20 integrals of rational functions
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Integrals of Rational Functions
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
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 TheoremGiven reduced P/Q where deg P < deg Q, then P/Q = F1 + F2 + .. + Fn where
Fi = or A
(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
Therefore finding the integrals of rational functions are reduced to finding integrals of or A
(ax + b)k Ax + B
(ax2 + bx + c)k
Integrals of Rational Functions
Therefore finding the integrals of rational functions are reduced to finding integrals of or A
(ax + b)k Ax + B
(ax2 + bx + c)k
Integrals of Rational Functions
The integrals ∫ are straight forward
with the substitution method by settingu = ax + b.
dx (ax + b)k
Therefore finding the integrals of rational functions are reduced to finding integrals of or A
(ax + b)k Ax + B
(ax2 + bx + c)k
Integrals of Rational Functions
The integrals ∫ are straight forward
with the substitution method by settingu = ax + b.
dx (ax + b)k
To integrate ∫ dx, we need to
complete the square for the denominator.
Ax + B(ax2 + bx + c)k
xx2 + 6x + 10
Example: Find ∫
dx
Integrals of Rational Functions
x2 + 6x + 10 - its irreducible.Complete the square on
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x ) + 10
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10
∫
dx
= ∫
x(x + 3)2 + 1
dx
Hence
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10
∫
dxSet u = x + 3 = ∫
x(x + 3)2 + 1
dx
substitution
Hence
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10
∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
Hence
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10
∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
Hence
Integrals of Rational Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10
∫
dxSet 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 Functionsx
x2 + 6x + 10
Example: Find ∫
dx
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10
∫
dxSet 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
xx2 + 6x + 10
Example: Find ∫
dx
Set w = u2 + 1
= 2u
substitution
dwdu
Integrals of Rational Functions
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
Integrals of Rational Functions
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
Integrals of Rational Functions
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
Integrals of Rational Functions
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
3xx2 + 7x + 10
Example: Find ∫
dx
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 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.
3xx2 + 7x + 10
Example: Find ∫
dx
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
3xx2 + 7x + 10
Example: 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)
Integrals of Rational Functions
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2) +
B(x + 5)
Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2) +
B(x + 5)
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)
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)
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)
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)
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
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2) +
B(x + 5)
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
3xx2 + 7x + 10
So ∫ dx
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2) +
B(x + 5)
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
3xx2 + 7x + 10
So ∫ dx
= ∫ dx -2
(x + 2) + 5
(x + 5)
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2) +
B(x + 5)
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
3xx2 + 7x + 10
So ∫ dx
= -2Ln(x + 2) + 5Ln(x + 5) + c
= ∫ dx -2
(x + 2) + 5
(x + 5)
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2) +
B(x + 5)
Integrals of Rational Functions1
(x2 + 1)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
1(x2 + 1)2x Example: Find ∫
dx
Integrals of Rational Functions1
(x2 + 1)2x Example: Find ∫
dx
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
Clear denominators
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
Clear denominators
1 = (Ax + B)(x2 + 1)
x + (Cx + D) x +E (x2 + 1)2
1(x2 + 1)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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)2x Example: Find ∫
dx
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+ Cx + D(x2 + 1)2
+ Ex
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've 1 = -x (x2 + 1)
x + (Cx + D) x +1(x2 + 1)2
1(x2 + 1)2x Example: Find ∫
dx
Integrals of Rational FunctionsThere is no x-term in
1 = -x (x2 + 1)
x + (Cx + D) x +1(x2 + 1)2
Integrals of Rational FunctionsThere is no x-term in
1 = -x (x2 + 1)
x + (Cx + D) x +1(x2 + 1)2 Hence Dx = 0, or D = 0.
Integrals of Rational FunctionsThere 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
Integrals of Rational FunctionsThere 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
Integrals of Rational FunctionsThere 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+
Integrals of Rational FunctionsThere 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
Integrals of Rational FunctionsThere 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
+ 1x
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
substitution
set u = x2 + 1
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
substitution
set u = x2 + 1 dudx = 2x
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
substitution
set u = x2 + 1 dudx = 2x
du2x dx =
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
substitution
set u = x2 + 1 dudx = 2x
du2x dx = = ∫
-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
substitution
set u = x2 + 1 dudx = 2x
du2x dx = = ∫
-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
= - ½ ∫
duu + Ln(x)½ ∫
duu2 –
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
substitution
set u = x2 + 1 dudx = 2x
du2x dx = = ∫
-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
= - ½ ∫
duu + Ln(x)½ ∫
duu2 –
= - ½ Ln(u) + ½ u-1 + Ln(x) + c
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫
dx
= ∫
-x dx(x2 + 1)
+ -x dx (x2 + 1)2
+ dxx ∫
∫
substitution
set u = x2 + 1 dudx = 2x
du2x dx = = ∫
-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
= - ½ ∫
duu + Ln(x)½ ∫
duu2 –
= - ½ Ln(u) + ½ u-1 + Ln(x) + c = - ½ Ln(x2 + 1) + + Ln(x) + c
1
2(x2 + 1)