\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{amsmath}

\begin{document}
\centerline{MATH 134 -- Calculus with Fundamentals 2}
\centerline{Practice Days on Partial Fractions}
\centerline{March 26 and 27, 2018}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
Every rational function (quotient of polynomials) can 
be written as {\it a polynomial plus a sum of one or more terms of the following 
forms}:
$$\frac{Ax+B}{(x^2+2bx+c)^k},\qquad \frac{C}{(x + a)^k}.$$
A rational function expressed this way is said to be {\it decomposed
into partial fractions}.  The process of finding this decomposition
is as follows:  Given a rational function $\frac{f(x)}{g(x)}$,
\vskip 10pt
\begin{enumerate}
\item  First, if the degree of $g$ is larger already, just
proceed to step 2 with $\frac{f(x)}{g(x)}$.  
Otherwise, if the degree of $f(x)$ is greater than or equal to the 
degree of $g(x)$, {\it divide} $g(x)$ into $f(x)$ using polynomial
long division and write $f(x) = q(x)g(x) + r(x)$ for some
quotient $q(x)$ and remainder $r(x)$.  Then 
$$\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}$$
and the degree of $r(x)$ is less than the degree of $g(x)$.  Continue 
to step 2 with $\frac{r(x)}{g(x)}$.  
\item  {\it Factor} $g(x)$ completely into a product of 
powers of linear polynomials and powers of quadratic
polynomials with no real roots (``irreducible'' quadratics).  
(The fact that this can always be done is one
form of the so-called ``Fundamental Theorem of Algebra''.  The
famous mathematician and physicist Carl Friedrich Gauss gave the first
complete proof of this result in 1799.)
\item  Assemble the partial fractions:  For each $(x + a)^m$ appearing 
in the factorization of $g(x)$, include a {\it group of terms}
$$\frac{C_1}{(x + a)} + \frac{C_2}{(x + a)^2} + \cdots + \frac{C_m}{(x + a)^m}$$
For each power $(x^2+bx+c)^n$ of an irreducible quadratic appearing 
in the factorization of $g(x)$, include a {\it group of terms}:
$$\frac{A_1x+B_1}{x^2+bx+c} + \frac{A_2x+B_2}{(x^2+bx+c)^2} + \cdots + \frac{A_n x+B_n}{(x^2+bx+c)^n}$$
\item  Set the rational function from step 1 (either the original
$f/g$ or $r/g$ as appropriate) equal to the sum of the partial fractions,
clear denominators, and solve for the coefficients.  
This last step can be done either by substituting well-chosen $x$-values,
or by equating coefficients of like powers of $x$ on both sides
and solving the resulting system of equations.
\item Integrate the polynomial quotient and the partial fractions.  The integrals of the partial fractions of the 
form $\frac{C}{(x + a)^k}$ are easy $u$-substitution forms; the integrals of the partial
fractions of the form $\frac{Ax + B}{(x^2 + bx + c)^k}$ can be done with a combination of $u$- and trigonometric (tangent) substitutions.  
The most general form after integration contains a logarithm term plus an inverse tangent term.
\end{enumerate}
\vskip 10pt
\noindent
{\it Questions}
\vskip 10pt
\begin{enumerate}
\item We'll work out $\displaystyle \int \frac{x^3+1}{x^3 + 25x}\ dx$ together
\item $\displaystyle \int \frac{x + 2}{x^2 + 4x + 3} \ dx$
\item $\displaystyle \int \frac{x^4+1}{x^3 + 3x^2 + 2x}\ dx$
\item $\displaystyle \int \frac{1}{x^4 + 5x^2 + 6}\ dx$  (The bottom factors here; you'll need to find the 
factorization!)
\end{enumerate}
\end{document}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{needleregion.png}
\end{center}
\caption{The region for problem 4.}
\end{figure}

\end{document}



