Strategies of Integration

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 $\displaystyle\frac{A_1 x + B_1}{x^2+bx+c} + \frac{A_2x+B_2}{(x^2+bx+c)^2}+\ldots + \frac{A_nx+B_n}{(x^2+bx+c)^n}.$

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?

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Partial Fractions

Process after factoring Q(x)Q(x)

Process after factoring $Q(x)$