Difference between revisions of "Partial fractions in integration"

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'''Integration by partial fractions''' is a [[Techniques of integration|technique]] in [[Calculus]] to facilitate the integration of a rational expression by partial fraction decomposition.
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{{Math-h}}
  
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'''Integration by partial fractions''' is a [[Techniques of integration|technique]] in [[Calculus]] to facilitate the [[integration]] of a rational expression by partial fraction decomposition. Given an integral
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Given an integral
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:<math>\int \frac {f(x)}{g(x)}dx</math>
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<br />
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where <math>f(x)</math> and <math>g(x)</math> are both [[polynomial]]s, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.
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<math>\int \frac {f(x)}{g(x)}dx</math>
 
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<br />
 
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where <math>f(x)</math> and <math>g(x)</math> are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.
 
  
 
==Integration by Partial Fractions==
 
==Integration by Partial Fractions==
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=== A 1st-Degree Denominator ===
 
=== A 1st-Degree Denominator ===
 
Given:
 
Given:
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<br /><math>\int \frac{1}{ax + b}dx</math>
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:<math>\int \frac{1}{ax + b}dx</math>
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<br />Substitute <math>u = ax + b</math>
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Substitute <math>u = ax + b</math>
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<br /><math>= \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C</math>
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:<math>= \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C</math>
  
 
This means that if given an integral such as:
 
This means that if given an integral such as:
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<br /><math>\int \frac{8}{3x+13}dx</math>
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:<math>\int \frac{8}{3x+13}dx</math>
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<br />The steps can be skipped by using the general formula above get:
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The steps can be skipped by using the general formula above get:
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<br /><math>= \frac{8}{3} \ln{|3x+13|}+C</math>
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:<math>= \frac{8}{3} \ln{|3x+13|}+C</math>
 
=== A Repeated 1st-Degree Denominator ===
 
=== A Repeated 1st-Degree Denominator ===
 
The formula for integrals where a first degree polynomial denominator is raised to a power greater than one is much different than the formula above.
 
The formula for integrals where a first degree polynomial denominator is raised to a power greater than one is much different than the formula above.
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Given
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Given:
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<br /><math>\int \frac{1}{(ax+b)^k}dx</math>
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:<math>\int \frac{1}{(ax+b)^k}dx</math>
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<br /><math>u = ax+b</math>
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:<math>u = ax+b</math>
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<br /><math>= \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-k} + C = {-1 \over kau^{k-1}} + C = {-1 \over {ka (ax+b)^{k-1}}} + C</math>
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:<math>= \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k)} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-(k-1)} + C = {-1 \over (k-1)au^{k-1}} + C = {-1 \over {(k-1)a (ax+b)^{k-1}}} + C</math>
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<br />Note that the above formula '''only''' works if <math>k>1</math>. It does not matter if <math>k</math> is a fraction or even irrational, just as long as it is '''greater than 1'''.
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Note that the above formula '''only''' works if <math>k \neq 1</math>.
 
<br />This means that integrals such as
 
<br />This means that integrals such as
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<br /><math>\int {5 \over {(3x+17)^{12}}}dx</math>
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:<math>\int {5 \over {(3x+17)^{12}}}dx</math>
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<br />are now very easy:
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are now very easy:
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<br /><math>= {-5 \over {(3)(12)(3x+17)^{11}}}+C = {-5 \over {36(3x+17)^{11}}}+C</math>
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:<math>= {-5 \over {(3)(11)(3x+17)^{11}}}+C = {-5 \over {33(3x+17)^{11}}}+C</math>
  
 
== A 2nd-Degree Denominator ==
 
== A 2nd-Degree Denominator ==
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:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>
 
:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>
  
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Therefore the problem can be restated,
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Therefore, the problem can be restated,
  
 
:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
 
:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
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:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>
 
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>
  
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==See Also==
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==See also==
 
*[[Calculus]]
 
*[[Calculus]]
 
*[[Integral]]
 
*[[Integral]]
 
*[[Methods of integration]]
 
*[[Methods of integration]]
  
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[[Category:Mathematics]]
 
 
[[Category:Calculus]]
 
[[Category:Calculus]]
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[[Category:Integration]]

Latest revision as of 18:11, December 19, 2018

This article/section deals with mathematical concepts appropriate for late high school or early college.

Integration by partial fractions is a technique in Calculus to facilitate the integration of a rational expression by partial fraction decomposition. Given an integral

where and are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.

Integration by Partial Fractions

A 1st-Degree Denominator

These are a few methods of solving integrals with first degree denominators.

A 1st-Degree Denominator

Given:

Substitute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = ax + b}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C}

This means that if given an integral such as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{8}{3x+13}dx}

The steps can be skipped by using the general formula above get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{8}{3} \ln{|3x+13|}+C}

A Repeated 1st-Degree Denominator

The formula for integrals where a first degree polynomial denominator is raised to a power greater than one is much different than the formula above. Given:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{(ax+b)^k}dx}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = ax+b}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k)} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-(k-1)} + C = {-1 \over (k-1)au^{k-1}} + C = {-1 \over {(k-1)a (ax+b)^{k-1}}} + C}

Note that the above formula only works if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \neq 1} .
This means that integrals such as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int {5 \over {(3x+17)^{12}}}dx}

are now very easy:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = {-5 \over {(3)(11)(3x+17)^{11}}}+C = {-5 \over {33(3x+17)^{11}}}+C}

A 2nd-Degree Denominator

2nd-Degree Denominators get more complicated, especially with those that do not factor.

A Reducible 2nd-Degree Polynomial Denominator

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx}

The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}}

This allows the denominator to split the fraction in to sums by cross multiplying the denominators,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}}

Therefore, the problem can be restated,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3x+11=A(x+2)+B(x-3)}

Now it is possible to solve for A and B by substituting x with a value that allows the term to go to 0. For example,

Let :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-2} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3(-2)+11=A(-2+2)+B(-2-3)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 5=A(0)+B(5)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=-1}

Let :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3(3)+11=A(3+2)+B(3-3)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 20=A(5)+B(0)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=4}

Then plug in the values of A and B and get,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4}{x-3}-\frac{1}{x+2}}

Now solve the integral.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c}

See also

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