This *Mathematica* notebook discusses the method of integration by partial-fraction decomposition. The examples it considers are simple enough to do by hand. Their purpose is to illustrate use of a computer-algebra system to carry out calculations that by hand can become tedious, time-consuming and prone to error. Once you master the basics of partial-fraction decomposition and practice enough to work problems by hand in 5 minutes or so, you can use this notebook to check your work, or to evaluate more complicated integrals. *Don*'*t* make the mistake of relying on *Mathematica* entirely, because it won't be available for quiz or exam questions! (And, as you'll see, it's *not* perfect either!)

**Note:** To see the result of a command, position the cursor (I-beam) to the right of the command and hit Enter on the numeric keypad, or type Shift-Return or Command (ýÿ)-Return on the main part of the keyboard, or select the command and choose **Evaluate Cells** from the **Evaluation** submenu of the **Kernel** menu.**1. The ****Apart**** Command. **Consider the problem of integrating the rational function *f(x)* = *p(x)/q(x)*, where *p(x)* = + 2-- *x* + 1 and *q(x)* = -- *x* -- 6. Defining a function in *Mathematica *requires the special colon/equal symbol **:=** as well as an underscore (_) subscript after the function's independent variable. (Failure to observe this rule can cause puzzling problems!) For example, the proper definition of the function *f *above is

**f[x_]:=(x^3 + 2 x^2 - x + 1)/(x^2 - x - 6)**

Since the degree of the numerator *p(x)* of *f(x)* is *not* less than the degree of its denominator *q(x)*, the function *f(x)* is not in *proper form* to decompose into a sum of partial fractions. Before proceeding by hand, you would have to divide the numerator by the denominator to obtain the quotient polynomial *s(x),* and then decompose the function *r(x)/q(x),* where *r(x)* is the remainder polynomial from the division process, for which deg *r(x)* < deg *q(x). *However, *Mathematica* will do that division in the background and provide a partial-fraction decomposition of *f(x)* in one step, by invoking its Apart command. To illustrate that, just execute the following command.

**Apart[f[x]]**

You don't even have to define the function *f(x),* as you can see by executing the following command.

**Apart[(x^3 + 2*x^2 - x + 1)/(x^2 - x - 6)]**

You can check the decomposition by asking *Mathematica* to add the partial fractions. Symmetric to Apart, the command to so is Together.

**Together[x + 3 + 43/(5*(x - 3)) - 3/(5*(x + 2))] **

You can check that the decomposition is exactly what you would get by hand by first dividing *p(x)* by *q(x)* to get *s(x)* and *r(x)*, and then using Apart on *r(x)/q(x)*. For that, use *Mathematica*'s built-in commands for finding the quotient and remainder when *p(x)* is divided by *q(x)*.

Note that *s(x)* = *x* + 3 added to the last expression is exactly what Apart gave when applied to *f(x). *Is *Mathematica* then foolproof? It would appear so, at least with regard to partial-fraction decomposition! Its partioal-decomposition makes it easy to integrate the given rational function *f(x)*:

∫ *f(x) dx* = ∫(3 + *x*) *dx* + ∫ *dx*

= 3*x* + + ln |*x* -- 3| -- ln |*x* + 2| + *C *What does

**Integrate[f[x], x]**

Some mathematicians claim that *Mathematica* could earn a B in a standard second-semester calculus course. It certainly doesn't get a *perfect* score on this problem: the answer has no constant of integration and no absolute-value signs around the variables in the ln function (which, like most computer-related products, *Mathematica* denotes by Log). Each of those omissions could lose one point on a 10-point question that asked for ∫ *f(x) dx, *so indeed *Mathematica* looks like no better than a B student on this! The moral: do enough partial-fraction decomposition by hand to be able to solve problems like this in 5 minutes without technological aids! Work most or all the homework set by hand, and use *Mathematica* to check your answers.

**2. Quadratic Denominator Factors. **Consider next an example in which the denominator function has an irreducible quadratic factor, ∫*dx*. The denominator factors as (-- 4)(+ 4) = (*x* -- 2)(*x* + 2)(* *+ 4), where the factor + 4 is irreducible (that is, is not the product of any linear factors) over the system **R** of real numbers. *Mathematica* gives this factorization easily. Try

**Factor[x^4 - 16]**

**The Apart command, however doesn't need this information, since it factors the polynomials in the background and then decomposes accordingly. It is enough to execute the following command directly.**

**Apart[x^2/(x^4 - 16)]**

Basic antidifferentiation formulas then give the integral with minimal additional work:

∫*dx* = ∫ -- ∫ + ∫

= ln |*x* -- 2| -- ln |*x* + 2| + arctan() + *C = * ln || + arctan() +

**Integrate[x^2/(x^4 - 16), x]**

At first glance, this is not even B work: besides the missing constant, the *arctan* term appears to be wrong. However, as you can verify for yourself, the answer actually is consistent with the result of the hand calculation. The only errors here are the same ones *Mathematica* made before: no constant of integration, and missing absolute-value bars in the Log functions. Another 8 points out of 10, probably!

It is a bit dangerous to rely heavily on the **Integrate** command, because you have no way to determine what method *Mathematica* uses to find an antiderivative. As the last example illustrates, it may well give a *correct* formula that on the surface looks quite *different* from an equally correct formula you (or the answer book) may derive by hand using a method from the course. So even as a checker, it is wise to treat *Mathematica* not as fountainhead of knowledge, but rather as a fellow student -- one with a pretty limited personality at that, and little potential to earn a grade better than B!

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