Integration by Partial Fractions

§ 01The Core Idea — Addition in Reverse

You already know how to add fractions: 1/(x−1) + 2/(x+3) = (3x+1)/((x−1)(x+3)). Partial fraction decomposition runs this process in reverse. Given a rational integrand, you split it back into simple pieces: each of which has an elementary antiderivative.

Without this technique, an integral like ∫ (3x+1)/((x−1)(x+3)) dx has no obvious approach. With it, the fraction splits into 1/(x−1) + 2/(x+3), and each piece integrates to a logarithm immediately. The heavy lifting is algebraic, not calculus.

The Goal

P(x)/Q(x) = (simple piece 1) + (simple piece 2) + …

Each simple piece has the form A/(linear factor) or (Bx+C)/(irreducible quadratic), all of which integrate to ln or arctan forms.

Two Prerequisites Before Starting Partial fractions only work cleanly when: (1) the degree of the numerator is strictly less than the degree of the denominator — if not, do polynomial long division first; and (2) the denominator is fully factored over the reals. Both conditions must be checked before writing down any decomposition.

§ 02Step 0 — Polynomial Long Division for Improper Fractions

A rational function P(x)/Q(x) is improper if deg(P) ≥ deg(Q). You must divide first, reducing it to a polynomial plus a proper remainder before any decomposition.

Example 01Simplify (x³ + 2x² − 1) / (x² − x) before decomposing

Check degree

Numerator degree 3 ≥ denominator degree 2 → improper. Must divide.

Long divide

x³ + 2x² − 1 ÷ x² − x = x + 3 + (3x − 1)/(x² − x)

Quotient: x + 3. Remainder: 3x − 1. Now decompose only the remainder fraction.

Result

(x³+2x²−1)/(x²−x) = x + 3 + (3x−1)/(x(x−1))

The integral of the polynomial part (x + 3) is immediate. Only the proper remainder fraction (3x−1)/(x(x−1)) needs decomposition, as shown in Case 1 below.

§ 03The Three Decomposition Cases

Every denominator, once factored, consists of some combination of three types of factor. The form of the partial fraction template depends entirely on which types appear.

Case 1Distinct Linear Factors

1/((x−a)(x−b)) → A/(x−a) + B/(x−b) One constant per distinct linear factor. Each integrates to A⋅ln|x−a|. Most common case in first-year calculus.

Case 2Repeated Linear Factors

1/(x−a)ⁿ → A/(x−a) + B/(x−a)² + … + N/(x−a)ⁿ One term for each power from 1 to n. Higher powers integrate to −1/((k−1)(x−a)^k−1), not logarithms.

Case 3Irreducible Quadratic Factors

1/(ax²+bx+c) → (Bx+C)/(ax²+bx+c) A linear numerator for each irreducible quadratic. These typically integrate to arctan or a combination of arctan and ln after completing the square.

How to Identify an Irreducible Quadratic A quadratic ax² + bx + c is irreducible over the reals if its discriminant b² − 4ac < 0. It cannot be factored into real linear factors. Examples: x² + 1, x² + x + 1, x² + 4. If the discriminant is ≥ 0, factor into linear factors and use Cases 1 or 2 instead.

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§ 04Case 1 — Distinct Linear Factors

The standard method: write the template, clear denominators by multiplying both sides by Q(x), then solve for constants by substituting the roots of each factor (the "cover-up" trick) or by comparing coefficients.

Example 02Evaluate ∫ (3x + 1) / ((x−1)(x+3)) dx

Template

(3x+1)/((x−1)(x+3)) = A/(x−1) + B/(x+3)

Clear denom.

3x + 1 = A(x+3) + B(x−1)

Cover-up

x = 1: 3(1)+1 = A(4) → A = 1 x = −3: 3(−3)+1 = B(−4) → B = 2

Integrate

∫ [1/(x−1) + 2/(x+3)] dx

Result

ln|x−1| + 2 ln|x+3| + C

Example 03Evaluate ∫ 5 / (x² − 9) dx

Factor denom.

x²−9 = (x−3)(x+3)

Template

5/((x−3)(x+3)) = A/(x−3) + B/(x+3)

Cover-up

x=3: 5 = A(6) → A = 5/6 x=−3: 5 = B(−6) → B = −5/6

Result

(5/6) ln|x−3| − (5/6) ln|x+3| + C = (5/6) ln|(x−3)/(x+3)| + C

Example 04Evaluate ∫ x / (x² + x − 6) dx

Factor

x²+x−6 = (x+3)(x−2)

Template

x/((x+3)(x−2)) = A/(x+3) + B/(x−2)

Cover-up

x=2: 2 = B(5) → B = 2/5 x=−3: −3 = A(−5) → A = 3/5

Result

(3/5) ln|x+3| + (2/5) ln|x−2| + C

§ 05Case 2 — Repeated Linear Factors

When a linear factor (x−a)ⁿ appears with multiplicity n > 1, the template must include one term for every power from 1 up to n. Using fewer terms gives a system with no solution, a common and frustrating error.

Example 05Evaluate ∫ (2x + 3) / (x(x−1)²) dx

Template

(2x+3)/(x(x−1)²) = A/x + B/(x−1) + C/(x−1)²

Clear denom.

2x + 3 = A(x−1)² + Bx(x−1) + Cx

Substitute

x=0: 3 = A(1) → A = 3 x=1: 5 = C(1) → C = 5

For B: expand and compare x² coefficients. A+B = 0 → B = −3.

Integrate

∫ [3/x − 3/(x−1) + 5/(x−1)²] dx

The last term: ∫ (x−1)⁻² dx = −(x−1)⁻¹ = −1/(x−1)

Result

3 ln|x| − 3 ln|x−1| − 5/(x−1) + C

Example 06Evaluate ∫ 1 / (x²(x+2)) dx

Template

1/(x²(x+2)) = A/x + B/x² + C/(x+2)

Clear denom.

1 = Ax(x+2) + B(x+2) + Cx²

Roots

x=0: 1 = 2B → B = 1/2 x=−2: 1 = 4C → C = 1/4

Compare x²: A+C = 0 → A = −1/4

Result

−(1/4) ln|x| − 1/(2x) + (1/4) ln|x+2| + C

§ 06Case 3 — Irreducible Quadratic Factors

When Q(x) contains an irreducible quadratic factor, the corresponding partial fraction has a linear numerator Bx + C, not just a constant. After decomposition, these terms typically require completing the square and integrating to arctan (and sometimes ln).

Example 07Evaluate ∫ (2x² + x + 1) / ((x−1)(x²+1)) dx

Check

x²+1 has discriminant 0−4 = −4 < 0. Irreducible. Proper fraction ✓.

Template

(2x²+x+1)/((x−1)(x²+1)) = A/(x−1) + (Bx+C)/(x²+1)

Clear denom.

2x²+x+1 = A(x²+1) + (Bx+C)(x−1)

Solve

x=1: 4 = 2A → A = 2

Expand: 2x²+x+1 = 2x²+2 + Bx²−Bx+Cx−C

x²: 2 = 2+B → B = 0 x: 1 = −B+C = C → C = 1 Check const: 1 = 2−C = 1 ✓

Integrate

∫ [2/(x−1) + 1/(x²+1)] dx

Result

2 ln|x−1| + arctan x + C

Example 08Evaluate ∫ (x + 3) / (x² + 2x + 5) dx — irreducible quadratic with no linear factor

Check disc.

b²−4ac = 4−20 = −16 < 0. Irreducible. Entire integrand is one irreducible quadratic term.

Complete square

x²+2x+5 = (x+1)² + 4

Split numerator

x+3 = (x+1) + 2 ∫ (x+3)/((x+1)²+4) dx = ∫ (x+1)/((x+1)²+4) dx + ∫ 2/((x+1)²+4) dx

First integral

Let u = (x+1)²+4, du = 2(x+1) dx:

∫ (x+1)/((x+1)²+4) dx = (1/2) ln((x+1)²+4)

Second integral

∫ 2/((x+1)²+4) dx = 2 ⋅ (1/2) arctan((x+1)/2) = arctan((x+1)/2)

Using ∫ du/(u²+a²) = (1/a) arctan(u/a) with u=x+1, a=2.

Result

(1/2) ln(x²+2x+5) + arctan((x+1)/2) + C

§ 07Definite Integrals by Partial Fractions

For definite integrals, decompose and integrate as usual, then evaluate at the limits. Be alert to any discontinuities in the domain: a factor like (x−2) in the denominator means the integrand is undefined at x = 2, so an interval containing that point would be an improper integral, not a standard definite integral.

Example 09Evaluate ∫₀² 4 / ((x+1)(x+3)) dx

Decompose

4/((x+1)(x+3)) = A/(x+1) + B/(x+3) x=−1: 4 = 2A → A=2 x=−3: 4=−2B → B=−2

Antiderivative

F(x) = 2 ln|x+1| − 2 ln|x+3| = 2 ln|(x+1)/(x+3)|

Evaluate

F(2) = 2 ln(3/5) F(0) = 2 ln(1/3) F(2)−F(0) = 2[ln(3/5) − ln(1/3)] = 2 ln(9/5)

Result

∫₀² 4/((x+1)(x+3)) dx = 2 ln(9/5) ≈ 1.176

Example 10Evaluate ∫₀¹ (x²+2) / (x³+x) dx

Factor denom.

x³+x = x(x²+1) (x²+1 is irreducible)

Template

(x²+2)/(x(x²+1)) = A/x + (Bx+C)/(x²+1)

Solve

x²+2 = A(x²+1) + (Bx+C)x x=0: 2=A x²: 1=A+B → B=−1 x: 0=C

Integrate

∫ [2/x − x/(x²+1)] dx = 2 ln|x| − (1/2) ln(x²+1)

Evaluate [0,1]

At x=1: 2 ln 1 − (1/2) ln 2 = 0 − (1/2) ln 2

As x→0⁺: 2 ln x − (1/2) ln(x²+1) → −∞ + 0: this is an improper integral at x=0! Need limit.

lim_x→0⁺ [2 ln x − (1/2)ln(x²+1)] = −∞

The integral diverges, the integrand has a non-integrable singularity at x = 0.

Result

∫₀¹ (x²+2)/(x³+x) dx diverges (singularity at x = 0)

§ 08Five Common Mistakes

Not dividing an improper fraction

Check deg(P) < deg(Q) firstIf the numerator degree equals or exceeds the denominator degree, long division is mandatory before any decomposition. Skipping this step gives a decomposition that cannot be solved — or worse, gives a wrong one.

Missing a power in repeated factors

(x−a)³ → 3 terms, not 1For a factor (x−a)ⁿ, the template must include A/(x−a), B/(x−a)², …, N/(x−a)ⁿ. Using only A/(x−a) will leave constants unsolvable.

Constant numerator on quadratic factor

(Bx+C)/(x²+1), not A/(x²+1)An irreducible quadratic always requires a linear numerator Bx+C in its partial fraction. Writing just A gives a system that is underdetermined and will produce a wrong decomposition.

Not fully factoring the denominator

x³−x = x(x−1)(x+1), not x(x²−1)The denominator must be completely factored before writing the template. An unfactored group like (x²−1) instead of (x−1)(x+1) leads to wrong term counts and wrong forms.

Integrating 1/(x−a)² as a logarithm

∫ (x−a)⁻²dx = −1/(x−a)The power rule for integration applies: ∫ (x−a)ⁿdx = (x−a)ⁿ⁺¹/(n+1) for n ≠ −1. The logarithm form only applies to n = −1. Writing ∫ 1/(x−a)² = ln|x−a| is a serious error.

§ 0910-Question Quiz

Test your decomposition skills: templates, constants, integration forms, and error identification.

§ 10Continue the Integration Series