a lecture on the partial fraction decomposition method

1 Integrating Rational Functions

A rational function is a function of the form$y=\frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials(with real coefficients). Here we are interested in how to solvethe integral

$$\int \frac{p(x)}{q(x)}𝑑x$$

We already know how to integrate some functions of this type. Aswe know, the chain rule tells us that the derivative of $y=\ln (g(x))$ is $y^{\prime }=\frac{g^{\prime }(x)}{g(x)}$, where $g(x)$ is any otherfunction. Therefore:

${\displaystyle \int \frac{g^{\prime }(x)}{g(x)}𝑑x}=\ln |g(x)|+C.$(1)

Example 1.1.

$$\int \frac{2x+1}{x^2+x+1}𝑑x=\ln |x^2+x+1|+C.$$

Example 1.2.

Equation (1) may also be used to integrate any function ofthe form $y=\frac{a}{bx+c}$, for any constants $a,b,c$. Indeed:

$$\int \frac{a}{bx+c}𝑑x=\frac{a}{b}\int \frac{b}{bx+c}𝑑x=\frac{a}{b}\ln |bx+c|+C$$

or alternatively use asubstitution $u=bx+c$.

Example 1.3.

Using a substitution we may also integrate any function of theform $y=\frac{a}{{(bx+c)}^n}$, namely $u=bx+c$ does the job. Forexample, using $u=2x+1$, $du=2dx$:

$$\int \frac{3}{{(2x+1)}^4}𝑑x=\frac{3}{2}\int \frac{1}{u^4}𝑑u=-\frac{1}{2}\frac{1}{u^3}+C=-\frac{1}{2{(2x+1)}^3}+C.$$

Example 1.4.

The derivative of the arc tangent function, $y=\arctan x$, is$y^{\prime }=\frac{1}{1+x^2}$. Therefore:

$$\int \frac{1}{1+x^2}𝑑x=\arctan x+C.$$

Thus, for any constant $a$, using a substitution $x=au$ one mayintegrate

$$\int \frac{1}{a^2+x^2}𝑑x=\frac{1}{a}\arctan \frac{x}{a}+C.$$

You may also use the arctangent function to integrate otherfunctions, by completing the square of the denominator. Forexample, $x^2+2x+2=1+{(x+1)}^2$, thus:

$$\int \frac{1}{x^2+2x+2}𝑑x=\int \frac{1}{1+{(x+1)}^2}𝑑x=\arctan (x+1)+C.$$

Example 1.5.

The arctangent also allows us to integrate another family offunctions, namely:

$$\int \frac{bx+c}{a^2+x^2}𝑑x.$$

The trick is to use the favorite strategy of Napoleon: divideand conquer, i.e. we break the fraction into a sum of two (herewe pick $a=1$ for simplicity):

$$\int \frac{bx+c}{1+x^2}𝑑x=\int \frac{bx}{1+x^2}𝑑x+\int \frac{c}{1+x^2}𝑑x=\frac{b}{2}\ln |1+x^2|+c\arctan x+C.$$

2 Partial Fraction Decomposition

The objective of this method is to reduce any integral of the type$\int \frac{p(x)}{q(x)}𝑑x$ to a sum of integrals of the typesdescribed in the previous sections. For example:

Example 2.1.

We would like to solve the following integral:

$$\int \frac{2}{x^2+3x-4}𝑑x.$$

First, we factor the denominator:

$$x^2+3x-4=(x+4)(x-1)$$

In orderto integrate, we are going to express the fraction in theintegrand as a sum:

$$\frac{2}{x^2+3x-4}=\frac{A}{x+4}+\frac{B}{x-1}$$

for some constants $A,B$ to be determined. The right hand side(after realizing a common denominator) adds up to$\frac{A(x-1)+B(x+4)}{(x+4)(x-1)}$. Therefore, in order to have anequality we need the numerators to be equal:

$A(x-1)+B(x+4)=2$(2)

for all values of $x$ (i.e. thisshould be an equality of polynomials). Thus, we can substitutevalues of $x$ and obtain equations relating $A$ and $B$ andhopefully we’ll be able to determine the value of the constants.The easiest values to pick are the roots of the denominator of theoriginal fraction. In this case, when we plug $x=1$ in Eq.(2) we get $5B=2$, and so $B=2/5$. When we plug $x=-4$ weobtain $-5A=2$ and so $A=-2/5$, and we are done! Now we can finishthe integral:

${\displaystyle \int \frac{2}{x^2+3x-4}𝑑x}$$=$${\displaystyle \int \frac{-2/5}{x+4}𝑑x}+{\displaystyle \int \frac{2/5}{x-1}𝑑x}$

$=$$-{\displaystyle \frac{2}{5}}\ln |x+4|+{\displaystyle \frac{2}{5}}\ln |x-1|+C={\displaystyle \frac{2}{5}}\ln |{\displaystyle \frac{x-1}{x+4}}|+C.$

The method. The goal of the method, as explained above, isto express any fraction $\frac{p(x)}{q(x)}$ as the sum of partial fractions of the types discussed in the previous section.

1. If the degree of $p(x)$ is larger than the degree of $q(x)$then use polynomial division to divide and obtain a quotient$t(x)$ and remainder $r(x)$ polynomials such that$p(x)=q(x)t(x)+r(x)$. Thus

   $$\frac{p(x)}{q(x)}=t(x)+\frac{r(x)}{q(x)}$$

   where the degree of $r$ is lower than the degree of $q$. Now usethe partial fraction decomposition with $r(x)/q(x)$.

2. Factor the denominator, $q(x)$, into irreducible polynomials(over $ℝ$). Thus, we may express $q(x)$ as a product of linearpolynomials (perhaps to a power other than $1$) and irreduciblequadratic polynomials.

3. If a linear factor $(x-a)$ to the first power appearsin the denominator of $q(x)$, the partial fraction decompositionshould have a term $\frac{A}{(x-a)}$, for some constant $A$ to bedetermined.

4. If a linear factor to the nth power, say ${(x-b)}^n$appears in the denominator of $q(x)$, the partial fractiondecomposition should have terms

   $$\frac{B_1}{(x-b)}+\frac{B_2}{{(x-b)}^2}+\mathrm{\dots }+\frac{B_n}{{(x-b)}^n}$$

   for some$n$ constants $B_1,B_2,\mathrm{\dots },B_n$ to be determined.

5. If a quadratic polynomial $a{x}^2+bx+c$ to the $n$th power appears in thefactorization of $q(x)$, i.e. ${(a{x}^2+bx+c)}^n$ is a factor of $q(x)$, then the partial fraction decompositionshould have terms

   $$\frac{C_{1}x+D_1}{a{x}^2+bx+c}+\frac{C_{2}x+D_2}{{(a{x}^2+bx+c)}^2}+\mathrm{\dots }+\frac{C_{n}x+D_n}{{(a{x}^2+bx+c)}^n}$$

   for some constants $C_i,D_i$ to be determined.

6. Once you have the sum of all appropriate partial fractions(see above), group together all these partial fractions into onefraction $\frac{P(x)}{q(x)}$ (use a minimum common multiple forthe denominator! The minimum common multiple will actually be$q(x)$). In order to have an equality you need to find appropriateconstants $A,B,\mathrm{\dots }$ such that $p(x)=P(x)$. For this, plugvalues of $x$ to obtain equations relating the constants.

Example 2.2.

Suppose we want to find the partial fraction decomposition of:

$$\frac{3x+2}{(x-1)(x-2){(x-3)}^2(1+x^2)}$$

Then, we need constants $A,B,C,D,E,F$ such that

$$\frac{3x+2}{(x-1)(x-2){(x-3)}^2(1+x^2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}+\frac{D}{{(x-3)}^2}+\frac{Ex+F}{1+x^2}.$$

The next step would be to add up all the partial fractions intoone big fraction

$$\frac{P(x)}{(x-1)(x-2){(x-3)}^2(1+x^2)}$$

andfind constants $A,B,C,D,E,F$ such that $P(x)=3x+2$ for all $x$.