Rational Functions and Basic Types

Section 4.1 Rational Functions and Basic Types

This entire week of the course is devoted to understanding the integrals of rational functions. Recall that a rational function is a function \(f(x) = \frac{p(x)}{q(x)}\) wher \(p(x)\) and \(q(x)\) are polynomials. The general approach to integrals of rational functions is twofold. First, I am going to describe the integral of some very specific rational functions. Second, I am going to introduce techniques that can break apart any rational function to the sum of many easier rational functions. The pieces of the sum will all be rational functions from the fisrt part. To integrate a rational function, I will break it up in these pieces and then apply the known results to the individual pieces.

Subsection 4.1.1 Basic Types

The first step is the integrals of certain simple rational functions. I am going to start with three integrals that have already been introducted in Math 200, in the rules or the table of integrals.

\begin{align*} \int \frac{1}{x^r} dx \amp = \frac{-1}{(r-1)x^{r-1}} \amp \amp r \neq 1\\ \int \frac{1}{x} dx \amp = \ln |x| + c \amp \amp \\ \int \frac{1}{1+x^2} dx \amp = \arctan x + c \amp \amp \end{align*}

This is the starting point: all of the integrals of rational functions will eventually boil down to something based on these three integrals. The first step is extending these into slightly more general forms. There are four integrals I want to do here.

\begin{equation*} \int \frac{1}{x+a} dx \end{equation*}

This can be sovled by substitution, using \(u = x + a\) with \(du = dx\text{.}\)

\begin{equation*} \int \frac{1}{x+a} dx = \int \frac{1}{u} du = \ln |u| = \ln |x + a| + c \end{equation*}
\begin{align*} \amp \int \frac{1}{(x+a)^n} dx \amp \amp n \geq 2 \end{align*}

The same substitution as the previous integral solves this one; I’ve not shown the work.

\begin{equation*} \int \frac{1}{(x+a)^n} = \frac{-1}{(n-1)(x-a)^{n-1}} + c \end{equation*}
\begin{equation*} \int \frac{2x + a}{x^2+ac+b} dx \end{equation*}

This can also be solved by a substitution, using \(u = x^2 + ax +b\) with \(du = (2x+a)dx\text{.}\)

\begin{equation*} \int \frac{2x + a}{x^2+ax+b} dx = \int \frac{1}{u} du = \ln |u| = \ln |x^2 + ax + b| + c \end{equation*}
\begin{equation*} \int \frac{1}{(x-\alpha)^2 + \beta^2} dx \end{equation*}

This integral is a bit trickier. It also uses a substitution, \(u = \frac{x-\alpha}{\beta}\text{,}\) with \(du = \frac{1}{\beta} dx\) and \(\beta du = dx\text{.}\) Before doing the substitution, I need one manipulation to put the integral into a suitable form. I divide numerator and denominator by \(\beta^2\text{.}\) Then I proceed to use the substitution.

\begin{align*} \int \frac{1}{(x-\alpha)^2 + \beta^2} dx \amp = \int \frac{\frac{1}{\beta^2}}{\frac{(x-\alpha)^2}{\beta^2} + \frac{\beta^2}{\beta^2}} dx \\ \amp = \frac{1}{\beta^2} \int \frac{1}{ \left(\frac{x-\alpha}{\beta} \right)^2 + 1} dx \\ \amp = \frac{1}{\beta^2} \int \frac{1}{u^2 + 1} \beta du \\ \amp = \frac{1}{\beta} \int \frac{1}{u^2 + 1} du \\ \amp = \frac{1}{\beta} \arctan u + c \\ \amp = \frac{1}{\beta} \arctan \left( \frac{x - \alpha}{\beta} \right) + c \end{align*}

Now I’ve solved the four basic types of rational function integrals that I am going to need. For easy reference, here are the four integrals I will return to in future sections.

\begin{equation*} \int \frac{1}{x + a} dx = \ln |x + a| + c \end{equation*}
\begin{equation*} \int \frac{1}{(x + a)^n} dx = \frac{-1}{n-1} \frac{1}{(x-a)^{n-1}} + c \end{equation*}
\begin{equation*} \int \frac{2x + a}{x^2 + ax + b} dx = \ln |x^2 + ax +b| + c \end{equation*}
\begin{equation*} \int \frac{1}{(x-\alpha)^2 + \beta^2} dx = \frac{1}{\beta} \arctan \left( \frac{x - \alpha}{\beta} \right) + c \end{equation*}

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