Home  18.013A  Chapter 19 


We know how to antidifferentiate a function of the form ${(xa)}^{b}$ for any $a$ and $b$ . This will allow us to find the antiderivative of a rational function $\frac{p(x)}{q(x)}$ if we can reduce it to a sum of terms of that form and possibly a polynomial instead.
If $p(x)$ has a higher degree than $q(x)$ we can extract a quotient polynomial $s(x)$ by a process akin to long division called synthetic division.
We may then be left with a remainder polynomial $r(x)$ . We know how to antidifferentiate $s(x)$ so the task of antidifferentiating $\frac{p(x)}{q(x)}$ reduces to antidifferentiating $\frac{r(x)}{q(x)}$ where the numerator has lower degree than the denominator.
Suppose now that we can factor $q$ into factors like $(xa)$ or ${(xb)}^{3}$ or ${({(xd)}^{2}+{c}^{2})}^{m}$ .
The wonderful fact is that the expression $\frac{r(x)}{q(x)}$ can be separated into terms each of which has the form ${(xa)}^{b}$ or ${\left({(xd)}^{2}+{c}^{2}\right)}^{b}$ or $\frac{x}{{\left({(xd)}^{2}+{c}^{2}\right)}^{b}}$ for some $a\text{'}s$ , $d\text{'}s$ and $c\text{'}s$ and integer values of $b$ , each of which can be antidifferentiated.
And here is a procedure for separating it.
Suppose the denominator $q$ can be factored into ${(xb)}^{k}t(x)$ such that $t(b)$ is not 0.
And suppose we find the first $k$ terms of the Taylor series expansion of $\frac{r(x)}{t(x)}$ about $x=b$
Then the terms in $\frac{r(x)}{q(x)}$ that involve inverse powers of $(xb)$ are given as follows:
If $k=1$ there is only one term, $A{(xb)}^{1}$ , and $A$ is given by $\frac{r(b)}{t(b)}$ ;
For
$k=2$
we have
$A{(xb)}^{2}+B(xb)$
where
$A$
is as before while
$B$
is
${{\left(\frac{r(x)}{t(x)}\right)}^{\text{'}}}_{x=b}$
; and so on.
There are similar rules for quadratic factors.
The process of separating the denominator $q$ in this manner is called "the method of partial fractions".
We review the various methods again in Section 27.1 and some integrals to practice on are given in Section 27.3 . We apologize for the redundancy.
