Partial Fractions
The integration technique of partial fractions is a way to
integrate rational functions of the form
$f(x)=\frac{P(x)}{Q(x)}$. (Recall that $f$ is a rational
function when $P(x)$ and $Q(x)$ are polynomials.) When
considering $\int\frac{P(x)}{Q(x)}\,dx$, first look for a simple
substitution, as with any integral. If you see a way to use
integration by parts, or even trig substitution, you should
probably try this first, as those methods can be a little
simpler. Sometimes partial fraction decomposition is the
obvious and only choice.
The key in the partial fractions
technique of integration is to decompose
$\displaystyle\frac{P(x)}{Q(x)}$ into a sum of simpler
fractions, whose denominators are related to the
factors of $Q(x)$.
For example, $\displaystyle\frac{2}{x^2-1} = \frac{1}{x-1} -
\frac{1}{x+1}$.
The process:
1) If the degree of $P(x)$ is greater or equal to the degree
of $Q(x)$, then we need to use long division to find
$\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}$, resulting in the degree
of the remainder $R$ being less than the degree of $Q$.
2) To decompose $\frac{P(x)}{Q(x)}$ or $\frac{R(x)}{Q(x)}$ (if
you did long division), we first factor $Q(x)$.
3) Use the following process.
Process after factoring $Q(x)$
- For every factor of $(x-a)$ in $Q(x)$, we have a
term $\displaystyle\frac{A}{x-a}$.
- For every repeated linear factor $(x-a)^n$, we have
$$\displaystyle\frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2}
+ \ldots + \frac{A_n}{(x-a)^n}.$$
- For every quadratic factor $x^2 + bx + c$, we have
$$\displaystyle\frac{Ax+B}{x^2+bx+c}.$$
- For every repeated quadratic factor $(x^2+bx+c)^n$,
we have $$\displaystyle\frac{A_1 x + B_1}{x^2+bx+c} +
\frac{A_2x+B_2}{(x^2+bx+c)^2}+\ldots +
\frac{A_nx+B_n}{(x^2+bx+c)^n}.$$
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Finally, we have to integrate the resulting terms. Linear
factors give logs. Substitution or trig substitution will
usually take care of the other factors.
Example: $\displaystyle\int
\frac{x^3}{x^2- 1}\,dx$
DO: Justify each equal sign below with the
work needed to get from the LHS to the RHS of the equal sign.
$\displaystyle \int \frac{x^3}{x^2- 1}\,dx = \int \left(x +
\frac{x}{x^2- 1}\right)\,dx= \int x\,dx+ \frac{1}{2}\int\left(
\frac{1}{x-1} + \frac{1}{x+1} \right)\,dx $
$\displaystyle\quad\qquad\qquad= \frac{x^2}{2}+
\frac{1}{2}\Bigl(\ln\lvert x-1\rvert +\ln\lvert x+1\rvert \Bigr)+ C$
DO: Just for practice, evaluate
this integral using trig substitution. Which method do you
prefer here?