52.1 Partial Fractions Concept Overview

PARTIAL FRACTIONS | CONCEPT OVERVIEW The topic of PARTIAL FRACTIONS can be referenced on page 28 and 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

CONCEPT INTRO: When asked to INTEGRATE a RATIONAL EXPRESSION, it is typical to look for the NUMERATOR to be a DERIVATIVE of the DENOMINATOR and then simply substitute and solve. However, when this is not possible, a process referred to as PARTIAL FRACTION DECOMPOSITION must be employed, allowing us to carry forth with any complex integration. This involves DECOMPOSING the rational expression into simpler RATIONAL EXPRESSIONS that can then be ADDED or SUBTRACTED. Most typically, we will encounter a rational fraction comprised of polynomials in the numerator and the denominator…which can be resolved into partial fractions. Generally stating, a rational fraction is given in the form:

𝑓 𝑥 =

𝑝 𝑥 𝑞(𝑥)

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Where 𝑝(𝑥) and 𝑞 𝑥 are polynomials and the degree of 𝑝(𝑥) is smaller than the degree of 𝑞(𝑥). This is important to note, that PARTIAL FRACTIONS can only be derived if the degree of the NUMERATOR is LESS than the degree of the DENOMINATOR. Once this CHECKS out, and it is determined that PARTIAL FRACTIONS can in fact be completed on the rational function, the first step is to tackle the DENOMINATOR. The DENOMINATOR will be factored as much as possible in to an equivalent expression determined by the form of the factor as it is originally presented in the denominator. These equivalent decomposed terms are highlighted in the following table: Original Factor in the Denominator

𝐴 𝑎𝑥 + 𝑏

𝑎𝑥 + 𝑏 𝑎𝑥 + 𝑏

Term(s) in Partial Fraction Decomposition

𝐴! 𝐴! + 𝑎𝑥 + 𝑏 𝑎𝑥 + 𝑏

!

𝑎𝑥 ! + 𝑏𝑥 + 𝑐 𝑎𝑥 ! + 𝑏𝑥 + 𝑥

!

+. . . + !

𝐴! 𝑎𝑥 + 𝑏

!

𝐴𝑥 + 𝐵 𝑎𝑥 ! + 𝑏𝑥 + 𝑐 𝐴! 𝑥 + 𝐵! 𝐴! 𝑥 + 𝐵! 𝐴! 𝑥 + 𝐵! + + ⋯ + 𝑎𝑥 ! + 𝑏𝑥 + 𝑐 𝑎𝑥 ! + 𝑏𝑥 + 𝑐 ! 𝑎𝑥 ! + 𝑏𝑥 + 𝑐

!

Once the rational expression is broken down in to Partial Fractions, we will bring the terms back together, we necessary, using our understanding of the COMMON Made with by Prepineer | Prepineer.com

DENOMINATOR. Doing this will result in equivalent DENOMINATORS on each side of the equation, allowing for us to set the NUMERATORS on each side equal to one another and solve for any unknown values. If consolidation of the DENOMINATOR is not necessary, then we would move forward as stated with the NUMERATORS, determining each of the unknowns one by one. This will be illustrated further through the concept example.

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PARTIAL FRACTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material to be presented. Evaluate the integral: 𝑥! − 𝑥 + 1 𝑑𝑥 𝑥+1 !

A. ln 𝑥 +

! (!!!)

−

! ! !!! !

!

B. ln 𝑥 + 1 + (!!!) − ! !

C. ln 𝑥 + 1 + (!!!) − ! D. ln 𝑥 + 1 +

! !!!

−

+𝐶 !

!!! ! ! !!! ! !

! !!! !

+𝐶 +𝐶 +𝐶

SOLUTION: In this problem, we are given the INTEGRAL: 𝑥! − 𝑥 + 1 𝑑𝑥 𝑥+1 !

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This is quite complex, however, we can confirm that we are indeed dealing with a rational fraction in the form:

𝑓 𝑥 =

𝑝 𝑥 𝑞(𝑥)

Where 𝑝(𝑥) and 𝑞 𝑥 are polynomials and the degree of 𝑝(𝑥) is smaller than the degree of 𝑞(𝑥). Because the degree of the NUMERATOR is LESS than the degree of the DENOMINATOR, we are able to break this function down in to PARTIAL FRACTIONS. Let’s focus in on the DENOMINATOR. We have: 𝑥+1

!

Taking a look at the table of equivalent decomposed expressions, we notice that the form of our DENOMINATOR takes that of the form highlighted in the third row, such that: Original Factor in the Denominator

Term(s) in Partial Fraction Decomposition 𝐴 𝑎𝑥 + 𝑏

𝑎𝑥 + 𝑏

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𝑎𝑥 + 𝑏

𝐴! 𝐴! + 𝑎𝑥 + 𝑏 𝑎𝑥 + 𝑏

!

𝐴! 𝑎𝑥 + 𝑏

!

𝐴𝑥 + 𝐵 𝑎𝑥 ! + 𝑏𝑥 + 𝑐

𝑎𝑥 ! + 𝑏𝑥 + 𝑐 𝑎𝑥 ! + 𝑏𝑥 + 𝑥

+. . . + !

!

𝐴! 𝑥 + 𝐵! 𝐴! 𝑥 + 𝐵! + 𝑎𝑥 ! + 𝑏𝑥 + 𝑐 𝑎𝑥 ! + 𝑏𝑥 + 𝑐

+ ⋯+ !

𝐴! 𝑥 + 𝐵! 𝑎𝑥 ! + 𝑏𝑥 + 𝑐

!

Therefore, our original expression can equivalently be rewritten as: 𝑥! − 𝑥 + 1 𝐴 𝐵 = + 𝑥+1 ! (𝑥 + 1) 𝑥+1

+ !

𝐶 𝑥+1

!

The next step is to consolidate the right side by bringing the terms together with a common denominator. Doing this, we get: 𝑥! − 𝑥 + 1 𝐴 𝑥 + 1 ! + 𝐵 𝑥 + 1 + 𝐶 = 𝑥+1 ! 𝑥+1 ! We will now set the NUMERATORS equal to one another and solve for each of the unknowns, 𝐴, 𝐵, and 𝐶: 𝑥! − 𝑥 + 1 = 𝐴 𝑥 + 1

!

+𝐵 𝑥+1 +𝐶

The first move here is to observe each factor and determine what values we can input to help us knock out each unknown one at a time. Made with

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The first two terms, associated with the unknown variables A and B, will be zero when we input the value -1 for x, therefore: For 𝑥 = −1: −1! + 1 + 1 = 𝐴 −1 + 1

!

+ 𝐵 −1 + 1 + 𝐶

And: 𝐶=3 Let’s plug this value in to our formula such that: 𝑥! − 𝑥 + 1 = 𝐴 𝑥 + 1

!

+𝐵 𝑥+1 +3

Unfortunately, that’s the only value of x that would allow us to completely eliminate an unknown to solve for the other unknowns, therefore, we will need to move forward with creating expressions that we can then solve to determine the final two variables. Let’s start with 𝑥 = 0: 0! − 0 + 1 = 𝐴 0 + 1

!

+𝐵 0+1 +3

Which establishes an expression: 𝐴 + 𝐵 = −2 Made with

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Plugging in 𝑥 = 1 1! − 1 + 1 = 𝐴 1 + 1

!

+𝐵 1+1 +3

Which gives us a second expression: 4𝐴 + 2𝐵 = −2 We now have two equations and two unknowns left, let’s solve. Taking our first expression, we can state that: 𝐴 = −2 − 𝐵 Substituting this in to our second equation we get: 4(−2 − 𝐵) + 2𝐵 = −2 Expanding this, we have: −8 − 4𝐵 + 2𝐵 = −2 Or: −2𝐵 − 8 = −2

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Rearranging and solving for the unknown variable B, we get: 𝐵 = −3 Taking this value, we now just need to plug it in to any of our expressions to determine the value for A. Taking our first expression, we get: 𝐴 − 3 = −2 Rearranging and solving for A, we get: 𝐴=1 Taking these values and plugging them back in to the original decomposed function, we can carry out the integration, giving us: 1 −3 + (𝑥 + 1) 𝑥+1

+ !

3 𝑥+1

!

𝑑𝑥

This creates a simple INTEGRATION, much more doable than that which we were originally given. Since this is a SUMMATION of multiple terms, we will lean on the ability to INTEGRATE each one INDEPENDENTLY and add them back together at the end.

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The final solution is then: 𝑥! − 𝑥 + 1 3 3 𝑑𝑥 = ln 𝑥 + 1 + − 𝑥+1 ! (𝑥 + 1) 2 𝑥 + 1 𝟑

!

+𝐶

The correct answer choice is D. 𝒍𝒏 𝒙 + 𝟏 + (𝒙!𝟏) − 𝟐

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𝟑 𝒙!𝟏 𝟐

+𝑪

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