3.4 Partial Fractions

Partial Fraction Decomposition

Partial fraction decomposition lets you break a complicated rational function into a sum of simpler fractions that you already know how to integrate. It's one of the most practical techniques in this course because many integrals you'll encounter have rational functions, and without this method, they'd be extremely difficult to handle.

The core idea: factor the denominator, set up a template of simpler fractions (one for each factor), then solve for the unknown constants.

When to Use Partial Fractions

Before decomposing, check two things:

1. The integrand must be a proper rational function, meaning the degree of the numerator $P(x)$ is strictly less than the degree of the denominator $Q(x)$.
2. If it's improper (numerator degree ≥ denominator degree), perform polynomial long division first, then apply partial fractions to the remainder.

General Setup

For a proper rational function $\frac{P(x)}{Q(x)}$, you factor $Q(x)$ completely, then write one partial fraction term for each factor. The form of each term depends on the type of factor (linear, repeated linear, or irreducible quadratic).

Once you've written the decomposition template, find the unknown constants by:

1. Multiplying both sides by $Q(x)$ to clear denominators.
2. Solving the resulting equation, either by substituting strategic values of $x$ or by equating coefficients of like powers of $x$.

Simple (Distinct) Linear Factors

When the denominator factors into distinct linear terms like $(x - a_1)(x - a_2)\cdots(x - a_n)$, the decomposition looks like:

$\frac{P(x)}{(x-a_1)(x-a_2)\cdots(x-a_n)} = \frac{A_1}{x-a_1} + \frac{A_2}{x-a_2} + \cdots + \frac{A_n}{x-a_n}$

Each constant $A_i$ can be found quickly using the cover-up method: multiply both sides by $(x - a_i)$ and then plug in $x = a_i$. This zeroes out every other term, leaving just $A_i$.

Worked example: Integrate $\displaystyle\int \frac{2x+1}{x^2 - x - 2}\,dx$.

1. Factor the denominator: $x^2 - x - 2 = (x-2)(x+1)$.

2. Set up the decomposition: $\frac{2x+1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$.

3. Multiply both sides by $(x-2)(x+1)$: $2x + 1 = A(x+1) + B(x-2)$.

4. Substitute $x = 2$: $5 = 3A$, so $A = \frac{5}{3}$.

5. Substitute $x = -1$: $-1 = -3B$, so $B = \frac{1}{3}$.

6. Integrate: $\int \frac{5/3}{x-2}\,dx + \int \frac{1/3}{x+1}\,dx = \frac{5}{3}\ln|x-2| + \frac{1}{3}\ln|x+1| + C$.

Repeated Linear Factors

When a linear factor $(x - a)$ appears with multiplicity $n$, you need n separate terms with increasing powers in the denominator:

$\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}$

To find the constants:

1. Multiply both sides by $(x-a)^n$.
2. Plug in $x = a$ to find $A_n$ directly.
3. Find the remaining constants by either equating coefficients or taking successive derivatives of both sides and evaluating at $x = a$.

Worked example: Decompose $\frac{x+2}{(x-1)^3}$.

1. Template: $\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$.

2. Multiply through by $(x-1)^3$: $x + 2 = A(x-1)^2 + B(x-1) + C$.

3. Set $x = 1$: $3 = C$.

4. Expand the right side and equate coefficients to find $A$ and $B$. Expanding: $A(x^2 - 2x + 1) + B(x - 1) + 3 = Ax^2 + (-2A+B)x + (A - B + 3)$.

5. Comparing to $0\cdot x^2 + 1\cdot x + 2$: $A = 0$, $-2(0) + B = 1$ so $B = 1$, and you can verify the constant term: $0 - 1 + 3 = 2$. ✓

The resulting integrals involve $\ln|x-a|$ for the first term and power rule (rewritten as $(x-a)^{-k}$) for the higher-power terms.

Irreducible Quadratic Factors

A quadratic factor $ax^2 + bx + c$ is irreducible when its discriminant $b^2 - 4ac < 0$, meaning it has no real roots and can't be factored over the reals. For each such factor, the numerator of the partial fraction must be linear (not just a constant):

$\frac{Ax + B}{ax^2 + bx + c}$

If the irreducible quadratic is repeated with multiplicity $n$, you need $n$ terms:

$\frac{A_1x + B_1}{ax^2+bx+c} + \frac{A_2x + B_2}{(ax^2+bx+c)^2} + \cdots + \frac{A_nx + B_n}{(ax^2+bx+c)^n}$

Find $A$ and $B$ by multiplying out and equating coefficients of like powers of $x$.

Integrating these terms typically requires splitting the fraction:

• A portion that looks like $\frac{Ax}{ax^2+bx+c}$ can often be handled with a $u$-substitution (since the derivative of the denominator is in the numerator, or close to it).
• A remaining constant portion like $\frac{B}{ax^2+bx+c}$ requires completing the square and produces an arctangent: $\frac{1}{a}\arctan\left(\frac{x+d}{e}\right)$.

Putting It All Together: The Full Process

Here's the complete workflow for any partial fractions problem:

1. Check degrees. If the numerator's degree ≥ the denominator's degree, do polynomial long division first.
2. Factor the denominator completely into linear and irreducible quadratic factors.
3. Write the decomposition template using the rules above (one term per distinct linear factor, $n$ terms per repeated linear factor of multiplicity $n$, linear numerator for each quadratic factor).
4. Clear the denominators by multiplying both sides by the full denominator.
5. Solve for the constants using strategic substitution (plug in roots of the denominator) and/or equating coefficients.
6. Integrate each term separately. Linear denominators give $\ln$; higher-power denominators give power-rule results; irreducible quadratics give $\ln$ and/or $\arctan$ terms.