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%TCIDATA{<META NAME="Title" CONTENT="Methods of Integration">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:06:39 1997}
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\begin{document}


\section{Methods \label{Methods of integration}of Integration}

Even though \textsl{Scientific Notebook} can evaluate many integrals
directly, several standard techniques of integration---such as integration%
\label{Techniques of integration} by parts, 
\index{Techniques of integration} 
\index{Methods of integration}substitution, and 
\index{Integration!techniques}partial fractions---are also available.

\subsection{\ Integration by Parts\label{Integration by parts}}

\begin{quotation}
The \emph{integration by parts}\ formula 
\index{Integration by parts}states that 
\[
\int u\,dv=uv-\int v\,du 
\]
\end{quotation}

\begin{quote}
$\blacktriangleright $ To use integration by parts on $\int x\ln x\,dx$
\end{quote}

\begin{enumerate}
\item[1.]  Leave the insertion point in the integral.

\item[2.]  From the \textsf{Calculus} submenu, choose \textsf{Integrate by
Parts}.

\item[3.]  Enter $\ln x$ (as the \textsf{Part to be Differentiated}) in the
dialog box.

\item[4.]  Choose \textsf{OK}.
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Calculus + Integrate by Parts}
\end{quote}

\begin{quotation}
$\dint x\ln x\,dx=\allowbreak 
\frac{1}{2}\left( \ln x\right) x^{2}-\dint \frac{1}{2}x\,dx$

Note that in this case, $u=\ln x$ and $dv=x\,dx$, so that $du=\frac{1}{x}%
\,dx $ and $v=\frac{1}{2}x^{2}$.
\end{quotation}

\strut

\subsection{\ Change of Variables\label{Change of variables}}

\begin{quotation}
The \emph{change of variables}\textsl{\ formula} 
\index{Change of variables}states that if $u=g(x)$, then $du=g^{\prime
}(x)\,dx$ and hence 
\[
\int f(g(x))\,g^{\prime }(x)\,dx=\int f(u)\,du 
\]
\end{quotation}

\begin{quote}
To perform a change of variables on $\int x\sin x^{2}dx$
\end{quote}

\begin{enumerate}
\item[1.]  Enter the integral $\int x\sin x^{2}\,dx.$

\item[2.]  From the \textsf{Calculus} submenu, choose \textsf{Change Variable%
}.

\item[3.]  Enter the substitution $u=x^{2}$ in the dialog box.

\item[4.]  Choose \textsf{OK}.
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Calculus + Change Variable}
\end{quote}

\begin{quotation}
$\dint x\sin x^{2}\,dx=\allowbreak \dint 
\frac{1}{2}\sin u\,\,du$

In this example, $u=x^2$. Note that $u=g(x)=x^2$, $f(u)=\sin u$, and $%
du=2x\,dx$.
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Calculus + Change Variable}
\end{quote}

\begin{quotation}
$\dint x^{2}\sqrt{x^{3}+1}dx=\allowbreak \dint \frac{2}{3}%
u^{2}\,du=\allowbreak \frac{2}{9}u^{3}=\allowbreak \frac{2}{9}\left( \sqrt{%
x^{3}+1}\right) ^{3}$

In this example, $u^{2}=x^{3}+1$.
\end{quotation}

\strut

\subsection{\ Partial Fractions\label{Partial fractions}}

\begin{quotation}
The \emph{method of partial fractions} is based 
\index{Partial fractions} on the fact that a factorable rational function
can be written as a sum of simpler fractions. Notice how evaluation of the
following integral gives the answer as a sum of terms.
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\dint 
\dfrac{3x^{2}+2x+4}{(x-1)^{2}(x^{2}+1)^{2}}dx=\allowbreak -\dfrac{9}{4\left(
x-1\right) }-\dfrac{5}{2}\ln \left( x-1\right) +\dfrac{5}{4}\ln \left(
x^{2}+1\right) -\dfrac{1}{4}\arctan x+\allowbreak \dfrac{1}{8}\dfrac{-4x-2}{%
x^{2}+1}$

To gain an appreciation for how this calculation might be done internally,
consider the method of partial fractions.
\end{quotation}

\begin{quote}
$\blacktriangleright $ To use the method of partial fractions on $\dint 
\dfrac{3x^{2}+2x+4}{(x-1)^{2}(x^{2}+1)^{2}}dx$
\end{quote}

\begin{enumerate}
\item[1.]  Enter the integral $\dint \dfrac{3x^{2}+2x+4}{%
(x-1)^{2}(x^{2}+1)^{2}}dx$.

\item[2.]  Select the rational expression $\dfrac{3x^{2}+2x+4}{%
(x-1)^{2}(x^{2}+1)^{2}}$ inside the integral.

\item[3.]  While pressing the \textsc{ctrl } key, choose \textsf{Partial
Fractions} from the \textsf{Calculus} submenu (or the \textsf{Polynomials}
submenu).

\item[4.]  While the result is still selected, click \FRAME{itbpF}{0.3009in}{%
0.3009in}{0.0701in}{}{}{parens.wmf}{\special{language "Scientific Word";type
"GRAPHIC";display "PICT";valid_file "F";width 0.3009in;height 0.3009in;depth
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"0";croptop "1";cropright "1";cropbottom "0";filename
'Parens.wmf';file-properties "XNPEU";}}to get 
\[
\dint \left( \frac{9}{4\left( x-1\right) ^{2}}-\frac{5}{2\left( x-1\right) }+%
\frac{1}{4}\frac{1+10x}{x^{2}+1}+\frac{1}{2}\frac{x-2}{\left( x^{2}+1\right)
^{2}}\right) dx 
\]

\item[5.]  The preceding integral can be written as a sum of four integrals. 
\[
\dint \frac{9}{4\left( x-1\right) ^{2}}dx-\int \frac{5}{2\left( x-1\right) }%
dx+\frac{1}{4}\int \frac{1+10x}{x^{2}+1}dx+\frac{1}{2}\int \frac{x-2}{\left(
x^{2}+1\right) ^{2}}dx 
\]
You can evaluate each of these integrals with \textsf{Evaluate}.
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\dint \dfrac{9}{4\left( x-1\right) ^{2}}dx=\allowbreak -\dfrac{9}{4\left(
x-1\right) }\vspace{6pt}$

$-\dint \dfrac{5}{2\left( x-1\right) }dx=\allowbreak -\frac{5}{2}\ln \left(
2x-2\right) \vspace{6pt}$

$\dfrac{1}{4}\dint \dfrac{1+10x}{x^{2}+1}dx=\allowbreak \dfrac{5}{4}\ln
\left( x^{2}+1\right) +\dfrac{1}{4}\arctan x\vspace{6pt}$

$\dfrac{1}{2}\dint \dfrac{-2+x}{\left( x^{2}+1\right) ^{2}}dx=\allowbreak 
\dfrac{1}{8}\dfrac{-4x-2}{x^{2}+1}-\dfrac{1}{2}\arctan x$
\end{quotation}

The original integral is the sum of the expressions on the right.

\begin{eqnarray*}
\dint \dfrac{3x^{2}+2x+4}{(x-1)^{2}(x^{2}+1)^{2}}dx &=&\allowbreak -\dfrac{9%
}{4\left( x-1\right) }-\dfrac{5}{2}\ln \left( 2x-2\right) \\
&&+\dfrac{5}{4}\ln \left( x^{2}+1\right) -\dfrac{1}{4}\arctan x+\allowbreak 
\dfrac{1}{8}\dfrac{-4x-2}{x^{2}+1}
\end{eqnarray*}

Applying \textsf{Simplify} to the difference between this answer and the
earlier answer computed directly with \textsf{Evaluate} gives a constant.$%
\allowbreak $%
\[
-\frac{5}{2}\ln \left( 2x-2\right) +\frac{5}{2}\ln \left( x-1\right)
=\allowbreak -\frac{5}{2}\ln 2 
\]

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Evaluating Calculus Expressions}{}{}{DM7-1.tex#Calculus
expressions}

\item  \hyperref{Limits}{}{}{DM7-2.tex#Limit}

\item  \hyperref{Differentiation}{}{}{DM7-3.tex#Derivative}

\item  \hyperref{Indefinite Integration}{}{}{DM7-4.tex#Indefinite integral}

\item  \hyperref{Definite Integrals}{}{}{DM7-6.tex#Definite integral}

\item  \hyperref{Exercises and Solutions}{}{}{DM7-9.tex#Exercises}
\end{itemize}

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"1";cropbottom "0";filename 'index.wmf';file-properties "XNPEU";}}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Change of
variable}{}{}{DMIndex.tex#Change of variable}, \hyperref{Integration by parts%
}{}{}{DMIndex.tex#Integration by parts}, \hyperref{Partial fractions}{}{}{%
DMIndex.tex#Partial fractions}}

\end{document}
