X20 integrals of rational functions

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.


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 = 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).


Therefore finding the integrals of rational

functions are reduced to finding integrals of

or A

(ax + b)kAx + B

(ax2 + bx + c)k




or A

(ax + b)kAx + B

(ax2 + bx + c)k


The integrals ∫ are straight forward

with the substitution method by setting

u = ax + b.

dx(ax + b)k



or A

(ax + b)kAx + B

(ax2 + bx + c)k


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 for the denominator.

Ax + B(ax2 + bx + c)k

xx2 + 6x + 10 Example: Find ∫ dx


x2 + 6x + 10 - its irreducible.Complete the square on

Integrals of Rational Functionsx

x2 + 6x + 10 Example: Find ∫ dx


x2 + 6x + 10 = (x2 + 6x ) + 10




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9

= (x + 3)2 + 1




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9

= (x + 3)2 + 1

xx2 + 6x + 10 ∫ dx

= ∫x

(x + 3)2 + 1 dx

Hence




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9

= (x + 3)2 + 1

xx2 + 6x + 10 ∫ dx

Set u = x + 3

= ∫x

(x + 3)2 + 1 dx

substitutionHence




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9

= (x + 3)2 + 1

xx2 + 6x + 10 ∫ dx

Set u = x + 3

x = u – 3

dx = du= ∫

x(x + 3)2 + 1

dx

substitutionHence




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9

= (x + 3)2 + 1

xx2 + 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




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9

= (x + 3)2 + 1

xx2 + 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




x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9

= (x + 3)2 + 1

xx2 + 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


Set w = u2 + 1

substitution

xx2 + 6x + 10 Example: Find ∫ dx

Set w = u2 + 1

= 2u

substitution

dwdu


Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u


Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u

= ∫u w

– 3 ∫1

u2 + 1 dudw2u


Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u

= ∫u w

– 3 ∫1

u2 + 1 dudw2u

= ½ ∫1 w – 3 tan-1(u) + c dw


Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u

= ∫u w

– 3 ∫1

u2 + 1 dudw2u

= ½ ∫1 w – 3 tan-1(u) + c dw

= ½ Ln(w) – 3 tan-1(x + 3) + c


Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u

= ∫u w

– 3 ∫1

u2 + 1 dudw2u

= 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


Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u

= ∫u w

– 3 ∫1

u2 + 1 dudw2u

= 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


3xx2 + 7x + 10 Example: Find ∫ dx

Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u

= ∫u w

– 3 ∫1

u2 + 1 dudw2u

= 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


Since x2 + 7x + 10 = (x + 2)(x + 5), we can

decompose the rational expression.


Set w = u2 + 1

= 2u

substitution

dwdu

du =dw2u

= ∫u w

– 3 ∫1

u2 + 1 dudw2u

= 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



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)


Clear denominators: 3x(x + 2)(x + 5) =

A(x + 2)

+B

(x + 5)


3x = A(x + 5) + B(x + 2)


A(x + 2)

+B

(x + 5)


3x = A(x + 5) + B(x + 2)

Evaluate at x = -5, we get -15 = -3B


A(x + 2)

+B

(x + 5)


3x = A(x + 5) + B(x + 2)

Evaluate at x = -5, we get -15 = -3B so B = 5


A(x + 2)

+B

(x + 5)


3x = A(x + 5) + B(x + 2)


Evaluate at x = -2, we get -6 = 3A


A(x + 2)

+B

(x + 5)


3x = A(x + 5) + B(x + 2)


Evaluate at x = -2, we get -6 = 3A so A = -2


A(x + 2)

+B

(x + 5)


Hencex

(x + 2)(x + 5) =-2

(x + 2) +

5(x + 5)

3x = A(x + 5) + B(x + 2)




A(x + 2)

+B

(x + 5)


Hencex

(x + 2)(x + 5) =-2

(x + 2) +

5(x + 5)

3x = A(x + 5) + B(x + 2)



3xx2 + 7x + 10 So ∫ dx


A(x + 2)

+B

(x + 5)


Hencex

(x + 2)(x + 5) =-2

(x + 2) +

5(x + 5)

3x = A(x + 5) + B(x + 2)



3xx2 + 7x + 10 So ∫ dx

= ∫ dx -2

(x + 2) +

5(x + 5)


A(x + 2)

+B

(x + 5)


Hencex

(x + 2)(x + 5) =-2

(x + 2) +

5(x + 5)

3x = A(x + 5) + B(x + 2)



3xx2 + 7x + 10 So ∫ dx

= -2Ln(x + 2) + 5Ln(x + 5) + c

= ∫ dx -2

(x + 2) +

5(x + 5)


A(x + 2)

+B

(x + 5)


1(x2 + 1)2x Example: Find ∫ dx


1(x2 + 1)2x =

Ax + B(x2 + 1)

+Cx + D(x2 + 1)2 +

Ex




1(x2 + 1)2x =

Ax + B(x2 + 1)

+Cx + D(x2 + 1)2 +

Ex

Clear denominators


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 =

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 =

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


Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2



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



There is no x4-term,



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



There is no x4-term, hence Ax4 + x4 = 0,



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



There is no x4-term, hence Ax4 + x4 = 0, so A = -1



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




There is no x3-term,



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




There is no x3-term, hence Bx3 = 0,



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




There is no x3-term, hence Bx3 = 0, so B = 0



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




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



There is no x-term in

1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2



1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2

Hence Dx = 0, or D = 0.



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



1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2


1 = -x2 (x2 + 1) + Cx2 +1(x2 + 1)2

Expand this



1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2


1 = -x2 (x2 + 1) + Cx2 +1(x2 + 1)2

Expand this

1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+



1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2


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



1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2


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


1(x2 + 1)2x Therefore ∫ dx

= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫



= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫

substitution

set u = x2 + 1



= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫

substitution

set u = x2 + 1 dudx = 2x



= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫

substitution


du2x

dx =



= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫

substitution


du2x

dx = = ∫-x u

du2x

+ ∫-x u2

du2x

+ Ln(x)



= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫

substitution


du2x

dx = = ∫-x u

du2x

+ ∫-x u2

du2x

+ Ln(x)

= - ½ ∫duu + Ln(x)½ ∫

duu2–



= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫

substitution


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



= ∫-x dx

(x2 + 1) +

-x dx (x2 + 1)2 +

dxx ∫ ∫

substitution


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) + c1

2(x2 + 1)

X20 integrals of rational functions

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