SubsectionUsing an Integral Table
Calculus has a long history, going back to Greek mathematicians in 400300 BC. Its main foundations were first investigated and understood independently by Isaac Newton and Gottfried Wilhelm Leibniz in the late 1600s, making the modern ideas of calculus well over 300 years old. It is instructive to realize that until the late 1980s, the personal computer did not exist, so calculus (and other mathematics) had to be done by hand for roughly 300 years. In the 21st century, however, computers have revolutionized many aspects of the world we live in, including mathematics. In this section we take a short historical tour to precede discussing the role computer algebra systems can play in evaluating indefinite integrals. In particular, we consider a class of integrals involving certain radical expressions.
As seen in the short table of integrals found in AppendixA, there are many forms of integrals that involve \(\sqrt{a^2 \pm w^2}\) and \(\sqrt{w^2  a^2}\text{.}\) These integral rules can be developed using the technique known as trigonometric substitution that we learned in the last section. To see how these rules are used, consider the differences among
\begin{equation*}
\int \frac{1}{\sqrt{1x^2}} \,dx, \ \ \ \int \frac{x}{\sqrt{1x^2}} \,dx, \ \ \ \text{and} \ \ \ \int \sqrt{1x^2} \,dx\text{.}
\end{equation*}
The first integral is a familiar basic one, and results in \(\arcsin(x) + C\text{.}\) The second integral can be evaluated using a standard \(u\)substitution with \(u = 1x^2\text{.}\) The third, however, is not familiar and does not lend itself to \(u\)substitution.
In AppendixA, we find the rule
\begin{equation*}
\text{(h)} ~ \int \sqrt{a^2  u^2} \, du = \frac{u}{2}\sqrt{a^2  u^2} + \frac{a^2}{2} \arcsin \frac{u}{a} + C\text{.}
\end{equation*}
Using the substitutions \(a = 1\) and \(u = x\) (so that \(du = dx\)), it follows that
\begin{equation*}
\int \sqrt{1x^2} \, dx = \frac{x}{2} \sqrt{1x^2} + \frac{1}{2} \arcsin x + C\text{.}
\end{equation*}
Whenever we are applying a rule in the table, we are doing a \(u\)substitution, particularly when the substitution is more complicated than setting \(u = x\) as in the last example.
Example5.75
Evaluate the integral
\begin{equation*}
\int \sqrt{9 + 64x^2} \, dx\text{.}
\end{equation*}
SolutionHere, we want to use Rule (c) from the table, but we now set \(a = 3\) and \(u = 8x\text{.}\) We also choose the \(+\)
option in the rule. With this substitution, it follows that \(du = 8dx\text{,}\) so \(dx = \frac{1}{8} du\text{.}\) Applying the substitution,
\begin{equation*}
\int \sqrt{9 + 64x^2} \, dx = \int \sqrt{9 + u^2} \cdot \frac{1}{8} \, du = \frac{1}{8} \int \sqrt{9+u^2} \, du\text{.}
\end{equation*}
By Rule (c), we now find that
\begin{align*}
\int \sqrt{9 + 64x^2} \, dx =\mathstrut \amp \frac{1}{8} \left( \frac{u}{2}\sqrt{u^2 + 9} + \frac{9}{2}\ln\leftu + \sqrt{u^2 + 9}\right + C \right)\\
=\mathstrut \amp \frac{1}{8} \left( \frac{8x}{2}\sqrt{64x^2 + 9} + \frac{9}{2}\ln\left8x + \sqrt{64x^2 + 9}\right + C \right)\text{.}
\end{align*}
Whenever we use a \(u\)subsitution in conjunction with AppendixA, it's important that we not forget to address any constants that arise and include them in our computations, such as the \(\frac{1}{8}\) that appeared in Example5.75.
Example5.76
For each of the following integrals, evaluate the integral using \(u\)substitution and/or an entry from the table found in AppendixA.
\(\int \sqrt{x^2 + 4} \, dx\)
\(\int \frac{x}{\sqrt{x^2 +4}} \, dx\)
\(\int \frac{2}{\sqrt{16+25x^2}}\, dx\)
\(\int \frac{1}{x^2 \sqrt{4936x^2}} \, dx\)
Hint
Compare to \(\int \sqrt{u^2 + a^2} \, du\text{.}\)
Try a straightforward \(u\)substitution; the table is unneeded.
Let \(a = 4\) and \(u = 5x\) and look for a similar integral in the table.
Let \(a = 7\) and \(u = 6x\text{;}\) find a related integral in the table.
Answer
\(\int \sqrt{x^2 + 4} \, dx = \frac{x}2\sqrt{x^2+4}+2\ln\bigx+\sqrt{x^2+4}\big+C\text{.}\)
\(\int \frac{x}{\sqrt{x^2 +4}} \, dx = \sqrt{x^2 + 4} + C\text{.}\)
\(\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln\big 5x + \sqrt{16+25x^2} \big + C\text{.}\)
\(\int \frac{1}{x^2 \sqrt{4936x^2}} \, dx =  \frac{\sqrt{4936x^2}}{49x} + C\text{.}\)
Solution

By Rule (c) in AppendixA with \(a=2\) and \(u = x\text{,}\)
\begin{equation*}
\int \sqrt{x^2 + 4} \, dx = \frac{x}2\sqrt{x^2+4}+\frac42\ln\leftx+\sqrt{x^2+4}\right+C\text{.}
\end{equation*}

Let \(u=x^2 + 4\text{,}\) so \(du = 2x\, dx\text{.}\) Thus
\begin{equation*}
\int \frac{x}{\sqrt{x^2 +4}} \, dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \frac{1}{2} \cdot 2u^{1/2} + C = \sqrt{x^2 + 4} + C\text{.}
\end{equation*}

Letting \(a = 4\) and \(u = 5x\text{,}\) we see \(du=5\,dx\text{.}\) Thus,
\begin{equation*}
\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \int \frac{du}{\sqrt{a^2 + u^2}}\text{.}
\end{equation*}
Now by Rule (b) in AppendixA, it follows
\begin{equation*}
\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln\left 5x + \sqrt{16+25x^2} \right + C\text{.}
\end{equation*}

Letting \(a = 7\) and \(u = 6x\text{,}\) it follows that \(x=\frac{1}{6}u\) and \(du=6\,dx\text{,}\) and therefore
\begin{equation*}
\int \frac{1}{x^2 \sqrt{4936x^2}} \, dx = \frac{1}{6} \int \frac{1}{\frac{1}{36}u^2 \sqrt{a^2u^2}} \, du\text{.}
\end{equation*}
Using Rule (k) in AppendixA and the fact that \(\frac{1}{6} \cdot 36 = 6\text{,}\) we see
\begin{equation*}
\int \frac{1}{x^2 \sqrt{4936x^2}} \, dx = 6\cdot \frac{\sqrt{4936x^2}}{49 \cdot 6x} + C\text{.}
\end{equation*}
SubsectionUsing Computer Algebra Systems
A computer algebra system (CAS) is a computer program that is capable of executing symbolic mathematics. For example, if we ask a CAS to solve the equation \(ax^2 + bx + c = 0\) for the variable \(x\text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are arbitrary constants, the program will return \(x = \frac{b \pm \sqrt{b^2  4ac}}{2a}\text{.}\) Research to develop the first CAS dates to the 1960s, and these programs became publicly available in the early 1990s. Two prominent examples are the programs Maple and Mathematica, which were among the first computer algebra systems to offer a graphical user interface. Today, Maple and Mathematica are exceptionally powerful professional software packages that can execute an amazing array of sophisticated mathematical computations. They are also very expensive, as each is a proprietary program. The CAS SAGE is an opensource, free alternative to Maple and Mathematica.
For the purposes of this text, when we need to use a CAS, we are going to turn instead to a similar, but somewhat different computational tool, the webbased computational knowledge engine
called WolframAlpha. There are two features of WolframAlpha that make it stand out from the CAS options mentioned above: (1) unlike Maple and Mathematica, WolframAlpha is free (provided we are willing to navigate some popup advertising); and (2) unlike any of the three, the syntax in WolframAlpha is flexible. Think of WolframAlpha as being a little bit like doing a Google search: the program will interpret what is input, and then provide a summary of options.
If we want to have WolframAlpha evaluate an integral for us, we can provide it syntax such as
integrate x^2 dx
to which the program responds with
\begin{equation*}
\int x^2 \, dx = \frac{x^3}{3} + \text{constant}\text{.}
\end{equation*}
To find the partial fraction decomposition of any rational function, in WolframAlpha, entering
partial fraction 5x/(x^2x2)
results in the output
\begin{equation*}
\frac{5x}{x^2x2} = \frac{10}{3(x2)} + \frac{5}{3(x+1)}\text{.}
\end{equation*}
While there is much to be enthusiastic about regarding CAS programs such as WolframAlpha, there are several things we should be cautious about: (1) a CAS only responds to exactly what is input; (2) a CAS can answer using powerful functions from very advanced mathematics; and (3) there are problems that even a CAS cannot do without additional human insight.
Although (1) likely goes without saying, we have to be careful with our input: if we enter syntax that defines the wrong function, the CAS will work with precisely the function we define. For example, if we are interested in evaluating the integral
\begin{equation*}
\int \frac{1}{165x^2} \, dx\text{,}
\end{equation*}
and we mistakenly enter
integrate 1/16  5x^2 dx
a CAS will (correctly) reply with
\begin{equation*}
\frac{1}{16}x  \frac{5}{3} x^3\text{.}
\end{equation*}
But if we are sufficiently wellversed in antidifferentiation, we will recognize that this function cannot be the one that we seek: integrating a rational function such as \(\frac{1}{165x^2}\text{,}\) we expect the logarithm function to be present in the result.
Regarding (2), even for a relatively simple integral such as \(\int \frac{1}{165x^2} \, dx\text{,}\) some CASs will invoke advanced functions rather than simple ones. For instance, if we use Maple to execute the command
int(1/(165*x^2), x);
the program responds with
\begin{equation*}
\int \frac{1}{165x^2} \, dx = \frac{\sqrt{5}}{20} \arctanh \left(\frac{\sqrt{5}}{4}x\right)\text{.}
\end{equation*}
While this is correct (save for the missing arbitrary constant, which Maple never reports), the inverse hyperbolic tangent function is not a common nor familiar one; a simpler way to express this function can be found by using the partial fractions method, and happens to be the result reported by WolframAlpha:
\begin{equation*}
\int \frac{1}{165x^2} \, dx = \frac{1}{8\sqrt{5}} \left(\log(4\sqrt{5}+5x)  \log(4\sqrt{5}5x)\right) + \text{constant}\text{.}
\end{equation*}
Using sophisticated functions from more advanced mathematics is sometimes the way a CAS says to the user I don't know how to do this problem.
For example, if we want to evaluate
\begin{equation*}
\int e^{x^2} \, dx\text{,}
\end{equation*}
and we ask WolframAlpha to do so, the input
integrate exp(x^2) dx
results in the output
\begin{equation*}
\int e^{x^2} \, dx = \frac{\sqrt{\pi}}{2}\erf (x) + \text{constant}\text{.}
\end{equation*}
The function erf\((x)\)
is the error function, which is actually defined by an integral:
\begin{equation*}
\erf (x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{t^2} \, dt\text{.}
\end{equation*}
So, in producing an output involving an integral, the CAS has basically reported back to us the very question we asked.
Finally, as remarked at (3) above, there are times that a CAS will actually fail without some additional human insight. If we consider the integral
\begin{equation*}
\int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx
\end{equation*}
and ask WolframAlpha to evaluate
int (1+x) * exp(x) * sqrt(1+x^2 * exp(2x)) dx
,
the program thinks for a moment and then reports
(no result found in terms of standard mathematical functions)
But in fact this integral is not that difficult to evaluate. If we let \(u = xe^{x}\text{,}\) then \(du = (1+x)e^x \, dx\text{,}\) which means that the preceding integral has the form
\begin{equation*}
\int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx = \int \sqrt{1+u^2} \, du\text{,}
\end{equation*}
which is a straightforward one for any CAS to evaluate.
So, we should proceed with some caution: while any CAS is capable of evaluating a wide range of integrals (both definite and indefinite), there are times when the result can mislead us. We must think carefully about the meaning of the output, whether it is consistent with what we expect, and whether or not it makes sense to proceed.