Partial fraction decomposition is one of the most important techniques of integration in Calculus 2. It is the method that turns one complicated rational function into several simpler pieces that can be integrated term by term.
Most students get stuck because they try to start solving too soon. The Woody Calculus method starts earlier:
Identify the denominator type before you build the setup.
Once you know the denominator type, the partial fractions setup becomes predictable. You stop guessing and start matching structure.
This lesson teaches the Woody Calculus 3-type system for partial fractions. It is designed for Calculus 2 students learning techniques of integration, especially students who need a reliable way to handle distinct linear factors, repeated factors, irreducible quadratic factors, solving for constants, and integrating each resulting term.
Partial fractions belongs to the larger Calculus 2 integration-techniques sequence. Students should also study integration by parts and trig substitution so they can recognize which method fits each integral.
Estimated read time: 14–18 minutes.
Last updated: July 8, 2026.
Quick Summary: Partial Fractions in Calculus 2
- Partial fractions are used to integrate rational functions.
- First check degrees. If the numerator degree is greater than or equal to the denominator degree, divide first.
- Type I: distinct linear factors get constants on top.
- Type II: repeated factors require every power of the repeated factor.
- Type III: irreducible quadratic factors get linear numerators.
- After building the setup, multiply through by the denominator.
- Use smart values and coefficient matching to solve for constants.
- Then integrate each simple piece term by term.
Partial Fractions Key Facts
- Partial fraction decomposition works on proper rational functions.
- A rational function is proper when \(\deg(\text{numerator})<\deg(\text{denominator})\).
- If the rational function is improper, use polynomial long division first.
- Distinct linear factors use constant numerators.
- Repeated linear factors require all powers from \(1\) through the repeated power.
- Irreducible quadratic factors use linear numerators of the form \(Bx+C\).
- Over the real numbers, \(x^2+1\) is irreducible because it has no real roots.
- Partial fractions are part of the broader Calculus 2 toolbox with integration by parts, trig substitution, improper integrals, sequences, series, and Taylor series.
What Are Partial Fractions?
Partial fractions are a Calculus 2 technique used to rewrite a rational function as a sum of simpler rational functions. This turns many rational-function integrals into logarithms, rational terms, and arctangent terms that can be integrated directly.
A rational function has the form:
\frac{P(x)}{Q(x)},
\]
where \(P(x)\) and \(Q(x)\) are polynomials.
Partial fraction decomposition breaks that rational function into simpler terms such as:
\frac{A}{x-a},\qquad
\frac{B}{(x-a)^2},\qquad
\frac{Cx+D}{x^2+bx+c}.
\]
Once the rational function is split into simple pieces, the integral becomes much easier. This is why partial fractions is one of the core Calculus 2 integration techniques.
Check Degrees First
Before doing any partial fraction setup, always check degrees.
If:
\deg(P)<\deg(Q), \]
then the rational function is proper, and you can start partial fractions.
If:
\deg(P)\ge \deg(Q),
\]
then the rational function is improper, and you must divide first.
Check degrees first. If degree on top is greater than or equal to degree on bottom, divide first.
This is one of the most common places students lose points. Partial fractions only start after the rational expression is proper.
The 3 Basic Partial Fraction Setups
The Woody Calculus system separates partial fractions into three basic denominator types:
| Type | Denominator Factor | Numerator Setup | Example |
|---|---|---|---|
| Type I | Distinct linear factors | Constants on top | \(\frac{A}{x+5}+\frac{B}{x-2}\) |
| Type II | Repeated linear factors | Every power appears | \(\frac{A}{x+5}+\frac{B}{(x+5)^2}\) |
| Type III | Irreducible quadratic factors | Linear numerator | \(\frac{Bx+C}{x^2+1}\) |
The denominator tells you the setup. The setup tells you how many constants you need to solve for.
Type I: Distinct Linear Factors
Type I happens when the denominator factors into distinct linear factors.
Consider:
\int \frac{x-9}{x^2+3x-10}\,dx.
\]
Factor the denominator:
x^2+3x-10=(x+5)(x-2).
\]
Because both factors are distinct and linear, the setup is:
\frac{x-9}{(x+5)(x-2)}
=
\frac{A}{x+5}
+
\frac{B}{x-2}.
\]
The rule is simple:
Distinct linear factors get constants on top.
Type I: Solve and Integrate
Start from:
\frac{x-9}{(x+5)(x-2)}
=
\frac{A}{x+5}
+
\frac{B}{x-2}.
\]
Multiply through by \((x+5)(x-2)\):
x-9=A(x-2)+B(x+5).
\]
Use smart values. First let \(x=2\):
2-9=A(0)+B(7),
\]
-7=7B,
\]
B=-1.
\]
Now let \(x=-5\):
-5-9=A(-7)+B(0),
\]
-14=-7A,
\]
A=2.
\]
So:
\frac{x-9}{(x+5)(x-2)}
=
\frac{2}{x+5}
–
\frac{1}{x-2}.
\]
Now integrate term by term:
\int \frac{x-9}{x^2+3x-10}\,dx
=
\int \left(\frac{2}{x+5}-\frac{1}{x-2}\right)\,dx.
\]
\boxed{
\int \frac{x-9}{x^2+3x-10}\,dx
=
2\ln|x+5|-\ln|x-2|+C.
}
\]
Plain-language formula: The integral of \((x-9)/(x^2+3x-10)\) is \(2\ln|x+5|-\ln|x-2|+C\).
Type II: Repeated Factors
Type II happens when a linear factor repeats.
Consider:
\int \frac{dx}{(x+5)^2(x-1)}.
\]
The repeated factor is:
(x+5)^2.
\]
The key rule is:
If a factor repeats, every power must appear.
So the setup is:
\frac{1}{(x+5)^2(x-1)}
=
\frac{A}{x+5}
+
\frac{B}{(x+5)^2}
+
\frac{C}{x-1}.
\]
Do not skip the lower power. The term \(\frac{A}{x+5}\) must be included before \(\frac{B}{(x+5)^2}\).
Type II: Find A, B, and C
Start from:
\frac{1}{(x+5)^2(x-1)}
=
\frac{A}{x+5}
+
\frac{B}{(x+5)^2}
+
\frac{C}{x-1}.
\]
Multiply through by \((x+5)^2(x-1)\):
1=A(x+5)(x-1)+B(x-1)+C(x+5)^2.
\]
Use smart values. Let \(x=1\):
1=36C,
\]
C=\frac{1}{36}.
\]
Let \(x=-5\):
1=-6B,
\]
B=-\frac{1}{6}.
\]
Then use \(x=0\):
1=A(5)(-1)+B(-1)+C(25).
\]
Substitute \(B=-\frac16\) and \(C=\frac1{36}\):
1=-5A+\frac16+\frac{25}{36}.
\]
1=-5A+\frac{31}{36}.
\]
\frac5{36}=-5A.
\]
A=-\frac1{36}.
\]
Therefore:
A=-\frac1{36},
\qquad
B=-\frac16,
\qquad
C=\frac1{36}.
\]
Type II: Integrate Each Piece
Using:
A=-\frac1{36},
\qquad
B=-\frac16,
\qquad
C=\frac1{36},
\]
we get:
\frac{1}{(x+5)^2(x-1)}
=
-\frac{1}{36(x+5)}
–
\frac{1}{6(x+5)^2}
+
\frac{1}{36(x-1)}.
\]
So:
\int \frac{dx}{(x+5)^2(x-1)}
=
\int
\left[
-\frac{1}{36(x+5)}
–
\frac{1}{6(x+5)^2}
+
\frac{1}{36(x-1)}
\right]dx.
\]
Integrate term by term:
\boxed{
\int \frac{dx}{(x+5)^2(x-1)}
=
-\frac1{36}\ln|x+5|
+
\frac{1}{6(x+5)}
+
\frac1{36}\ln|x-1|
+C.
}
\]
Plain-language formula: The integral of \(1/((x+5)^2(x-1))\) is \(-\frac1{36}\ln|x+5|+\frac{1}{6(x+5)}+\frac1{36}\ln|x-1|+C\).
Type III: Irreducible Quadratic
Type III happens when the denominator contains an irreducible quadratic factor.
Consider:
\int
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
\,dx.
\]
The denominator has one linear factor and one irreducible quadratic factor:
x-1,
\qquad
x^2+1.
\]
The setup is:
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
=
\frac{A}{x-1}
+
\frac{Bx+C}{x^2+1}.
\]
The key rule is:
Linear factor gets a constant numerator. Irreducible quadratic gets a linear numerator.
Over the real numbers, \(x^2+1\) is irreducible because it has no real roots.
Type III: Find A, B, and C
Start from:
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
=
\frac{A}{x-1}
+
\frac{Bx+C}{x^2+1}.
\]
Multiply through by \((x-1)(x^2+1)\):
3x^2-4x+5
=
A(x^2+1)+(Bx+C)(x-1).
\]
Expand:
3x^2-4x+5
=
A(x^2+1)+Bx^2-Bx+Cx-C.
\]
Collect terms:
3x^2-4x+5
=
(A+B)x^2+(-B+C)x+(A-C).
\]
Match coefficients:
A+B=3,
\]
-B+C=-4,
\]
A-C=5.
\]
Solving gives:
A=2,
\qquad
B=1,
\qquad
C=-3.
\]
Therefore:
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
=
\frac{2}{x-1}
+
\frac{x-3}{x^2+1}.
\]
Type III: Integrate the Result
Using:
A=2,
\qquad
B=1,
\qquad
C=-3,
\]
the integrand becomes:
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
=
\frac{2}{x-1}
+
\frac{x-3}{x^2+1}.
\]
Split carefully:
\int
\left[
\frac{2}{x-1}
+
\frac{x-3}{x^2+1}
\right]dx
=
\int \frac{2}{x-1}\,dx
+
\int \frac{x}{x^2+1}\,dx
–
3\int \frac{1}{x^2+1}\,dx.
\]
Now integrate:
\int \frac{2}{x-1}\,dx=2\ln|x-1|,
\]
\int \frac{x}{x^2+1}\,dx=\frac12\ln(x^2+1),
\]
\int \frac{1}{x^2+1}\,dx=\arctan(x).
\]
Therefore:
\boxed{
\int
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
\,dx
=
2\ln|x-1|
+
\frac12\ln(x^2+1)
–
3\arctan(x)
+C.
}
\]
Plain-language formula: The integral of \((3x^2-4x+5)/((x-1)(x^2+1))\) is \(2\ln|x-1|+\frac12\ln(x^2+1)-3\arctan(x)+C\).
Partial Fractions Checklist
Multiple denominator types can appear in the same problem. For example:
\frac{P(x)}{(x-1)(x+2)^2(x^2+4)^2},
\qquad
\deg(P)<7. \]
This denominator contains:
- a distinct linear factor: \(x-1\),
- a repeated linear factor: \((x+2)^2\),
- a repeated irreducible quadratic factor: \((x^2+4)^2\).
The correct setup is:
\frac{P(x)}{(x-1)(x+2)^2(x^2+4)^2}
=
\frac{A}{x-1}
+
\frac{B}{x+2}
+
\frac{C}{(x+2)^2}
+
\frac{Dx+E}{x^2+4}
+
\frac{Fx+G}{(x^2+4)^2}.
\]
The memory rules are:
- Distinct linear factor \(\to\) constant on top.
- Repeated factor \(\to\) every power gets a fraction.
- Irreducible quadratic factor \(\to\) linear numerator.
The checklist is the bridge between this partial fractions lesson and the rest of the Woody Calculus integration system. Once students can identify the denominator structure, they can decide whether the integral needs partial fractions, integration by parts, or trigonometric substitution.
Worked Examples
Partial Fractions Example: Distinct Linear Factors
Evaluate:
\int \frac{x-9}{x^2+3x-10}\,dx.
\]
Factor:
x^2+3x-10=(x+5)(x-2).
\]
Set up:
\frac{x-9}{(x+5)(x-2)}
=
\frac{A}{x+5}
+
\frac{B}{x-2}.
\]
Solving gives \(A=2\) and \(B=-1\), so:
\boxed{
\int \frac{x-9}{x^2+3x-10}\,dx
=
2\ln|x+5|-\ln|x-2|+C.
}
\]
Partial Fractions Example: Repeated Factor
Evaluate:
\int \frac{dx}{(x+5)^2(x-1)}.
\]
Set up:
\frac{1}{(x+5)^2(x-1)}
=
\frac{A}{x+5}
+
\frac{B}{(x+5)^2}
+
\frac{C}{x-1}.
\]
Solving gives:
A=-\frac1{36},
\qquad
B=-\frac16,
\qquad
C=\frac1{36}.
\]
Therefore:
\boxed{
\int \frac{dx}{(x+5)^2(x-1)}
=
-\frac1{36}\ln|x+5|
+
\frac{1}{6(x+5)}
+
\frac1{36}\ln|x-1|
+C.
}
\]
Partial Fractions Example: Irreducible Quadratic
Evaluate:
\int
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
\,dx.
\]
Set up:
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
=
\frac{A}{x-1}
+
\frac{Bx+C}{x^2+1}.
\]
Solving gives:
A=2,
\qquad
B=1,
\qquad
C=-3.
\]
Therefore:
\boxed{
\int
\frac{3x^2-4x+5}{(x-1)(x^2+1)}
\,dx
=
2\ln|x-1|
+
\frac12\ln(x^2+1)
–
3\arctan(x)
+C.
}
\]
Common Mistakes
Mistake 1: Forgetting to check degrees first
Partial fractions only applies directly to proper rational functions. If the numerator degree is greater than or equal to the denominator degree, divide first.
Mistake 2: Skipping powers of a repeated factor
If the denominator contains \((x+a)^2\), the setup must include both:
\frac{A}{x+a}
\qquad
\text{and}
\qquad
\frac{B}{(x+a)^2}.
\]
Do not skip the lower power.
Mistake 3: Putting a constant over an irreducible quadratic
For an irreducible quadratic factor, the numerator must be linear:
\frac{Bx+C}{x^2+1}.
\]
A constant numerator is not general enough.
Mistake 4: Forgetting absolute values in logarithms
Integrals such as \(\int \frac{1}{x-a}\,dx\) produce:
\ln|x-a|+C.
\]
Mistake 5: Forgetting arctangent from irreducible quadratics
The integral:
\int \frac{1}{x^2+1}\,dx
\]
produces:
\arctan(x)+C.
\]
Partial Fractions Cheat Sheet
| Denominator Type | Setup Rule | Example Setup |
|---|---|---|
| Distinct linear factor | Constant numerator | \(\frac{A}{x-a}\) |
| Repeated linear factor | Include every power | \(\frac{A}{x-a}+\frac{B}{(x-a)^2}\) |
| Irreducible quadratic | Linear numerator | \(\frac{Bx+C}{x^2+bx+c}\) |
| Repeated irreducible quadratic | Every power gets a linear numerator | \(\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}\) |
Partial Fractions FAQ
What are partial fractions?
Partial fractions are a method for rewriting a rational function as a sum of simpler rational functions. This makes many Calculus 2 integrals easier to compute.
When should I use partial fraction decomposition?
Use partial fraction decomposition when integrating a rational function whose denominator can be factored into linear or irreducible quadratic factors.
What should I check before using partial fractions?
Check degrees first. If the numerator degree is greater than or equal to the denominator degree, use polynomial long division before partial fractions.
What is the setup for distinct linear factors?
Distinct linear factors get constant numerators. For example, \((x+5)(x-2)\) gives \(\frac{A}{x+5}+\frac{B}{x-2}\).
What is the setup for repeated factors?
If a linear factor repeats, every power must appear. For example, \((x+5)^2\) requires terms over \(x+5\) and \((x+5)^2\).
What is the setup for an irreducible quadratic?
An irreducible quadratic gets a linear numerator. For example, \(x^2+1\) gets a numerator of the form \(Bx+C\).
Why does an irreducible quadratic produce arctangent?
Terms involving \(\frac{1}{x^2+1}\) integrate to \(\arctan(x)\). That is why irreducible quadratic factors often produce logarithms plus arctangent terms.
What is the biggest mistake students make with partial fractions?
The biggest mistake is writing the setup incorrectly. The denominator type determines the correct numerator form, so identify the type before solving for constants.
Master Partial Fractions with Woody Calculus
Partial fractions is not about guessing constants. It is about building the correct algebraic structure before solving.
The Woody Calculus method is:
- Check degrees first.
- Divide first if needed.
- Factor completely.
- Identify the denominator type.
- Build the correct setup.
- Multiply through by the denominator.
- Solve for constants.
- Integrate each piece term by term.
To make the system automatic, rely on strict formula memorization, rewrite perfect solutions, and say each step out loud until the denominator type tells you the setup instantly. This builds the kind of peak performance study habits students need for Calculus 2 and advanced mathematics.
For students who want structured Calculus 2 help, join the Woody Calculus Mastery Lab. The Mastery Lab helps students build skill through repetition, step-by-step lessons, written solutions, exam-focused systems, and direct support.
You can also visit Woody Calculus on Skool for the learning community.
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