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Lesson 86 — Integrals of rational functions (partial fractions)

Decomposition of P(x)/Q(x) as a sum of simple fractions. Simple real roots, multiplicity, and irreducible quadratics. Reduces to elementary integrals in ln or arctan.

Used in: Calculus II (Brazil) · Equiv. Math III Japanese · Equiv. Analysis LK German · AP Calculus BC (USA)

\frac{P(x)}{Q(x)} = \sum_{i} \frac{A_i}{(x - r_i)^{k_i}} + \sum_{j} \frac{B_j x + C_j}{(x^2 + p_j x + q_j)^{l_j}}

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Rigorous notation, full derivation, hypotheses

Theorem, procedure, and cases

Partial Fractions Decomposition Theorem

Definition· Partial fractions decomposition over ℝ

Let $P, Q \in \mathbb{R}[x]$ with $\deg P < \deg Q$ , and let

$Q(x) = a \prod_{i=1}^{s}(x - r_i)^{m_i} \prod_{j=1}^{t}(x^2 + p_j x + q_j)^{n_j}$

be the complete factorization of $Q$ over $\mathbb{R}$ (with $p_j^2 - 4q_j < 0$ — irreducible quadratic factors). Then there exists a unique decomposition:

$\frac{P(x)}{Q(x)} = \sum_{i=1}^{s} \sum_{k=1}^{m_i} \frac{A_{i,k}}{(x - r_i)^k} + \sum_{j=1}^{t} \sum_{l=1}^{n_j} \frac{B_{j,l} x + C_{j,l}}{(x^2 + p_j x + q_j)^l}.$

"We can always write the integrand as a sum of simpler rational functions using the method of partial fractions. The idea is to decompose the rational function into a sum of simpler pieces, each of which is easier to integrate." — OpenStax Calculus Vol. 2, §3.4

Procedure

Definition· Algorithm for integration by partial fractions

Check the degree: if $\deg P \geq \deg Q$ , perform polynomial division first: $P/Q = S(x) + R(x)/Q(x)$ with $\deg R < \deg Q$ .
Factor $Q(x)$ completely over $\mathbb{R}$ .
Write the generic form with constants $A_{i,k}$ , $B_{j,l}$ , $C_{j,l}$ .
Determine the constants:
- Cover-up (simple roots): substitute $x = r_i$ directly.
- Coefficients: expand and equate degree by degree.
- Derivatives (high multiplicity): differentiate the numerator identity.
Integrate each term according to the rules:

\int \frac{A}{x - r}\, dx = A \ln\lvert x - r\rvert + C.

what this means · Integral of fraction with simple root — results in natural logarithm.

\int \frac{A}{(x - r)^k}\, dx = \frac{-A}{(k-1)(x-r)^{k-1}} + C, \quad k > 1.

what this means · Integral of fraction with repeated root of multiplicity k greater than 1.

For the irreducible quadratic $(x^2 + px + q)$ with $p^2 - 4q < 0$ : complete the square $(x + p/2)^2 + \beta^2$ and decompose the numerator $Bx + C = B(x + p/2) + (C - Bp/2)$ . The first term integrates in $\ln$ and the second in $\arctan$ .

"If the degree of the numerator is less than the degree of the denominator, the rational function is called proper, and partial fractions works directly. If not, perform polynomial division first to reduce to a proper fraction." — APEX Calculus §6.5

Heaviside Formula

For simple roots $r_1, \ldots, r_n$ of $Q$ :

A_i = \frac{P(r_i)}{Q'(r_i)}.

what this means · Direct formula of residues for simple roots — result of cover-up.

Worked Examples

Example— 1· Simple roots — cover-up (basic)

Problem. Calculate $\int \dfrac{1}{x^2 - 1}\, dx$ .

Strategy. Factor denominator into simple roots, apply cover-up, integrate each $\ln$ .

Solution.

$x^2 - 1 = (x-1)(x+1)$ . Decomposition: $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$ .

Multiplying by $(x-1)(x+1)$ : $1 = A(x+1) + B(x-1)$ .

Cover-up: $x = 1 \Rightarrow 1 = 2A \Rightarrow A = 1/2$ . $x = -1 \Rightarrow 1 = -2B \Rightarrow B = -1/2$ .

$\int \frac{1}{x^2 - 1}\, dx = \frac{1}{2}\int \frac{dx}{x-1} - \frac{1}{2}\int \frac{dx}{x+1} = \frac{1}{2}\ln\left\lvert\frac{x-1}{x+1}\right\rvert + C.$

Check. Differentiating the result: $\frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) = \frac{1}{2}\cdot\frac{2}{x^2 - 1} = \frac{1}{x^2 - 1}$ . Correct.

Source. APEX Calculus §6.5, Example 6.5.1 — CC-BY-NC.

Example— 2· Root of multiplicity 2 (intermediate)

Problem. Calculate $\int \dfrac{x + 1}{(x - 2)^2}\, dx$ .

Strategy. Factor with multiplicity 2: two terms in the decomposition.

Solution.

$\dfrac{x+1}{(x-2)^2} = \dfrac{A}{x-2} + \dfrac{B}{(x-2)^2}$ .

Multiplying by $(x-2)^2$ : $x + 1 = A(x-2) + B$ .

At $x = 2$ : $3 = B$ . Coefficient of $x$ : $1 = A$ . Therefore $A = 1$ , $B = 3$ .

$\int \frac{x+1}{(x-2)^2}\, dx = \int \frac{1}{x-2}\, dx + \int \frac{3}{(x-2)^2}\, dx = \ln\lvert x - 2\rvert - \frac{3}{x - 2} + C.$

Check. Differentiate $\ln\lvert x-2\rvert - 3(x-2)^{-1}$ : $\frac{1}{x-2} + \frac{3}{(x-2)^2} = \frac{x-2+3}{(x-2)^2} = \frac{x+1}{(x-2)^2}$ . Correct.

Source. OpenStax Calculus Vol. 2, §3.4, Example 3.40 — CC-BY-NC-SA.

Example— 3· Irreducible quadratic (intermediate)

Problem. Calculate $\int \dfrac{x}{(x^2 + 1)(x - 1)}\, dx$ .

Strategy. Linear factor + irreducible quadratic factor.

Solution.

$\dfrac{x}{(x^2+1)(x-1)} = \dfrac{A}{x-1} + \dfrac{Bx + C}{x^2 + 1}$ .

Multiplying by $(x^2+1)(x-1)$ : $x = A(x^2+1) + (Bx+C)(x-1)$ .

At $x = 1$ : $1 = 2A \Rightarrow A = 1/2$ .

Coefficient of $x^2$ : $0 = A + B \Rightarrow B = -1/2$ .

Coefficient of $x^0$ : $0 = A - C \Rightarrow C = 1/2$ .

$\int \frac{x}{(x^2+1)(x-1)}\, dx = \frac{1}{2}\ln\lvert x-1\rvert - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\arctan x + C.$

Check. Differentiating the result confirms the integrand (algebra).

Source. APEX Calculus §6.5, Example 6.5.3 — CC-BY-NC.

Example— 4· High degree — polynomial division first (intermediate)

Problem. Calculate $\int \dfrac{x^3 + 1}{x^2 - 1}\, dx$ .

Strategy. $\deg P = 3 \geq \deg Q = 2$ : divide first.

Solution.

Division: $\dfrac{x^3 + 1}{x^2 - 1} = x + \dfrac{x + 1}{x^2 - 1}$ .

Partial fractions of $\dfrac{x+1}{x^2-1} = \dfrac{x+1}{(x-1)(x+1)} = \dfrac{1}{x-1}$ (simplifies directly).

$\int \frac{x^3+1}{x^2-1}\, dx = \int x\, dx + \int \frac{1}{x-1}\, dx = \frac{x^2}{2} + \ln\lvert x-1\rvert + C.$

Check. Differentiating: $x + \frac{1}{x-1} = \frac{x(x-1)+1}{x-1} = \frac{x^2 - x + 1}{x-1}$ . Multiply by $(x+1)/(x+1)$ : $\frac{(x^2-x+1)(x+1)}{x^2-1} = \frac{x^3+1}{x^2-1}$ . Correct.

Source. OpenStax Calculus Vol. 2, §3.4, Example 3.41 — CC-BY-NC-SA.

Example— 5· Logistic equation — partial fractions in ODE (challenge)

Problem. Solve $\int \dfrac{dN}{N(1 - N/K)}$ using partial fractions (key step in solving the logistic equation).

Strategy. Rewrite the denominator as a product; cover-up; integrate.

Solution.

$\dfrac{1}{N(1 - N/K)} = \dfrac{1}{N\cdot\frac{K-N}{K}} = \dfrac{K}{N(K-N)}$ .

Decomposition: $\dfrac{K}{N(K-N)} = \dfrac{A}{N} + \dfrac{B}{K-N}$ .

Multiplying: $K = A(K-N) + BN$ .

$N = 0$ : $K = AK \Rightarrow A = 1$ . $N = K$ : $K = BK \Rightarrow B = 1$ .

$\int \frac{dN}{N(1-N/K)} = \int \frac{dN}{N} + \int \frac{dN}{K - N} = \ln\lvert N\rvert - \ln\lvert K - N\rvert + C_0 = \ln\left\lvert\frac{N}{K-N}\right\rvert + C_0.$

Equating to $rt + C_1$ and exponentiating: $\dfrac{N}{K-N} = Ae^{rt}$ , so $N(t) = \dfrac{K}{1 + B e^{-rt}}$ .

Check. Confirm that $N(t) = K/(1 + Be^{-rt})$ satisfies $\dot{N} = rN(1 - N/K)$ (direct differentiation).

Source. Active Calculus §5.5, Activity 5.5.2 — CC-BY-NC-SA.

Exercise list

35 exercises · 8 with worked solution (25%)

Application 26Modeling 4Challenge 3Proof 2

Ex. 86.1Application
Decompose $\dfrac{1}{x^2 - 1}$ into partial fractions.
Solve online
Ex. 86.2Application
Decompose $\dfrac{1}{x(x+1)}$ into partial fractions.
Solve online
Ex. 86.3Application
Decompose $\dfrac{x}{(x-1)(x-2)}$ into partial fractions.
Solve online
Ex. 86.4Application
Decompose $\dfrac{2x + 1}{x^2 - x - 6}$ into partial fractions.
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Ex. 86.5Application
Decompose $\dfrac{1}{x^3 - x}$ into partial fractions.
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Ex. 86.6ApplicationAnswer key
Decompose $\dfrac{x + 1}{(x - 2)^2}$ into partial fractions.
Solve online
Ex. 86.7Application
Decompose $\dfrac{1}{x^2(x + 1)}$ into partial fractions.
Solve online
Ex. 86.8Application
Show that $\dfrac{1}{x^2 + 1}$ is already a simple fraction (irreducible quadratic denominator) and calculate its integral.
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Ex. 86.9ApplicationAnswer key
Decompose $\dfrac{x}{(x^2 + 1)(x - 1)}$ into partial fractions.
Solve online
Ex. 86.10ApplicationAnswer key
Calculate $\int \dfrac{1}{x^2 - 1}\, dx$ .
Solve online
Ex. 86.11Application
Calculate $\int \dfrac{1}{x(x+1)}\, dx$ .
Solve online
Ex. 86.12ApplicationAnswer key
Calculate $\int \dfrac{x}{(x-1)(x-2)}\, dx$ .
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Ex. 86.13Application
Calculate $\int \dfrac{1}{x^2 - 4}\, dx$ .
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Ex. 86.14Application
Calculate $\int \dfrac{1}{x^2 - 9}\, dx$ .
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Ex. 86.15ApplicationAnswer key
Calculate $\int \dfrac{x + 4}{x^2 + 5x + 6}\, dx$ .
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Ex. 86.16Application
Calculate $\int \dfrac{3}{x^2 + x - 2}\, dx$ .
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Ex. 86.17ApplicationAnswer key
Calculate $\int \dfrac{1}{x^3 - x}\, dx$ .
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Ex. 86.18Application
Calculate $\int \dfrac{1}{(x - 1)^2}\, dx$ .
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Ex. 86.19Application
Calculate $\int \dfrac{x + 1}{(x - 2)^2}\, dx$ .
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Ex. 86.20Application
Calculate $\int \dfrac{1}{x^2(x + 1)}\, dx$ .
Solve online
Ex. 86.21ApplicationAnswer key
Calculate $\int \dfrac{1}{x^2 + 4}\, dx$ .
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Ex. 86.22Application
Calculate $\int \dfrac{1}{x^2 + 2x + 5}\, dx$ .
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Ex. 86.23Application
Calculate $\int \dfrac{2x + 3}{x^2 + 2x + 5}\, dx$ .
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Ex. 86.24Application
Calculate $\int \dfrac{x}{(x^2 + 1)(x - 1)}\, dx$ .
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Ex. 86.25ApplicationAnswer key
Calculate $\int \dfrac{1}{x^4 - 1}\, dx$ . Hint: factor as $(x^2-1)(x^2+1)$ .
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Ex. 86.26Application
Calculate $\int \dfrac{x^3 + 1}{x^2 - 1}\, dx$ . Divide first.
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Ex. 86.27Modeling
Logistic equation $\dot{N} = rN(1 - N/K)$ . Separate and integrate $\int \dfrac{dN}{N(1 - N/K)}$ to find $N(t)$ .
Solve online
Ex. 86.28Modeling
Inverse Laplace: given $H(s) = \dfrac{1}{s(s+1)}$ , use partial fractions to find $h(t) = \mathcal{L}^{-1}\{H(s)\}$ .
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Ex. 86.29Modeling
Cauchy distribution: determine the constant $c$ such that $f(x) = c/(1+x^2)$ is a probability density on $\mathbb{R}$ .
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Ex. 86.30Modeling
Chemical reaction $\dot{c} = k(a - c)(b - c)$ with $a \neq b$ . Separate and integrate via partial fractions.
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Ex. 86.31Challenge
Calculate $\int \dfrac{1}{x^4 + 1}\, dx$ . Hint: factor as $(x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$ .
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Ex. 86.32Challenge
Calculate $\int \dfrac{1}{x^3 + 1}\, dx$ . Factor the denominator first.
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Ex. 86.33Challenge
Calculate $\int \dfrac{x^2}{(x^2 + 1)^2}\, dx$ .
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Ex. 86.34Proof
Proof. Prove the Heaviside formula $A_i = P(r_i)/Q'(r_i)$ for simple roots $r_i$ of $Q$ .
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Ex. 86.35Proof
Proof. Prove that the partial fractions decomposition is unique for $P/Q$ with $\deg P < \deg Q$ .
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Sources

APEX Calculus v5 — Hartman et al. · 2024 · CC-BY-NC · §6.5. Primary source.
Calculus Volume 2 (OpenStax) — OpenStax · 2016 · CC-BY-NC-SA · §3.4.
Active Calculus 2.0 — Boelkins · 2024 · CC-BY-NC-SA · §5.5.