Integrals · Lesson 5 of 9

Integration by Partial Fractions

“Break rational functions into simpler pieces that integrate through familiar rules.”

Learning Objectives

• Distinguish proper and improper rational functions and use division when needed. • Choose a complete partial-fraction form from the denominator factors. • Determine coefficients using polynomial identities. • Integrate terms involving linear factors and irreducible quadratics. • Combine a genuine change of variable with partial fractions when useful.

Why decomposition helps

A rational function is a quotient of polynomials. A complicated denominator may be difficult to integrate as one object but straightforward after the quotient is written as a sum of simpler fractions. Partial fractions reverse the familiar process of adding fractions over a common denominator.

Definition

Proper rational function

A rational function $P (x) / Q (x)$ is proper if the degree of $P$ is smaller than the degree of $Q$ . It is improper when the numerator degree is greater than or equal to the denominator degree. In either case $Q (x) \neq = 0$ on the domain.

For an improper quotient, divide the polynomials first. The quotient polynomial can be integrated directly, leaving a proper remainder for decomposition. This step prevents a partial-fraction form with too few terms from being forced onto the wrong kind of expression.

Polynomial division before decompositionLaTeX

\frac{P ( x )}{Q ( x )} = T (x) + \frac{R ( x )}{Q ( x )}, de g R < de g Q

Here

T

is the polynomial quotient and

R

the remainder. The equality holds wherever the original denominator is nonzero.

Choosing a complete decomposition

The denominator factors dictate the fractions to include. A repeated factor requires every power up to its multiplicity. An irreducible quadratic needs a linear numerator because a constant alone cannot represent every possible remainder. The coefficients are unknown constants, determined after the correct structure is chosen.

Denominator structure	Form for a proper quotient
$(x - a) (x - b)$ , $a \neq = b$	$A / (x - a) + B / (x - b)$
$(x - a)^{2}$	$A / (x - a) + B / (x - a)^{2}$
$(x - a) (x - b) (x - c)$ , distinct roots	$A / (x - a) + B / (x - b) + C / (x - c)$
$(x - a)^{2} (x - b)$ , $a \neq = b$	$A / (x - a) + B / (x - a)^{2} + C / (x - b)$
$(x - a) (x^{2} + b x + c)$ , quadratic irreducible over the reals	$A / (x - a) + (B x + C) / (x^{2} + b x + c)$

Multiply by the common denominator to obtain a polynomial identity, meaning an equality true for all real values of the polynomial variable. Compare coefficients or substitute convenient values to find the constants. Values excluded from the original fraction can be substituted into this polynomial identity because its denominators have already been cleared; this does not redefine the original function at those points.

Example — Two distinct linear factors

Problem
Evaluate $\int d x / [(x + 1) (x + 2)]$ .

1.Write $1/ [(x + 1) (x + 2)] = A / (x + 1) + B / (x + 2)$ . Clearing denominators gives $1 = A (x + 2) + B (x + 1)$ .
2.Matching coefficients gives $A + B = 0$ and $2 A + B = 1$ , so $A = 1$ and $B = - 1$ .
3.Integrate the two pieces: $ln ∣ x + 1∣ - ln ∣ x + 2∣ + C = ln ∣ (x + 1) / (x + 2) ∣ + C$ .

Example — Divide before decomposing

Problem
Evaluate $\int (x^{2} + 1) / (x^{2} - 5 x + 6) d x$ .

1.The degrees are equal. Division gives $1 + (5 x - 5) / (x^{2} - 5 x + 6)$ , and the denominator factors as $(x - 2) (x - 3)$ .
2.Write $(5 x - 5) / [(x - 2) (x - 3)] = A / (x - 2) + B / (x - 3)$ . Clearing denominators gives $5 x - 5 = A (x - 3) + B (x - 2)$ .
3.Thus $A + B = 5$ and $3 A + 2 B = 5$ . Solve to get $A = - 5$ and $B = 10$ .
4.The integral is $x - 5 ln ∣ x - 2∣ + 10 ln ∣ x - 3∣ + C$ , on intervals avoiding $2$ and $3$ .

Example — A repeated factor

Problem
Evaluate $\int (3 x - 2) / [(x + 1)^{2} (x + 3)] d x$ .

1.Use $A / (x + 1) + B / (x + 1)^{2} + C / (x + 3)$ . Clearing denominators gives $3 x - 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)^{2}$ .
2.Putting $x = - 1$ gives $- 5 = 2 B$ , hence $B = - 5/2$ . Putting $x = - 3$ gives $- 11 = 4 C$ , hence $C = - 11/4$ . The coefficient of $x^{2}$ gives $A = - C = 11/4$ .
3.The three antiderivatives are $(11/4) ln ∣ x + 1∣$ , $5/ [2 (x + 1)]$ , and $- (11/4) ln ∣ x + 3∣$ .
4.Combining gives $\frac{11}{4} ln ∣ (x + 1) / (x + 3) ∣ + 5/ [2 (x + 1)] + C$ . Omitting the first-power term in the decomposition would lose the needed logarithm.

Example — Three distinct factors

Problem
Evaluate $\int d x / [x (x - 1) (x + 1)]$ .

1.Write the fraction as $A / x + B / (x - 1) + D / (x + 1)$ . The polynomial identity is $1 = A (x^{2} - 1) + B x (x + 1) + D x (x - 1)$ .
2.Evaluate at $x = 0, 1, - 1$ to get $A = - 1$ , $B = 1/2$ , and $D = 1/2$ .
3.Integrate to obtain $- ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 1∣ + \frac{1}{2} ln ∣ x + 1∣ + C$ . The permitted intervals avoid all three roots.

Quadratic factors and two different uses of substitution

Some denominators contain quadratic factors that do not split into real linear factors. Their integrals use the standard forms from the previous lesson. A temporary algebraic letter may simplify coefficient matching without changing the integration variable; a genuine substitution, by contrast, must also transform the differential.

Example — An algebraic abbreviation is not a change of variable

Problem
Evaluate $\int x^{2} / [(x^{2} + 1) (x^{2} + 4)] d x$ .

1.Temporarily write $y = x^{2}$ only in the fraction identity: $y / [(y + 1) (y + 4)] = A / (y + 1) + B / (y + 4)$ .
2.The identity $y = A (y + 4) + B (y + 1)$ gives $A + B = 1$ and $4 A + B = 0$ , so $A = - 1/3$ and $B = 4/3$ .
3.Return to $x$ before integrating: the integral is $- \frac{1}{3} \int d x / (x^{2} + 1) + \frac{4}{3} \int d x / (x^{2} + 4)$ . There is no replacement of $d x$ by $d y$ .
4.The answer is $- \frac{1}{3} tan^{- 1} x + \frac{2}{3} tan^{- 1} (x /2) + C$ .

Example — A linear factor and an irreducible quadratic

Problem
Evaluate $\int (x^{2} + x + 1) / [(x + 2) (x^{2} + 1)] d x$ .

1.Write the integrand as $A / (x + 2) + (B x + D) / (x^{2} + 1)$ . Clearing denominators gives $x^{2} + x + 1 = A (x^{2} + 1) + (B x + D) (x + 2)$ .
2.Matching coefficients gives $A + B = 1$ , $2 B + D = 1$ , and $A + 2 D = 1$ . Solving yields $A = 3/5$ , $B = 2/5$ , and $D = 1/5$ .
3.The numerator $2 x /5 + 1/5$ above $x^{2} + 1$ separates into a logarithmic-derivative term and an inverse-tangent term.
4.The answer is $\frac{3}{5} ln ∣ x + 2∣ + \frac{1}{5} ln (x^{2} + 1) + \frac{1}{5} tan^{- 1} x + C$ .

Example — A genuine substitution followed by partial fractions

Problem
Evaluate $\int (3 sin ϕ - 2) cos ϕ / [5 - cos^{2} ϕ - 4 sin ϕ] d ϕ$ .

1.Put $y = sin ϕ$ and $d y = cos ϕ d ϕ$ . Using $cos^{2} ϕ = 1 - y^{2}$ , the integral becomes $\int (3 y - 2) / (y - 2)^{2} d y$ .
2.Write $(3 y - 2) / (y - 2)^{2} = A / (y - 2) + B / (y - 2)^{2}$ . Since $3 y - 2 = A (y - 2) + B$ , we get $A = 3$ and $B = 4$ .
3.Integration gives $3 ln ∣ y - 2∣ - 4/ (y - 2) + C$ .
4.Substitute back: $3 ln (2 - sin ϕ) + 4/ (2 - sin ϕ) + C$ . The logarithm needs no absolute value in this final form because $2 - sin ϕ$ is always positive.

Common mistake

A repeated linear factor needs a whole chain of fractions, not just the highest power. Also, do not replace a quadratic numerator by a constant in advance. Finally, using a symbol such as $y = x^{2}$ to organise an algebraic identity does not justify replacing $d x$ by $d y$ during integration.

Quick check

When is a rational function proper?

Quick check

Which structure is needed for a proper fraction with denominator $(x - 1)^{2} (x + 2)$ ?

Quick check

What numerator is generally required above the irreducible quadratic $x^{2} + 1$ ?

Quick check

What is $\int d x / (x - 2)^{2}$ ?

Quick check

Why is division the first step for $(x^{2} + 1) / (x^{2} - 5 x + 6)$ ?

Practice Problems

Find $\int x / [(x + 1) (x + 2)] d x$ .
Find $\int (3 x - 1) / [(x - 1) (x - 2) (x - 3)] d x$ .
Find $\int x / [(x - 1) (x + 2)^{2}] d x$ .
Find $\int (x^{3} + x + 1) / (x^{2} - 1) d x$ after polynomial division.
Find $\int 2/ [(1 - x) (1 + x^{2})] d x$ .
Find $\int d x / [x (x^{n} + 1)]$ for $x > 0$ and a positive integer $n$ . Use $t = x^{n}$ after multiplying by $x^{n - 1} / x^{n - 1}$ .
Find $\int cos x / [(1 - sin x) (2 - sin x)] d x$ .
Find $\int d x / (e^{x} - 1)$ by putting $t = e^{x}$ and decomposing the resulting rational function.

Key Takeaways

Divide improper rational functions before decomposing the proper remainder. Denominator factors determine the full partial-fraction form. Find coefficients from a polynomial identity, then integrate each simple term. Irreducible quadratics may contribute both logarithms and inverse tangents. An algebraic abbreviation and a genuine integration substitution have different effects on the differential.

Previous · Lesson 4

Standard Integrals Involving Quadratic Expressions

Next · Lesson 6

Integration by Parts and Radical Forms

Why decomposition helps

Choosing a complete decomposition

Quadratic factors and two different uses of substitution

Quiz

Practice Problems

Key Takeaways