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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{Integrals of Exponential and Logarithmic functions}

Remember that integral formulas are simply derivative formulas in reverse. So,
\begin{eqnarray*}
\int e^{x}\, dx & = & e^{x}+c\\
\int a^{x}\, dx & = & \frac{1}{\ln (a)}a^{x}+c\\
\int \frac{1}{x}dx & = & \ln \left( \left| x\right| \right) +c\\
\int \frac{1}{x\, \ln (a)}dx & = & \log _{a}\left( \left| x\right| \right) +c.
\end{eqnarray*}
In practice, only the first and third of these formulas are worth bothering
about. 

What is most glaring is what was left out. The section heading talkes about
integrals of logarithmic functions, but that is not listed. Instead, the integral
of that which gives you the natural logarithm is listed. That's the formula
that is the derivative rule, backwards, so that is to be expected, but it seems
odd to not have the integral \emph{of} the natural logarithm. 

It's not impossible to figure out that integral:
\[
\int \ln (x)\, dx=x\, \ln (x)-x+c.\]
As to how this was found, it's a lot easier to show that it indeed is the right
integral. The reason why it's true is a trick called integration by parts. Here,
at least, is a check to see that the integral is as I claimed it:
\begin{eqnarray*}
\left( x\, \ln (x)-x\right) ' & = & 1\, \ln (x)+x\, \frac{1}{x}-1\, (\textrm{by}\, \textrm{the}\, \textrm{product}\, \textrm{rule})\\
 & = & \ln (x).
\end{eqnarray*}


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