The Central Idea
Integration by partial fractions follows the pattern
\[ \boxed{ \text{complicated rational function} \longrightarrow \text{sum of simple rational pieces} \longrightarrow \text{integrate term by term} }. \]For example,
\[ \frac{1}{(x-1)(x+2)} \]can be rewritten as
\[ \frac{A}{x-1} + \frac{B}{x+2}. \]Each new term integrates using a logarithm.
What Is a Rational Function?
A rational function has the form
\[ \boxed{ \frac{P(x)}{Q(x)} }, \]where \(P\) and \(Q\) are polynomials and \(Q(x)\neq0\).
Partial fractions applies primarily when the denominator can be factored over the real numbers into linear and irreducible quadratic factors.
Proper Rational Functions
A rational function is proper when
\[ \boxed{ \deg P<\deg Q }. \]Partial fraction decomposition assumes this condition.
If the numerator degree is greater than or equal to the denominator degree, use polynomial division first.
Improper Rational Functions
Suppose
\[ \deg P\geq\deg Q. \]Divide:
\[ \frac{P(x)}{Q(x)} = S(x)+\frac{R(x)}{Q(x)}, \]where
\[ \deg R<\deg Q. \]Then integrate the polynomial \(S(x)\) directly and decompose only the proper remainder.
Factor the Denominator First
The structure of the decomposition is determined entirely by the factorization of the denominator.
Before writing any partial fractions:
- make the rational function proper;
- factor the denominator completely over the real numbers;
- identify repeated factors;
- identify irreducible quadratic factors.
Case 1: Distinct Linear Factors
If
\[ Q(x)=(x-a)(x-b), \]then write
\[ \boxed{ \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} }. \]There is one constant numerator for each distinct linear factor.
Example: Distinct Linear Factors
Decompose
\[ \frac{5x+1}{(x-1)(x+2)}. \]Write
\[ \frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}. \]Clear the Denominators
Multiply by
\[ (x-1)(x+2). \]Then
\[ 5x+1 = A(x+2)+B(x-1). \]The problem is now an ordinary polynomial identity.
Solving Coefficients by Convenient Substitution
Set
\[ x=1. \]Then
\[ 6=3A, \]so
\[ A=2. \]Set
\[ x=-2. \]Then
\[ -9=-3B, \]so
\[ B=3. \]Therefore,
\[ \boxed{ \frac{5x+1}{(x-1)(x+2)} = \frac2{x-1} + \frac3{x+2} }. \]Integrating the Decomposition
\[ \int \frac{5x+1}{(x-1)(x+2)}\,dx = \int\frac2{x-1}\,dx + \int\frac3{x+2}\,dx. \]Therefore,
\[ \boxed{ 2\ln|x-1| + 3\ln|x+2| + C }. \]Solving Coefficients by Matching Terms
Starting from
\[ 5x+1 = A(x+2)+B(x-1), \]expand:
\[ 5x+1 = (A+B)x+(2A-B). \]Match coefficients:
\[ A+B=5, \] \[ 2A-B=1. \]Solving gives
\[ A=2,\qquad B=3. \]Coefficient matching is useful when convenient substitution does not solve every constant immediately.
Case 2: Repeated Linear Factors
If the denominator contains
\[ (x-a)^m, \]include one term for every power:
\[ \boxed{ \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_m}{(x-a)^m} }. \]Omitting intermediate powers produces an incomplete decomposition.
Example: Repeated Linear Factor
Decompose
\[ \frac{3x+5}{(x-1)^2}. \]Write
\[ \frac{3x+5}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}. \]Multiply through by \((x-1)^2\):
\[ 3x+5 = A(x-1)+B. \]Solving the Repeated-Factor Example
Set \(x=1\):
\[ 8=B. \]Therefore,
\[ B=8. \]Compare the coefficient of \(x\):
\[ A=3. \]Thus,
\[ \boxed{ \frac{3x+5}{(x-1)^2} = \frac3{x-1} + \frac8{(x-1)^2} }. \]Integrating Repeated Linear Terms
\[ \int \frac{3x+5}{(x-1)^2}\,dx = 3\int\frac1{x-1}\,dx + 8\int(x-1)^{-2}\,dx. \]Therefore,
\[ \boxed{ 3\ln|x-1| - \frac8{x-1} + C }. \]Case 3: Irreducible Quadratic Factors
Suppose the denominator contains an irreducible quadratic
\[ ax^2+bx+c \]that cannot be factored over the real numbers.
The numerator must be linear:
\[ \boxed{ \frac{Ax+B}{ax^2+bx+c} }. \]A constant numerator alone is not general enough.
Example: Linear and Quadratic Factors
Decompose
\[ \frac{x^2+3x+5}{(x-1)(x^2+1)}. \]Since \(x^2+1\) is irreducible over the reals, write
\[ \boxed{ \frac{x^2+3x+5}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} }. \]Clearing Denominators
Multiply through by
\[ (x-1)(x^2+1). \]Then
\[ x^2+3x+5 = A(x^2+1) +(Bx+C)(x-1). \]Expand the right side:
\[ A x^2+A + Bx^2-Bx+Cx-C. \]Therefore,
\[ x^2+3x+5 = (A+B)x^2 +(-B+C)x +(A-C). \]Matching Coefficients
Equate coefficients:
\[ A+B=1, \] \[ -B+C=3, \] \[ A-C=5. \]Solving gives
\[ A=3,\qquad B=-2,\qquad C=1. \]Thus,
\[ \boxed{ \frac{x^2+3x+5}{(x-1)(x^2+1)} = \frac3{x-1} + \frac{-2x+1}{x^2+1} }. \]Integrating the Linear-over-Quadratic Piece
Split:
\[ \int \frac{-2x+1}{x^2+1}\,dx = -2\int\frac{x}{x^2+1}\,dx + \int\frac1{x^2+1}\,dx. \]The first term gives a logarithm:
\[ -\ln(x^2+1). \]The second gives
\[ \arctan x. \]Therefore,
\[ \boxed{ 3\ln|x-1| - \ln(x^2+1) + \arctan x + C }. \]Case 4: Repeated Irreducible Quadratic Factors
If the denominator contains
\[ (x^2+1)^m, \]include a linear numerator over every power:
\[ \boxed{ \frac{A_1x+B_1}{x^2+1} + \frac{A_2x+B_2}{(x^2+1)^2} + \cdots + \frac{A_mx+B_m}{(x^2+1)^m} }. \]General Decomposition Template
For distinct linear factors:
\[ \frac{A}{x-a}. \]For repeated linear factors:
\[ \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} +\cdots. \]For irreducible quadratic factors:
\[ \frac{Ax+B}{x^2+px+q}. \]For repeated irreducible quadratics, include a linear numerator over each power.
Why Quadratic Factors Need Linear Numerators
A numerator over a quadratic factor must have degree lower than the quadratic.
The most general polynomial of degree less than \(2\) is
\[ Ax+B. \]That is why
\[ \frac{Ax+B}{x^2+px+q} \]is required.
Completing the Square after Decomposition
An irreducible quadratic may not initially look like the standard arctangent form.
For example,
\[ x^2+4x+5 \]becomes
\[ (x+2)^2+1. \]Completing the square often reveals an integral of the form
\[ \int\frac{du}{u^2+a^2}. \]The Arctangent Form
Recall
\[ \boxed{ \int \frac{1}{u^2+a^2}\,du = \frac1a \arctan\left(\frac{u}{a}\right) + C }. \]This form appears frequently after partial fractions with irreducible quadratic factors.
Look for the Derivative of the Denominator
After decomposition, a term such as
\[ \frac{Ax+B}{x^2+px+q} \]can often be split into:
- a multiple of the denominator's derivative;
- a leftover constant term.
The derivative part integrates logarithmically, while the remainder often leads to arctangent after completing the square.
Example: Splitting a Linear Numerator
Consider
\[ \int \frac{3x+5}{x^2+4x+5}\,dx. \]The derivative of the denominator is
\[ 2x+4. \]Write
\[ 3x+5 = \frac32(2x+4)-1. \]Therefore,
\[ \int \frac{3x+5}{x^2+4x+5}\,dx = \frac32 \int \frac{2x+4}{x^2+4x+5}\,dx - \int \frac1{(x+2)^2+1}\,dx. \]Hence,
\[ \boxed{ \frac32\ln(x^2+4x+5) - \arctan(x+2) + C }. \]The Cover-Up Shortcut
For distinct linear factors, coefficients can sometimes be found quickly.
If
\[ \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}, \]then
\[ A= \left. \frac{P(x)}{x-b} \right|_{x=a}, \]and similarly for \(B\).
This shortcut works cleanly for distinct linear factors but should not be used blindly for repeated or quadratic factors.
Example: Polynomial Division First
Evaluate
\[ \int \frac{x^2+1}{x-1}\,dx. \]Since the numerator degree is larger, divide:
\[ \frac{x^2+1}{x-1} = x+1+\frac2{x-1}. \]Therefore,
\[ \int \frac{x^2+1}{x-1}\,dx = \int(x+1)\,dx + 2\int\frac1{x-1}\,dx. \]Hence,
\[ \boxed{ \frac{x^2}{2} + x + 2\ln|x-1| + C }. \]Partial Fractions in Definite Integrals
The same decomposition method works for definite integrals, provided the integrand is continuous on the interval.
For example,
\[ \int_0^1 \frac{1}{(x+1)(x+2)}\,dx. \]First decompose:
\[ \frac1{(x+1)(x+2)} = \frac1{x+1} - \frac1{x+2}. \]Then integrate term by term and evaluate the endpoints.
Example: Definite Integral
\[ \int_0^1 \left( \frac1{x+1} - \frac1{x+2} \right)\,dx. \]An antiderivative is
\[ \ln(x+1)-\ln(x+2). \]Therefore,
\[ [\ln(x+1)-\ln(x+2)]_0^1. \]This gives
\[ (\ln2-\ln3)-(\ln1-\ln2). \]Since \(\ln1=0\),
\[ \boxed{ 2\ln2-\ln3 }. \]Connection to Improper Integrals
If a partial-fraction decomposition reveals a denominator factor that vanishes inside the interval, the definite integral is improper.
For example,
\[ \int_0^2\frac1{x-1}\,dx \]must be split at \(x=1\).
Algebraic decomposition never removes the need to check domain and singularities.
Verify the Decomposition
Before integrating, recombine the partial fractions mentally or algebraically.
A quick verification can catch:
- sign errors;
- missing repeated-factor terms;
- incorrect coefficients;
- incorrect numerator degree over quadratics.
Common Error: Skipping Polynomial Division
If
\[ \deg P\geq\deg Q, \]do not begin partial fractions immediately.
Divide first so the remainder is proper.
Common Error: Not Factoring Completely
The denominator determines the entire template.
If a factorization is incomplete, the partial fraction form will also be incomplete or incorrect.
Common Error: Omitting Powers of a Repeated Factor
For
\[ (x-a)^3, \]you need
\[ \boxed{ \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3} }. \]Including only the first and third terms is not sufficient.
Common Error: Using a Constant Numerator over a Quadratic
Over an irreducible quadratic factor, use
\[ \boxed{Ax+B}, \]not merely \(A\).
Common Error: Forgetting Absolute Values in Logarithms
For a linear factor,
\[ \int\frac1{x-a}\,dx = \boxed{\ln|x-a|+C}. \]The absolute value is required in the general indefinite integral.
Common Error: Treating Every Quadratic Term as a Logarithm
A term such as
\[ \frac1{x^2+1} \]does not have logarithmic antiderivative.
Instead,
\[ \boxed{ \int\frac1{x^2+1}\,dx = \arctan x+C }. \]Common Error: Solving Coefficients before Clearing Denominators
The clean coefficient equation comes after multiplying through by the common denominator.
This converts the rational identity into a polynomial identity that can be solved by substitution or coefficient matching.
A Reliable Partial-Fractions Procedure
- Check whether the rational function is proper.
- If not, perform polynomial division.
- Factor the denominator completely over the reals.
- Write the correct decomposition template.
- Clear denominators.
- Solve for the unknown coefficients.
- Rewrite the original rational function.
- Integrate each simpler term.
- Use logarithms, power rules, and arctangent forms as appropriate.
- Differentiate or recombine to verify if needed.
Template Summary
| Denominator factor | Required partial-fraction terms |
|---|---|
| \(x-a\) | \(\frac{A}{x-a}\) |
| \((x-a)^m\) | \(\frac{A_1}{x-a}+\cdots+\frac{A_m}{(x-a)^m}\) |
| \(x^2+px+q\) irreducible | \(\frac{Ax+B}{x^2+px+q}\) |
| \((x^2+px+q)^m\) | linear numerator over every power |
Practice 1: Distinct Linear Factors
Decompose
\[ \frac{7x+1}{(x-1)(x+3)}. \]Practice 2: Integrate Distinct Linear Factors
Evaluate
\[ \int \frac{7x+1}{(x-1)(x+3)}\,dx. \]Practice 3: Repeated Linear Template
Write the correct decomposition form for
\[ \frac{P(x)}{(x-2)^3(x+1)}. \]Practice 4: Quadratic Template
Write the correct decomposition form for
\[ \frac{P(x)}{(x-1)(x^2+4)}. \]Practice 5: Repeated Quadratic Template
Write the correct decomposition form for
\[ \frac{P(x)}{(x^2+1)^2}. \]Practice 6: Polynomial Division
Rewrite
\[ \frac{x^2+3}{x+1} \]as a polynomial plus a proper rational function.
Practice 7: Simple Integral
Evaluate
\[ \int \frac1{x^2-1}\,dx. \]Practice 8: Irreducible Quadratic
Evaluate
\[ \int \frac{2x+3}{x^2+4}\,dx. \]Practice 9: Complete the Square
Rewrite
\[ x^2+6x+13 \]in completed-square form.
Practice 10: Identify the Error
A student writes
\[ \frac{1}{(x-1)^2} = \frac{A}{(x-1)^2} \]as the full repeated-factor template.
What term is missing?
Practice 11: Quadratic Numerator Error
A student writes
\[ \frac{P(x)}{x^2+1} = \frac{A}{x^2+1}. \]What is the correct general numerator?
Practice 12: Domain Check
Why must
\[ \int_0^2\frac1{x-1}\,dx \]be treated as an improper integral?
Practice Answers
Practice 1
Write
\[ \frac{7x+1}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3}. \]Clearing denominators:
\[ 7x+1=A(x+3)+B(x-1). \]Set \(x=1\):
\[ 8=4A, \]so
\[ A=2. \]Set \(x=-3\):
\[ -20=-4B, \]so
\[ B=5. \]Therefore,
\[ \boxed{ \frac{7x+1}{(x-1)(x+3)} = \frac2{x-1} + \frac5{x+3} }. \]Practice 2
Using the decomposition above,
\[ \int \frac{7x+1}{(x-1)(x+3)}\,dx = 2\int\frac1{x-1}\,dx + 5\int\frac1{x+3}\,dx. \]Hence,
\[ \boxed{ 2\ln|x-1| + 5\ln|x+3| + C }. \]Practice 3
\[ \boxed{ \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{x+1} }. \]Practice 4
\[ \boxed{ \frac{A}{x-1} + \frac{Bx+C}{x^2+4} }. \]Practice 5
\[ \boxed{ \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} }. \]Practice 6
Divide:
\[ x^2+3=(x+1)(x-1)+4. \]Therefore,
\[ \boxed{ \frac{x^2+3}{x+1} = x-1+\frac4{x+1} }. \]Practice 7
Factor:
\[ x^2-1=(x-1)(x+1). \]Decompose:
\[ \frac1{x^2-1} = \frac{1/2}{x-1} - \frac{1/2}{x+1}. \]Therefore,
\[ \boxed{ \frac12\ln|x-1| - \frac12\ln|x+1| + C }. \]Practice 8
Split:
\[ \int \frac{2x+3}{x^2+4}\,dx = \int\frac{2x}{x^2+4}\,dx + 3\int\frac1{x^2+4}\,dx. \]Therefore,
\[ \boxed{ \ln(x^2+4) + \frac32\arctan\left(\frac{x}{2}\right) + C }. \]Practice 9
\[ x^2+6x+13 = (x+3)^2+4. \]Therefore,
\[ \boxed{ (x+3)^2+4 }. \]Practice 10
The missing term is
\[ \boxed{ \frac{B}{x-1} }. \]A repeated factor of power \(2\) requires terms over both the first and second powers.
Practice 11
The correct general numerator is linear:
\[ \boxed{Ax+B}. \]Practice 12
The denominator is zero at
\[ x=1, \]which lies inside the interval \([0,2]\).
Therefore, the integrand has an interior singularity and the integral must be split and interpreted through one-sided limits.
What to Remember
Partial fractions applies to rational functions after they are made proper and the denominator is completely factored.
Distinct linear factor:
\[ \boxed{ \frac{A}{x-a} }. \]Repeated linear factor:
\[ \boxed{ \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} +\cdots }. \]Irreducible quadratic factor:
\[ \boxed{ \frac{Ax+B}{x^2+px+q} }. \]The method is:
\[ \boxed{ \text{divide} \rightarrow \text{factor} \rightarrow \text{decompose} \rightarrow \text{solve coefficients} \rightarrow \text{integrate} }. \]Once decomposition is complete, the complicated rational integral becomes a collection of familiar logarithmic, reciprocal-power, and arctangent integrals.
Where We Go Next
Partial fractions is one of the major algebraic techniques for integration. It works alongside substitution and integration by parts, and it is especially useful for rational expressions whose denominators factor into manageable algebraic pieces.