Difference between revisions of "Partial Fractions"

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Suppose we want to find <math>\int\frac{3x+1}{x^2+x}dx</math> . One way to do this is to simplify the integrand by finding constants <math>A</math> and <math>B</math> so that
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:<math>\frac{3x+1}{x^2+x}=\frac{3x+1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}</math> .
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This can be done by cross multiplying the fraction which gives
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:<math>\frac{3x+1}{x(x+1)}=\frac{A(x+1)+Bx}{x(x+1)}</math>
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As both sides have the same denominator we must have
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:<math>3x+1=A(x+1)+Bx</math>
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This is an equation for <math>x</math> so it must hold whatever value <math>x</math> is. If we put in <math>x=0</math> we get
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<math>A=1</math>
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and putting <math>x=-1</math> gives <math>=-B=-2</math> so <math>B=2</math> .
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So we see that
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:<math>\frac{3x+1}{x^2+x}=\frac{1}{x}+\frac{2}{x+1}</math>
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Returning to the original integral
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:{|
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|<math>\int\frac{3x+1}{x^2+x}dx</math>
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|<math>=\int\frac{dx}{x}+\int\frac{2}{x+1}dx</math>
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|-
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|
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|<math>=\int\frac{dx}{x}+2\int\frac{dx}{x+1}</math>
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|-
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|
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|<math>=\ln|x|+2\ln\Big|x+1\Big|+C</math>
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|}
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Rewriting the integrand as a sum of simpler fractions has allowed us to reduce the initial integral to a sum of simpler integrals. In fact this method works to integrate any rational function.
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==Method of Partial Fractions==
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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To decompose the rational function <math>\frac{P(x)}{Q(x)}</math>:
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*'''Step 1''' Use long division (if necessary) to ensure that the degree of <math>P(x)</math> is less than the degree of <math>Q(x)</math>.
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*'''Step 2''' Factor Q(x) as far as possible.
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*'''Step 3''' Write down the correct form for the partial fraction decomposition (see below) and solve for the constants. 
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</blockquote>
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To factor Q(x) we have to write it as a product of linear factors (of the form <math>ax+b</math>) and irreducible quadratic factors (of the form <math>ax^2+bx+c</math> with <math>b^2-4ac<0</math>).
 +
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Some of the factors could be repeated. For instance if <math>Q(x) = x^3-6x^2+9x</math> we factor <math>Q(x)</math> as
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:<math>Q(x)=x(x^2-6x+9)=x(x-3)(x-3)=x(x-3)^2</math>
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It is important that in each quadratic factor we have <math>b^2-4ac<0</math> , otherwise it is possible to factor that quadratic piece further. For example if <math>Q(x)=x^3-3x^2+2x</math> then we can write
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:<math>Q(x)=x(x^2-3x+2)=x(x-1)(x-2)</math>
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We will now show how to write <math>\frac{P(x)}{Q(x)}</math> as a sum of terms of the form
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:<math>\frac{A}{(ax+b)^k}</math> and <math>\frac{Ax+B}{(ax^2+bx+c)^k}</math>
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Exactly how to do this depends on the factorization of <math>Q(x)</math> and we now give four cases that can occur.
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===Q(x) is a product of linear factors with no repeats===
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This means that <math>Q(x)=(a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)</math> where no factor is repeated and no factor is a multiple of another.
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For each linear term we write down something of the form <math>\frac{A}{(ax+b)}</math> , so in total we write
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:<math>\frac{P(x)}{Q(x)}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{a_2x+b_2}+\cdots+\frac{A_n}{a_nx+b_n}</math>
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====Example Problem====
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Find <math>\int\frac{1+x^2}{(x+3)(x+5)(x+7)}dx</math>
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Here we have <math>P(x)=1+x^2\ ,\ Q(x)=(x+3)(x+5)(x+7)</math> and ''Q(x)'' is a product of linear factors. So we write
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<math>\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{A}{x+3}+\frac{B}{x+5}+\frac{C}{x+7}</math>
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Multiply both sides by the denominator
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<math>1+x^2=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)</math>
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Substitute in three values of ''x'' to get three equations for the unknown constants,
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<math>\begin{matrix} x=-3 & 1+3^2=2\cdot 4 A \\  x=-5 & 1+5^2=-2\cdot 2 B \\ 
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x=-7 & 1+7^2=(-4)\cdot (-2)C \end{matrix}</math>
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so <math>A=\tfrac{5}{4}\ ,\ B=-\tfrac{13}{2}\ ,\ C=\tfrac{25}{4}</math> , and
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<math>\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{5}{4x+12}-\frac{13}{2x+10}+\frac{25}{4x+28}</math>
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We can now integrate the left hand side.
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<math>\int\frac{1+x^2}{(x+3)(x+5)(x+7)}dx=\tfrac{5}{4}\ln\Big|x+3\Big|-\tfrac{13}{2}\ln\Big|x+5\Big|+\tfrac{25}{4}\ln\Big|x+7\Big|+C</math>
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===Exercises===
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Evaluate the following by the method partial fraction decomposition.
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1. <math>\int\frac{2x+11}{(x+6)(x+5)}dx</math>
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2. <math>\int\frac{7x^2-5x+6}{(x-1)(x-3)(x-7)}dx</math>
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===Q(x) is a product of linear factors some of which are repeated===
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If <math>(ax+b)</math> appears in the factorisation of <math>Q(x)</math> ''k''-times then instead of writing the piece <math>\frac{A}{ax+b}</math> we use the more complicated expression
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<math>\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\frac{A_3}{(ax+b)^3}+\cdots+\frac{A_k}{(ax+b)^k}</math>
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====Example 2====
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Find <math>\int\frac{dx}{(x+1)(x+2)^2}</math>
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Here <math>P(x)=1</math> and <math>Q(x)=(x+1)(x+2)^2</math> We write
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<math>\frac{1}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}</math>
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Multiply both sides by the denominator
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<math>1=A(x+2)^2+B(x+1)(x+2)+C(x+1)</math>
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Substitute in three values of <math>x</math> to get 3 equations for the unknown constants,
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<math>\begin{matrix} x=0 & 1=2^2A+2B+C \\  x=-1 & 1=A \\ 
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x=-2 & 1=-C \end{matrix}</math>
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so <math>A=1\ ,\ B=-1\ ,\ C=-1</math> and
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<math>\frac{1}{(x+1)(x+2)^2}=\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{(x+2)^2}</math>
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We can now integrate the left hand side.
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:<math>\int\frac{dx}{(x+1)(x+2)^2}=\ln\left|x+1\right|-\ln\left|x+2\right|+\frac{1}{x+2}+C</math>
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We now simplify the fuction with the property of Logarithms.
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:<math>\ln\left|x+1\right|-\ln\left|x+2\right|+\frac{1}{x+2}+C=\ln\left|\frac{x+1}{x+2}\right|+\frac{1}{x+2}+C</math>
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====Exercise====
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3. Evaluate <math>\int\frac{x^2-x+2}{x(x+2)^2}dx</math> using the method of partial fractions.
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===Q(x) contains some quadratic pieces which are not repeated===
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If <math>ax^2+bx+c</math> appears we use <math>\frac{Ax+B}{ax^2+bx+c}</math> .
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====Exercises====
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Evaluate the following using the method of partial fractions.
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4. <math>\int\frac{2}{(x+2)(x^2+3)}dx</math>
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5. <math>\int\frac{dx}{(x+2)(x^2+2)}</math>
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===Q(x) contains some repeated quadratic factors===
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If <math>ax^2+bx+c</math> appears k-times then use
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:<math>\frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+\frac{A_3x+B_3}{(ax^2+bx+c)^3}+\cdots+\frac{A_kx+B_k}{(ax^2+bx+c)^k}</math>
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====Exercise====
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Evaluate the following using the method of partial fractions.
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6. <math>\int\frac{dx}{(x-1)(x^2+1)^2}</math>
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===Exercies Solutions===
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# <math>\ln\Big|x+6\Big|+\ln\Big|x+5\Big|+C</math>
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# <math>\tfrac{2}{3}\ln\Big|x-1\Big|-\tfrac{27}{4}\ln\Big|x-3\Big|+\tfrac{157}{12}\ln\Big|x-7\Big|+C</math>
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# <math>\frac{\ln\Big|x(x+2)\Big|}{2}+\frac{4}{x+2}+C</math>
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# <math>\ln\left(\sqrt[7]{\frac{(x+2)^2}{x^2+3}}\right)+\frac{4\arctan\Big(\tfrac{x}{\sqrt3}\Big)}{7\sqrt3}+C</math>
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# <math>\ln\left(\sqrt[12]{\frac{(x+2)^2}{x^2+2}}\right)+\frac{\sqrt2\arctan\Big(\tfrac{x}{\sqrt2}\Big)}{6}+C</math>
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# <math>\frac{1-x}{4(x^2+1)}+\tfrac{1}{8}\ln\left(\frac{(x-1)^2}{x^2+1}\right)-\frac{\arctan(x)}{2}+C</math>
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==Resources==
 
[https://youtu.be/8MMALX7mUqQ Integration Using Partial Fraction Decomposition Part 1] by James Sousa
 
[https://youtu.be/8MMALX7mUqQ Integration Using Partial Fraction Decomposition Part 1] by James Sousa
  
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[https://youtu.be/7GcSOjLitF8 Partial Fraction Decompositions] by patrickJMT
 
[https://youtu.be/7GcSOjLitF8 Partial Fraction Decompositions] by patrickJMT
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 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Calculus/Integration_techniques/Partial_Fraction_Decomposition Partial Fraction Decomposition, Wikibooks: Calculus/Integration techniques] under a CC BY-SA license

Latest revision as of 11:09, 29 October 2021

Suppose we want to find . One way to do this is to simplify the integrand by finding constants and so that

.

This can be done by cross multiplying the fraction which gives

As both sides have the same denominator we must have

This is an equation for so it must hold whatever value is. If we put in we get and putting gives so . So we see that

Returning to the original integral

Rewriting the integrand as a sum of simpler fractions has allowed us to reduce the initial integral to a sum of simpler integrals. In fact this method works to integrate any rational function.

Method of Partial Fractions

To decompose the rational function :

  • Step 1 Use long division (if necessary) to ensure that the degree of is less than the degree of .
  • Step 2 Factor Q(x) as far as possible.
  • Step 3 Write down the correct form for the partial fraction decomposition (see below) and solve for the constants.

To factor Q(x) we have to write it as a product of linear factors (of the form ) and irreducible quadratic factors (of the form with ).

Some of the factors could be repeated. For instance if we factor as

It is important that in each quadratic factor we have , otherwise it is possible to factor that quadratic piece further. For example if then we can write

We will now show how to write as a sum of terms of the form

and

Exactly how to do this depends on the factorization of and we now give four cases that can occur.

Q(x) is a product of linear factors with no repeats

This means that where no factor is repeated and no factor is a multiple of another.

For each linear term we write down something of the form , so in total we write

Example Problem

Find

Here we have and Q(x) is a product of linear factors. So we write

Multiply both sides by the denominator

Substitute in three values of x to get three equations for the unknown constants,

so , and

We can now integrate the left hand side.

Exercises

Evaluate the following by the method partial fraction decomposition.

1.

2.

Q(x) is a product of linear factors some of which are repeated

If appears in the factorisation of k-times then instead of writing the piece we use the more complicated expression

Example 2

Find

Here and We write

Multiply both sides by the denominator

Substitute in three values of to get 3 equations for the unknown constants,

so and

We can now integrate the left hand side.

We now simplify the fuction with the property of Logarithms.

Exercise

3. Evaluate using the method of partial fractions.

Q(x) contains some quadratic pieces which are not repeated

If appears we use .

Exercises

Evaluate the following using the method of partial fractions.

4.

5.

Q(x) contains some repeated quadratic factors

If appears k-times then use

Exercise

Evaluate the following using the method of partial fractions.

6.

Exercies Solutions

Resources

Integration Using Partial Fraction Decomposition Part 1 by James Sousa

Integration Using Partial Fraction Decomposition Part 2 by James Sousa

Partial Fraction Decomposition Part 1 (Linear) by James Sousa

Partial Fraction Decomposition - Part 2 of 2 by James Sousa

Partial Fraction Decomposition - Example 1 by patrickJMT

Partial Fraction Decomposition - Example 2 by patrickJMT

Partial Fraction Decomposition - Example 4 by patrickJMT

Partial Fraction Decomposition - Example 5 by patrickJMT

Partial Fraction Decomposition - Example 6 by patrickJMT

Partial Fraction Decompositions by patrickJMT

Licensing

Content obtained and/or adapted from: