62. Integrals of Rational Functions
1
(x2 + 1)2xTherefore ∫ dx
= ∫
-x dx
(x2 + 1)
+ -x dx
(x2 + 1)2 + dx
x∫ ∫
substitution
set u = x2 + 1
du
dx = 2x
du
2xdx == ∫
-x
u
du
2x
+ ∫
-x
u2
du
2x
+ Ln(lxl)
= - ½ ∫
du
u + Ln(lxl)½ ∫
du
u2–
63. Integrals of Rational Functions
1
(x2 + 1)2xTherefore ∫ dx
= ∫
-x dx
(x2 + 1)
+ -x dx
(x2 + 1)2 + dx
x∫ ∫
substitution
set u = x2 + 1
du
dx = 2x
du
2xdx == ∫
-x
u
du
2x
+ ∫
-x
u2
du
2x
+ Ln(x)
= - ½ ∫
du
u + Ln(x)½ ∫
du
u2–
= - ½ Ln(u) + ½ u-1 + Ln(x) + c
64. Integrals of Rational Functions
1
(x2 + 1)2xTherefore ∫ dx
= ∫
-x dx
(x2 + 1)
+ -x dx
(x2 + 1)2 + dx
x∫ ∫
substitution
set u = x2 + 1
du
dx = 2x
du
2xdx == ∫
-x
u
du
2x
+ ∫
-x
u2
du
2x
+ Ln(x)
= - ½ ∫
du
u + Ln(x)½ ∫
du
u2–
= - ½ Ln(u) + ½ u-1 + Ln(x) + c
= - ½ Ln(x2 + 1) + + Ln(x) + c1
2(x2 + 1)