Rational functions

Section 1.21 Rational functions

Definition 1.21.1. Rational function.

A rational function is a function that can be expressed as a quotient of two polynomials.

Theorem 1.21.2. Polynomial facts.

Degree.
Suppose \(f(x)=\anpoly\) with \(a_n\ne 0\text{.}\) We call \(n\) the degree of \(f\text{,}\) denoted \(\deg f\text{.}\)
Roots of polynomials.
A polynomial of degree \(n\) has at most \(n\) distinct roots.
Equating coefficients.

Given polynomials \(f(x)=\anpoly\) and \(g(x)=\bmpoly\) where \(a_n\ne 0\) and \(b_m\ne 0\text{,}\) we have

\begin{equation*} f(x)=g(x) \iff n=m \text{ and } a_i=b_i \text{ for all \(i\). } \end{equation*}
Irreducible.
A nonzero polynomial is irreducible if it cannot be factored into two polynomials of smaller degree. If \(f(x)\) is an irreducible polynomial with real coefficients, then \(\deg f=1\) or \(\deg f=2\text{.}\)

Theorem 1.21.3. Partial fraction decomposition.

Let \(p(x)/q(x)\) be a rational function where \(\deg p < \deg q\text{.}\)

If \(q\) factors into non-repeated linear factors as

\begin{equation*} q(x)=D(x-a_1)(x-a_2)\cdots (x-a_r)\text{,} \end{equation*}

then there is a unique choice of constants \(A_1, A_2,\dots, A_r\) such that

\begin{equation} \frac{p(x)}{q(x)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_r}{x-a_r}\text{.}\tag{1.21.1} \end{equation}
If \(q(x)\) factors into non-repeated irreducible linear and quadratic factors as

\begin{equation*} q(x)=D(x-a_1)(x-a_2)\cdots (x-a_r)(x^2+b_1x+c_1)(x^2+b_2x+c_2)\cdots (x^2+b_sx+c_s)\text{,} \end{equation*}

there there is a unique choice of constants \(A_1,A_2,\dots, A_r, B_1,B_2,\dots, B_s, C_1, C_2,\dots, C_s\) such that

\begin{equation} \frac{p(x)}{q(x)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_r}{x-a_r}+\frac{B_1x+C_1}{x^2+b_1x+c_1}+\frac{B_2x+C_2}{x^2+b_2x+c_2}+\cdots +\frac{B_sx+C_s}{x^2+b_sx+c_s}\text{.}\tag{1.21.2} \end{equation}

Procedure 1.21.4. Partial fraction decomposition.

Let \(f(x)\) be a rational function. To compute the partial fraction decomposition of \(f\text{,}\) proceed as follows.

Polynomial division.
Using polynomial division if necessary, write \(f(x)=g(x)+p(x)/q(x)\text{,}\) where \(g\text{,}\) \(p\) and \(q\) are polynomials, and \(\deg p < \deg q\text{.}\)
Factor.
Factor \(q(x)\) into powers of distinct irreducible polynomials.

Factoring trick. If \(q(x)\) has integer coefficients and a leading coefficient equal to 1, then any integer roots of \(q(x)\) will be factors of the constant term.
Polynomial equation with undetermined constants.
Set up a partial fraction decomposition equation for \(p(x)/q(x)\) of form (1.21.1) or (1.21.2). Clear the denominators of both sides of the equation, resulting in an identity between two polynomials. The polynomial on the right will be expressed in terms of the unknowns (\(A_i\text{,}\) \(B_i\text{,}\) etc.).
Solve for undetermined constants.
To solve for the undetermined constants (\(A_i\text{,}\) \(B_i\text{,}\) etc.) set up and solve a linear system of equations using one of the following techniques.
1. Evaluate equality at choices of \(x\). Evaluate the polynomial equation at various explicit choices of \(x\text{.}\) When \(q\) factors as \(q(x)=D(x-a_1)(x-a_2)\cdots (x-a_m)\) where the roots \(a_k\) are distinct, then evaluating the polynomial equality at each of the choices \(x=a_k\) will allow you to solve directly for each undetermined constant in turn.
2. Equate coefficients. Express the polynomial on right in “standard form” by collecting like terms. For the left and right polynomials to be equal, their corresponding coefficients must all be equal. This yields a system of equations in the unknowns (\(A_i\text{,}\) \(B_i\text{,}\) etc.) that you must now solve.

Remark 1.21.5.

There is a more general statement of partial fraction decomposition for the case where the irreducible factors of \(q\) include repeated linear and/or quadratic terms, but we will not cover it. Consult the textbook if you are curious.

Example 1.21.6. Rational function.

Compute \(\displaystyle\int \frac{x+2}{x^2-1}\, dx\)

Solution.

We have

\begin{align*} \frac{x+2}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}\amp \implies x+2=A(x+1)+B(x-1) \text{.} \end{align*}

Evaluating the right-hand side at \(x=1\) and \(x=-1\) yields

\begin{align*} 3 \amp = 2A\\ 1 \amp = -2B\text{,} \end{align*}

and thus \(A=3/2\text{,}\) \(B=-1/2\text{.}\) We conclude

\begin{align*} \int \frac{x+2}{x^2-1}\, dx \amp= \int \frac{3}{2}\frac{1}{x+1}-\frac{1}{2}\frac{1}{x-1}\, dx \\ \amp = \frac{3}{2}\ln\abs{x+1}-\frac{1}{2}\ln\abs{x-1}+C\text{.} \end{align*}

Example 1.21.7. Long division first.

Compute \(\displaystyle\int \frac{x^2}{x^2+2x-1}\, dx\)

Solution.

First perform polynomial division with remainder to write

\begin{equation*} \frac{x^2}{x^2+2x-1}=1-\frac{2x-1}{x^2+2x-1}\text{.} \end{equation*}

Next, we factor \(x^2+2x-1=(x-\alpha)(x-\beta)\text{,}\) where

\begin{align*} \alpha \amp = -1+\sqrt{2} \amp \beta\amp =-1-\sqrt{2}\text{.} \end{align*}

Now perform partial fraction decomposition

\begin{align*} \frac{2x-1}{(x-\alpha)(x-\beta)} \amp = \frac{A}{x-\alpha}+\frac{B}{x-\beta}\\ 2x-1 \amp = A(x-\beta)+B(x-\alpha)\text{.} \end{align*}

Evaluating the last identity at \(x=\alpha\) and \(x=\beta\text{,}\) we conclude

\begin{align*} 2\alpha-1 \amp = A(\alpha-\beta)\\ 2\beta-1 \amp = B(\beta-\alpha)\text{,} \end{align*}

or equivalently,

\begin{align*} -3+2\sqrt{2} \amp = 2\sqrt{2}A\\ -3-2\sqrt{2} \amp= -2\sqrt{2}B\text{.} \end{align*}

We conclude that \(A=\frac{1}{2\sqrt{2}}(-3+2\sqrt{2})=\frac{4-3\sqrt{2}}{4}\) and \(B=\frac{1}{2\sqrt{2}}(-3-2\sqrt{2})=\frac{4+3\sqrt{2}}{4}\text{,}\) and thus

\begin{align*} \int \frac{x^2}{x^2+2x-1}\, dx \amp = \int 1-\frac{4-3\sqrt{2}}{4}\frac{1}{x-\alpha}-\frac{4+3\sqrt{2}}{4}\frac{1}{x-\beta}\, dx\\ \amp = x-\frac{4-3\sqrt{2}}{4}\ln\abs{x-(-1+\sqrt{2})}-\frac{4+3\sqrt{2}}{4}\ln\abs{x-(-1-\sqrt{2})}+C \end{align*}

Example 1.21.8. Three distinct roots.

Compute \(\displaystyle\int\frac{x^2+1}{x^3+2x^2-x-2}\, dx\)

Solution.

First we factor \(q(x)=x^3+2x^2-x-2\text{.}\) An integer root of \(q(x)\) must divide \(2\text{,}\) thus we investigate \(\pm 1, \pm 2\) as potential roots. Since \(q(1)=1+1-1-2=0\text{,}\) we see that \(x=1\) is a root of \(q(x)\) and thus \((x-1)\) is a factor of \(q(x)\text{.}\) Performing polynomial division with remainder, we see that \(q(x)=(x-1)(x^2+3x+2)\text{.}\) Factoring the quadratic term further, we see that

\begin{equation*} q(x)=(x-1)(x+1)(x+2)\text{.} \end{equation*}

Now perform partial fractions:

\begin{align*} \frac{x^2+1}{x^3+2x^2-x-2} \amp =\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}\\ x^2+1 \amp = A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)\text{.} \end{align*}

Evaluating this identity at \(x=1, -1,-2\text{,}\) we see that

\begin{align*} 2 \amp = 6A \\ 2 \amp = -2B\\ 5 \amp = 3C \end{align*}

and thus \(A=\frac{1}{3}, B=-1, C=\frac{5}{3}\text{.}\) Thus

\begin{align*} \int\frac{x^2+1}{x^3+2x^2-x-2}\, dx \amp = \frac{1}{3}\frac{1}{x-1}-\frac{1}{x+1}+\frac{5}{3}\frac{1}{x+2}\, dx \\ \amp = \frac{1}{3}\ln\abs{x-1}-\ln\abs{x+1}+\frac{5}{3}\ln\abs{x+2}+C\text{.} \end{align*}

Example 1.21.9. Two irreducible quadratics.

Compute \(\displaystyle\int\frac{1}{x^4+3x^2+2}\)

Solution.

First note that the denominator \(q(x)=x^4+3x^2+2=(x^2)^2+3x^2+2=(x^2+1)(x^2+2)\text{,}\) where \(x^2+1\) and \(x^2+2\) are both irreducible. Now perform partial fractions:

\begin{align*} \frac{1}{x^4+3x^2+2} \amp = \frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}\\ 1 \amp = (Ax+B)(x^2+2)+(Cx+D)(x^2+1)=(A+C)x^3+(B+D)x^2+(A+2C)x+B+2D\text{.} \end{align*}

Note that in this case we cannot evaluate the identity above at the roots of \(x^2+1\) and \(x^2+2\) since they have none (in the reals)! Instead we use (3) from Theorem 1.21.2 and produce a linear system by equating the coefficients of two polynomials \((A+B)x^3+(B+D)x^2+(A+2C)x+(B+2D)\) and \(0x^3+0x^2+0x+1\text{:}\)

\begin{align} A+C \amp = 0\tag{1.21.3}\\ B+D \amp = 0\tag{1.21.4}\\ 2A+C \amp = 0\tag{1.21.5}\\ 2B+D \amp = 1\text{.}\tag{1.21.6} \end{align}

Equations (1.21.3) and (1.21.5) together imply \(A=C=0\text{.}\) Equation (1.21.4) implies \(B=-D\text{;}\) equation (1.21.6) then implies \(D=-1\) and \(B=1\text{.}\) We conclude that

\begin{equation*} \frac{1}{x^4+3x^2+2}=\frac{1}{x^2+1}-\frac{1}{x^2+2} \end{equation*}

and thus

\begin{align*} \int \frac{1}{x^4+3x^2+2}\, dx \amp = \int \frac{1}{x^2+1}-\frac{1}{x^2+2}\, dx \\ = \int \frac{1}{x^2+1}-\frac{1}{2(x/\sqrt{2})^2+1)}\, dx \\ \amp = \arctan(x)-\frac{\sqrt{2}}{2}\arctan(x/\sqrt{2})+C \end{align*}

Example 1.21.10. With substitution.

Compute \(\displaystyle \frac{\sin \theta}{\cos^2\theta +\cos \theta-2}\, d\theta\)

Solution.

Resist the temptation to apply partial fraction decomposition directly to the integrand! Since the integrand is not a rational function, Theorem 1.21.3 does not apply. Instead we first do an obvious substitution:

\begin{align*} \int \frac{\sin \theta}{\cos^2\theta +\cos \theta-2}\, d\theta \amp = -\int \frac{1}{u^2 +u-2}\, du \amp (u=\cos\theta, du=-\sin\theta d\theta) \\ \amp = -\int \frac{1}{3}\frac{1}{u-1}-\frac{1}{3}\frac{1}{u+2}\, du \amp (\text{part. frac. decomp.})\\ \amp =\frac{1}{3}(\ln\abs{\cos\theta+2}-\ln\abs{\cos\theta-1})+C\\ \amp = \ln\sqrt[3]{\abs{\frac{\cos\theta+2}{\cos\theta-1}}}+C \amp (\text{log. props.}) \end{align*}

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