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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Techniques of integration}


\subsubsection{Partial fractions}

The technique of partial fractions helps solve integrals of nasty rational functions,
by splitting up the fraction itself. What we will do first is talk about the
basic rational functions we \textit{can} integrate. Then we'll deal with more
complicated expressions, reducing them in terms of these. 

The general thing here is to integrate a rational expression: \[
\int \frac{P(x)}{Q(x)}dx,\]
where \( P(x) \) and \( Q(x) \) are polynomials.    

The basic rational functions we can always integrate are of the following two
kinds. Note that they have powers of linear terms, or powers of \textit{irreducible}
quadratic terms in the denominator. Note also that the numerator is 1 in the
first case, and at most linear in the second. We will reduce all other integrals
of rational functions to these types.  

\begin{enumerate}
\item \[
\int \frac{dx}{(ax+b)^{n}}.\]
 
\item \[
\int \frac{(Ax+B)dx}{ax^{2}+bx+c}.\]

\end{enumerate}
The first is actually a simple substitution integral. These cause no trouble. 

\begin{example}
\[
\int \frac{dx}{2x+3}.\]
 Here the substitution is \( u=2x+3 \), so that \( du=2dx \) or \( dx=du/2 \).
Then: \[
\int \frac{dx}{2x+3}=\int \frac{du/2}{u}\]

\end{example}
%

\begin{example}
\[
\int \frac{dx}{(4x-5)^{3}}.\]
 
\end{example}
The second type can cause some trouble. There is a preliminary step, completion
of the square, that must be taken for the most complicated ones, and sometimes
the problem splits into more than one integral. We'll start with an easy example: 

\begin{example}
\[
\int \frac{dx}{x^{2}+3}.\]
 The trick here is to reduce that integral to a standard one, and hopefully
\[
\int \frac{dx}{x^{2}+1}\]
 comes to mind. The first thing is to turn the 3 into a 1, which you do by simply
factoring it out (out of \textbf{both} terms!). Then, substitute as indicated:
\begin{eqnarray*}
\int \frac{dx}{x^{2}+3} & = & \frac{1}{3}\int \frac{dx}{(x^{2}/3)+1}\\
 & = & \frac{1}{3}\int \frac{dx}{(x/\sqrt{3})^{2}+1}\; \mathrm{Now},\, \mathrm{substitute}:\, u=x/\sqrt{3}\\
 & = & \frac{1}{3}\int \frac{\sqrt{3}du}{u^{2}+1}\\
 & = & \frac{1}{\sqrt{3}}\arctan (u)+C=\frac{1}{\sqrt{3}}\arctan (x/\sqrt{3})+C.
\end{eqnarray*}

\end{example}
%

\begin{example}
\[
\int \frac{dx}{3x^{2}+4}\]
 
\end{example}
\noindent Now a harder one: 

\begin{example}
\[
\int \frac{(2x+1)dx}{x^{2}+2x+3}.\]
 The denominator can be reduced to the form \( u^{2}+a^{2} \) by completing
the square. It then can be dealt with as above. Use the substitution to deal
with the linear term in the numerator as well, and split the integral in two.
\( x^{2}+2x+3=(x+1)^{2}+2 \) by completing the square, so our \textit{first}
substitution is to set \( u=x+1 \). We will make a second substitution later.
\begin{eqnarray*}
\int \frac{(2x+1)dx}{x^{2}+2x+3} & = & \int \frac{(2x+1)dx}{(x+1)^{2}+2}\\
 & = & \int \frac{(2(u-1)+1)du}{u^{2}+2}\\
 & = & \frac{1}{2}\int \frac{(2(u-1)+1)du}{(u/\sqrt{2})^{2}+1}.\; \mathrm{Now},\, \mathrm{set}\, v=u/\sqrt{2}\\
 & = & \frac{1}{2}\int \frac{(2(\sqrt{2}v-1)+1)\sqrt{2}dv}{v^{2}+1}\\
 & = & \frac{1}{\sqrt{2}}\int \frac{(2\sqrt{2}v-1)dv}{v^{2}+1}\\
 & = & 2\int \frac{vdv}{v^{2}+1}-\frac{1}{\sqrt{3}}\int \frac{dv}{v^{2}+1}\\
 & = & \ln (v^{2}+1)-\frac{1}{\sqrt{2}}\arctan (v)+C\\
 & = & \ln ((u/\sqrt{2})^{2}+1)-\frac{1}{\sqrt{2}}\arctan (u/\sqrt{2})+C\\
 & = & \ln (((x+1)/\sqrt{2})^{2}+1)-\frac{1}{\sqrt{2}}\arctan ((x+1)/\sqrt{2})+C
\end{eqnarray*}

\end{example}
There was even a \textit{third} substitution that I slid in there without mentioning
it: \( t=v^{2}+1 \), so \( dt=2vdv \), which happened to be the numerator
of the first integral. 

\begin{example*}
\[
\int \frac{xdx}{2x^{2}+3x+6}.\]
 
\end{example*}

\subsubsection{Real Partial Fractions }

In a sense, this isn't even a technique of \textit{integration}, but a fact
from algebra that we use in integration. This technique helps solve integrals
of nasty rational functions, by splitting up the fraction itself. There are
several parts: 

The general thing here is to integrate a rational expression: \[
\int \frac{P(x)}{Q(x)}dx,\]


where \( P(x) \) and \( Q(x) \)

are polynomials.


\subsubsection*{\noindent Part 1, reduce the fraction. }

\noindent If \( deg(P(x))\geq deg(Q(x)) \), the rest of the technique won't
work. You gotta divide out. Example: \[
\frac{3x^{3}+4x^{2}+5x+1}{x^{2}+1}=3x+4+\frac{2x-3}{x^{2}+1}.\]
You should be able to verify that, and recall how to do the long division. 


\subsubsection*{\noindent Part 2, General form: }

\noindent If you have a fraction \[
\frac{P(x)}{Q(x)},\]
 with \( deg(P(x))<deg(Q(x)) \) --- divide out if you have to --- then we set
up the general form of the \textit{decomposition}. For every linear factor,
\( ax+b \), in the denominator, put an unknown constant over \( ax+b \) on
the right. If there is a square, or higher power, of \( ax+b \), then you need
terms with denominators \( ax+b \), \( (ax+b)^{2} \), \( (ax+b)^{3} \),...,
up to the power that \( (ax+b) \) has. Each gets a constant numerator. Any
irreducible quadratic term, \( ax^{2}+bx+c \) with no roots, has a linear term
\( Ax+B \) as numerator. Repeats get more terms, up to the power occurring,
each with a linear numerator. 

This is just a recipe for setting up the problem. 

\begin{example*}
\begin{eqnarray*}
\frac{7x+3}{(x-1)(x+2)} & = & \frac{A}{x-1}+\frac{B}{x+2}\\
\frac{7x^{2}+3}{(x-1)(x^{2}+2)} & = & \frac{A}{x-1}+\frac{Bx+C}{x^{2}+2}\\
\frac{7x^{2}+3}{(x-1)^{2}(x+2)} & = & \frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2}\\
\frac{7x^{2}+3}{(x-1)(x^{2}+2)^{2}} & = & \frac{A}{x-1}+\frac{Bx+C}{x^{2}+2}+\frac{Dx+E}{(x^{2}+2)^{2}}.
\end{eqnarray*}

\end{example*}
The deal is, these always have solutions. You just follow the recipe, like in
a cookbook. There is a reason why that will always work, but we will keep that
from you --- it's best explained in a course on complex analysis. 


\subsubsection*{\noindent Part 3. Find the unknown numbers. }

\noindent This is an exercise in algebraic manipulation. Let's look at the second
one. The others are similar --- and maybe the first one is too easy. If \[
\frac{7x^{2}+3}{(x-1)(x^{2}+2)}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+2},\]
 what are the numbers \( A \), \( B \), and \( C \)? You find them by putting
the right-hand side back under the common denominator: \begin{eqnarray*}
\frac{7x^{2}+3}{(x-1)(x^{2}+2)} & = & \frac{A}{x-1}+\frac{Bx+C}{x^{2}+2}\\
 & = & \frac{A(x^{2}+2)}{(x-1)(x^{2}+2)}+\frac{(Bx+C)(x-1)}{(x^{2}+2)(x-1)}\\
 & = & \frac{A(x^{2}+2)+(Bx+C)(x-1)}{(x^{2}+2)(x-1)},
\end{eqnarray*}
then you compare numerators. The numerators gotta be the same, since the denominators
are. And, they gotta be the same for \textbf{all} \( x \). So, \[
7x^{2}+3=A(x^{2}+2)+(Bx+C)(x-1).\]
The easiest way to solve this is to say that the two sides are the same for
all \( x \), so they are at the \( x \)'s that are easy. Like, when \( x=1 \),
you get: \( 10=A3 \)

since the \( (x-1) \) becomes \( 0 \). So, \( A=10/3 \). One down. 

Then, try --- hmm, no more really easy ones. Well, \( x=0 \) is still not too
bad, then: \[
3=A(2)+(C)(-1)=20/3-C.\]
So, \( C=11/3 \). Finally, take \( x=-1 \) --- it keeps the arithmetic simple.
That gives: \[
10=7+3=A(1+2)+(-B+C)(-1-1)=(10/3)(3)+(-B+11/3)(-2),\]
so \( B=11/3 \).  

Some people get concerned when we set \( x=1 \) in this problem, since that
is a zero of the denominator. However, the point is that the numerators, after
you put the right-hand side back under the common denominator, must be equal
\textit{as polynomials}, so they are the same even at \( x=1 \). 


\subsubsection*{\noindent Part 4 Actually solve the integral. }

\noindent This will usually be the easy part. Let's try it for the example above.
\begin{eqnarray*}
\int \frac{7x^{2}+3}{(x-1)(x^{2}+2)}dx & = & \frac{10}{3}\int \frac{dx}{x-1}+\frac{11}{3}\int \frac{x}{x^{2}+2}dx+\frac{11}{3}\int \frac{1}{x^{2}+2}dx\\
 & = & \frac{10}{3}\ln (x-1)+\frac{11}{6}\ln (x^{2}+2)+\frac{11\sqrt{2}}{6}\arctan (\sqrt{2}x)+C,
\end{eqnarray*}
by the techniques we went over earlier. 
\end{document}
