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Notes for Math 220, Calculus 2

Brenton LeMesurier

Section 3.4 Partial Fractions

References.

Introduction.

The integration of rational functions can be based on a few examples that we already know how to handle:
\begin{align} \int \frac{1}{x-a} dx \amp=\amp \ln|x-a| + C\tag{3.4.1}\\ \int \frac{1}{x^2+a^2} dx \amp=\amp \frac{1}{a} \arctan \frac{x}{a} + C\tag{3.4.2}\\ \int \frac{x}{x^2+a^2} dx \amp=\amp \frac{1}{2} \ln(x^2+a^2) +C\tag{3.4.3}\\ \amp\amp \text{and for } n > 1:\tag{3.4.4}\\ \int \frac{1}{(x-a)^n} dx \amp=\amp \frac{-1}{n-1} \frac{1}{(x-a)^{n-1}} + C\tag{3.4.5}\\ \int \frac{x}{(x^2+a^2)^n} dx \amp=\amp \frac{-1}{2(n-1)} \frac{1}{(x^2+a^2)^{n-1}} + C.\tag{3.4.6}\\ \amp\amp\text{ using substitution } u = x^2 + a^2.\tag{3.4.7}\\ \int \frac{1}{(x^2+a^2)^n} dx \amp\dots\amp \text{ Use substitution } x = a\tan\theta.\tag{3.4.8} \end{align}

Simplifying to a Sum of the Above Forms: Proper Rational Functions.

The rest of the task is simplifying any rational function to a sum of terms of the above forms.
The first stage of simplification is writing a rational function as the sum of a proper rational function and a polynomial. A Proper Rational Function is a ratio of polynomials, \(\displaystyle \frac{P(x)}{Q(x)}\text{,}\) with the top, \(P(x)\text{,}\) of lower degree than the bottom, \(Q(x)\text{.}\) This simplification is done by synthetic division of polynomials.
  • Simplify \(\displaystyle \frac{2x^3+2x+3}{x^2+1}\) to a sum of a polynomial and a proper rational function.
  • Use this to evaluate \(\displaystyle\int \frac{2x^3+2x+3}{x^2+1} dx.\)
The rest of this section deals with integrating the resulting proper rational function.

Case I: Partial Fractions Expansion With All Distinct Linear Factors Below.

The next step is to factorize the denominator, if possible.
We will deal first with the case that the denominator can be factorized, with no factor repeated. That is, the rational function is \(\displaystyle f(x)=\frac{P(x)}{Q(x)}\) with denominator \(Q(x)\) of degree \(n\text{,}\) and \(Q(x)=C(x-r_1)(x-r_2) \cdots (x-r_n),\) with all the roots \(r_i\) different, \(C\) a constant.
Then the wonderful fact is that for certain numbers \(A_1, \dots A_n\text{,}\) the function has a Partial Fractions Expansion: it can be put in the form
\begin{equation} f(x) = \frac{P(x)}{Q(x)} = \frac{A_1}{x-r_1} + \frac{A_2}{x-r_2} +\cdots + \frac{A_n}{x-r_n}\tag{3.4.9} \end{equation}
and then each piece can be integrated using Equation (3.4.1)
But we need to find those coefficients \(A_i\)
The first step is to first clear the denominator in the partial fractions expansion of Equation (3.4.9) by multiplying both sides by all of the factors \((x-r_1) \dots (x-r_n)\text{.}\)
Then several methods can be used.
My favorite is The Method of Strategic Substitution: successively evaluate for each of the roots: \(x = r_1,\, x = r_2,\, \dots,\, x = r_n\text{.}\) This gives a string of simple equations, each giving one of the coefficients \(A_1 , \dots , A_n\text{.}\)
An alternatve is The Method of Equating Coefficients, which however requires solving simultaneous equations. This approach is to expand both sides of the equation into powers of \(x\) and require that the coefficients of each power are the same on both sides.
It gives \(n\) simultaneous equations to be solved for the \(n\) unknown coefficients \(A_1 \dots A_n\text{.}\)
  • Derive the partial fractions expansion of \(\displaystyle \frac{7x+1}{x^2-4}\text{.}\)
  • Use this to evaluate \(\displaystyle \int \frac{7x+1}{x^2-4} dx\text{.}\)
The above strategy fails when a factor in denominator \(Q(x)\) is repeated: for example, we cannot express \(\displaystyle\frac{1}{(x-1)^2}\) as \(\displaystyle \frac{A_1}{x-1} + \frac{A_2}{x-1}\text{.}\)
But nor do we need to, because we can integrate \(\displaystyle\frac{1}{(x-1)^2}\text{,}\) and indeed any term \(\displaystyle\frac{1}{(x-r)^n}\text{,}\) using Equation (3.4.4). So we can work with negative powers of \((x-r)\text{:}\)

Case II: Partial Fractions Expansion With All Linear Factors, Some Repeated.

For \(Q(x) = C(x-r_1)^{p_1}(x-r_2)^{p_2} \cdots (x-r_m)^{p_m}\) we modify the partial fractions form to have a term for each power of \(\displaystyle\frac{1}{x-r_i}\) up to the one appearing in \(Q(x)\text{,}\) \(\displaystyle\frac{1}{(x-r_i)^{p_i}}\text{:}\)
\begin{equation} \frac{P(x)}{Q(x)} = \frac{A_1}{(x-r_1)^{p_1}} + \cdots + \frac{A_{p_1}}{(x-r_1)} + \cdots\tag{3.4.10} \end{equation}
For example, the partial fractions form for \(\displaystyle \frac{2x^2+4x+1}{(x-2)^2(x+3)}\) is
\begin{equation*} \frac{2x^2+4x+1}{(x-2)^2(x+3)} = \frac{A}{(x-2)^2} + \frac{B}{x-2} + \frac{C}{x+3}. \end{equation*}

Determining the Coefficients in the Partial Fractions Expansion with Repeated Linear Factors.

As before, the first step is always to clear the denominator in the partial fractions expansion by multiplying both sides by all of the factors \((x-r_1)^{p_1} \dots (x-r_m)^{p_m}\text{.}\) However, the method of successively evaluating for each of the roots only gets some of the coefficients: one for each different root, so \(m\) of them.

Determining the Rest of the Coefficients with Repeated Linear Factors.

To determine the other coefficients, one can substitute these values into the expansion, and then use one of several methods:
  1. Insert \(n-m\) other simple values for \(x\) like \(x=0\text{,}\) \(x=1\text{,}\) \(x=-1\) etc. to get \(n-m\) simultaneous equations for the remaining \(n-m\) coefficients.
  2. Use the Method of Equating Coefficients described above: expand into various powers of \(x\text{,}\) and require that the coefficient of each power is the same on each side.
    This again gives \(n-m\) simultaneous equations for the remaining coefficients.
In fact, these strategies work for all cases, including those described next.
  • Derive the partial fractions expansion of \(\displaystyle \frac{2x^2+4x+1}{x^3-x^2-x+1}\text{.}\)
  • Use this to evaluate \(\displaystyle \int \frac{2x^2+4x+1}{x^3-x^2-x+1} dx\text{.}\)

Case III: Partial Fractions Expansion With Irreducible Quadratic Factors, None Repeated.

  • Sometimes the denominator \(Q(x)\) cannot be written completely in terms of linear factors given by roots (at least not using only real numbers); for example, if \(Q(x)=x^2+1\text{.}\)
  • But again the integrals in Equations (3.4.2) and (3.4.3) should help.
  • So we seek a partial fractions form using terms like this.

Irreducible Quadratic Factors: \(x^2+bx+c\text{,}\) \(b^2 < 4c\).

  • This situation arises when factorizing the the denominator leads to a quadratic factor \(x^2+bx+c\) that has no real roots because it has negative discriminant, \(b^2-4c < 0\text{.}\)
  • The quadratic formulas then involves the square root of this negative quantity. (Note that there is no need for a coefficient on \(x^2\text{;}\) the denominator and each factor can always be written with coefficient 1 on the highest power of \(x\text{.}\))
  • Another way to characterize such quadratics is that when you complete the square, the form is \((x+k)^2+a^2\text{:}\) in fact \(k=b/2\text{,}\) \(a=\sqrt{c-b^2/4}\text{.}\)
  • The new rule for finding the partial fractions form is that for each irreducible quadratic factor \(x^2+bx+c\text{,}\) the form has a factor
    \begin{equation} \frac{Ax+B}{x^2+bx+c}\tag{3.4.11} \end{equation}
    in addition to the terms as above for each linear factor.

Determining the Coefficients: Same as Above!

As mentioned above, the coefficients can then be determined by the same methods as in Case II of all linear factors, some repeated.
Find the partial fractions form for
\begin{equation*} \frac{2x^2+3x+1}{(x+1)(x-2)^2(x^2+3)(x^2+x+1)}. \end{equation*}

Case IV: Anything Goes (Repeated Irreducible Quadratic Factors, etc.).

Finally, we can deal with the case where anything can happen, including repeated irreducible quadratic factors \((x^2+bx+c)^p\text{,}\) \(b^2 < 4c\text{.}\)
For each such factor, the partial fractions form uses terms that combine the ideas seen above for repeated (linear) factors and irreducible factors:
\begin{equation} \frac{A_1x+B_1}{x^2+bx+c} + \cdots + \frac{A_px+B_p}{(x^2+bx+c)^p}\tag{3.4.12} \end{equation}
Note that, as always, the number of coefficients (here \(2p\)) is the same as the degree of the factor that they go with: here, \((x^2+bx+c)^p, = x^{2p}+ \cdots\text{.}\)

Integrating The Irreducible Quadratic Terms: Substitution \(u = x+k = x+b/2\).

The new type of terms arising in Equation (3.4.12) can be integrated with the help of the substitution \(u=x+k\) where \(k=b/2\text{.}\)
This turns the denominator into the form \(u^2 + a^2\text{,}\) \(a = \sqrt{c-b^2/4}\text{,}\) while the numerators stay linear, so all terms are of the forms in Equations (3.4.2), (3.4.3), (3.4.5) and (3.4.6).

Summary of the Terms Used in the Partial Fractions Form.

To summarize, the partial fractions form for a proper rational function \(\displaystyle\frac{P(x)}{Q(x)}\) is a sum of pieces determined by the factors of the denominator \(Q(x)\) as follows:
  • For a factor \(x-r\) appearing only once, a term of the form
    \begin{equation*} \frac{A}{x-r} \end{equation*}
  • For a repeated factor \((x-r)^p\text{,}\) \(p\) terms of the form
    \begin{equation*} \frac{A_1}{(x-r)^p} + \cdots + \frac{A_p}{x-r} \end{equation*}
  • For an irreducible quadratic factor \(x^2+bx+c\) [\(4c>b^2\)] appearing only once, two coefficients in a term of the form
    \begin{equation*} \frac{Ax+B}{x^2+bx+c} \end{equation*}
  • For an irreducible quadratic factor \(x^2+bx+c\) appearing as \((x^2+bx+c)^p\text{,}\) \(2p\) coefficients in \(p\) terms of the form
    \begin{equation*} \frac{A_1x+B_1}{(x^2+bx+c)^p}+ \cdots + \frac{A_px+B_p}{x^2+bx+c} \end{equation*}

Recap of Methods for Determining the Values of The Coefficients, and Thus the Partial Fractions Expansion.

  • Once you have the Partial Fractions Form, multiply both sides through by the denominator \(Q(x)\text{,}\) to get an equation with polynomials on each side.
  • I then recommend substituting in turn each root of \(Q(x)\) as the value of \(x\text{,}\) which easily determines the values of at least some of the coefficients.
  • If this does not determine all the coefficients, put these values into the above polynomial equation, to reduce the number of unknown coefficients remaining.
  • Insert various other simple values for \(x\) like \(0\text{,}\) \(1\text{,}\) \(-1\text{:}\) one \(x\) value for each coefficient still to be determined, or
  • expand out the polynomials on each side as a sum of different powers of \(x\text{,}\) and write the equations that equate each power of \(x\) on the two sides.
  • Either way, then solve these linear equations for the remaining coefficients.
  • These coefficients can then be inserted into the partial fractions form (the rational function form before clearing the denominator) to express the original integrand \(\displaystyle\frac{P(x)}{Q(x)}\) as a sum of simpler rational functions, each of which can be integrated.

Section Study Guide.

Study Calculus Volume 2, Section 3.4 2 ; in particular
  • The Problem Solving Strategy under heading The General Method
  • Examples 28–34
  • All Checkpoints
  • and one or several exercises from each of the following groups: 182–185, 186& 187, 188–191, 196–200, 202–204, and a few from 207, 209, 210, 211 and 212.
openstax.org/books/calculus-volume-2/pages/3-4-partial-fractions
openstax.org/books/calculus-volume-2/pages/3-4-partial-fractions
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