Integration by Partial Fraction Decomposition

An Integration Technique used to rewrite Proper Rational Functions into equivalent partial fractions.

Process

Given a Proper Rational Function in the form $f(x)=\frac{P(x)}{Q(x)}$

1. Factor $Q(x)$ into irreducible factors
2. Write the fraction decomposition in the form of a Quotient Function sum: $f(x)=\frac{p_1(x)}{q_1(x)}+\cdots +\frac{p_n(x)}{q_n(x)}$ with:
   • $q_1,\dots ,q_n$ adhering to rule 1 or rule 2 of the irreducible factors of $Q(x)$
   • $p_1,\dots p_n$ being functions one Degree lower than its respective $q_1,\dots ,q_n$
3. Solve for the unknown constants introduced in part 2 by:
   1. Equating $P(x)=\frac{p_1(x)Q(x)}{q_1(x)}+\cdots +\frac{p_n(x)Q(x)}{q_n(x)}$
   2. Factoring $P(x)$, then finding what values of $p(x)$ allow $P(x)$ to be 0
4. Integrate

Fraction Decomposition Rules

Linear (Rule 1)

This is if:

• $p_1,\dots ,p_n$ have degrees of 0
• $q_1$ has degree 1
• $q_2$ has degree 2
• …
• $q_n$ has degree n

Quadratic (Rule 2)

• $p_1,\dots ,p_n$ have degrees of 1
• $q_1$ has degree 2
• $q_2$ has degree 4
• …
• $q_n$ has degree 2n

Examples

Example 1

Find the integral of: $\int \frac{1}{(x+2)(x-2)}dx$

1. We can represent the decomposition as: $\int \frac{A}{x+2}dx+\int \frac{B}{x-2}dx$
2. Solving for constants: $1=A(x-2)+B(x+2)$
3. With $x=2⟹1=A(0)+B(4)⟹B=\frac{1}{4}$
4. With $x=-2⟹1=A(-4)+B(0)⟹A=-\frac{1}{4}$
5. Thus, $=\int \frac{-\frac{1}{4}}{x+2}dx+\int \frac{\frac{1}{4}}{x-2}dx$
6. $=-\frac{1}{4}\int \frac{1}{x+2}dx+\frac{1}{4}\int \frac{1}{x-2}dx$
7. $=-\frac{1}{4}\ln |x+2|+c+\frac{1}{4}\ln |x-2|+c$
8. $=\frac{1}{4}\ln \frac{|x-2|}{|x+2|}+c$

Example 3

Find the integral of: $\int \frac{x}{(x-1)(x-2{)}^2}dx$

1. We can represent the decomposition as: $\int \frac{A}{x-1}dx+\int \frac{B}{x-2}dx+\int \frac{C}{(x-2{)}^2}dx$
2. Then, $x=A(x-2{)}^2+B(x-2)(x-1)+C(x-1)$
3. $x=2⟹C=2$
4. $x=1⟹A=1$
5. $x=3⟹B=-1$
6. Then, $=\ln (\frac{|x-1|}{|x-2|})-\frac{2}{x-2}+c$ Note we used Integration by Substitution for this one aswell.