Partial fractions resurfaces in this course at a specific, load-bearing moment: turning an inverse Laplace transform's rational function of s back into a sum of pieces the transform table actually lists. The decomposition itself doesn't change — only what the split pieces get used for does.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
The "cover-up" shortcut for finding a partial-fraction constant works cleanly because:
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Cover-up specifically needs each root to be a genuine root of a DISTINCT linear factor -- it doesn't extend cleanly to every case (e.g. every constant on a repeated factor).
Cover-up works for a linear numerator too -- it isolates whichever constant sits over the factor being covered.
Denominators still need to be cleared first -- cover-up is a shortcut for solving the resulting equation, not a replacement for it.
Why
Plugging in x=r_i makes every term except the one over (x-r_i) vanish (since each other term still has that factor), isolating that one constant instantly.
From Calculus II — Integral Calculus & Infinite Series
Evaluate \displaystyle\int_2^3 \frac{4x}{x^2-1}\,dx (give a decimal rounded to 3 places).
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Why
u=x^2-1, du=2x\,dx gives 2\ln|x^2-1|\Big|_2^3 = 2\ln8-2\ln3 = 2\ln(8/3)\approx1.962.
Correct answer
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Why
Recognize this as a direct u-substitution, u=x^2-1, rather than partial fractions -- x^2-1 already appears as one block.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
What is the correct partial-fraction setup for \dfrac{p(x)}{(x-2)^2(x+1)}?
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A repeated linear factor of multiplicity 2 needs BOTH powers represented, \frac1{x-2} and \frac1{(x-2)^2}, not just the highest one.
This is the setup for a factor appearing only once -- it's missing the second power the repeat requires.
A linear (not quadratic) repeated factor never needs a linear numerator -- constants suffice for every linear term.
Why
Multiplicity 2 on (x-2) contributes two terms, one per power from 1 to 2, plus the ordinary term for (x+1).
From Calculus II — Integral Calculus & Infinite Series
Evaluate \displaystyle\int_0^2 \frac{4}{x^2+4}\,dx.
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Why
\int\frac{4}{x^2+4}\,dx = 2\arctan(x/2), so 2\arctan(1)-2\arctan(0) = 2\cdot\frac\pi4-0 = \frac\pi2\approx1.5708.
Correct answer
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Why
This is a pure arctangent form with a=2; the antiderivative is 2\arctan(x/2).
A single complicated fraction is really several simple fractions added together, wearing one denominator as a disguise. Partial fractions is just the unmasking. Pull the disguise apart, and every piece underneath is something you already know how to handle.
A rational function \dfrac{P(x)}{Q(x)} splits into a sum of simpler pieces determined entirely by Q(x)'s factorization: a distinct linear factor (x-a) contributes a single term \dfrac{A}{x-a}; a repeated linear factor (x-a)^k contributes one term for every power up to k; an irreducible quadratic factor contributes a term with a linear numerator. Later in this course, Q(s) will be the denominator left behind by a Laplace transform, and the whole point of decomposing it is that each resulting piece matches a transform-table entry directly — the decomposition is the translation step, not an end in itself.
Set up the decomposition
Match each factor to the correct number of unknown-coefficient terms before solving for them.
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Worked example
Distinct linear factors
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1
Decompose \dfrac{5x-1}{(x-2)(x+1)}.
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2
Write \dfrac{5x-1}{(x-2)(x+1)} = \dfrac{A}{x-2}+\dfrac{B}{x+1} and clear denominators: 5x-1 = A(x+1)+B(x-2).
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3
x=2 gives 9=3A, so A=3. x=-1 gives -6=-3B, so B=2. \boxed{\dfrac{3}{x-2}+\dfrac{2}{x+1}}
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Worked example
A repeated linear factor
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1
Decompose \dfrac{3x+1}{x^2}.
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2
The repeated factor x^2 contributes two terms: \dfrac{3x+1}{x^2} = \dfrac{A}{x}+\dfrac{B}{x^2}, so 3x+1 = Ax+B.
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3
Matching coefficients: A=3, B=1. \boxed{\dfrac{3}{x}+\dfrac{1}{x^2}}
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Where this goes next
Inverting a Laplace transform
The Laplace transform turns a differential equation into an algebraic one in s, and solving that algebraic equation typically leaves a rational function of s behind. Getting back to a function of t means running the transform table in reverse — which only works one simple piece at a time, exactly the pieces partial fractions produces.
Check your understanding (counts toward your progress)
A rational function's denominator factors as (x-3)(x+5), two distinct linear factors. How many unknown constants does its partial-fraction decomposition need?
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Each distinct linear factor contributes its own separate term with its own unknown constant.
There are exactly two distinct linear factors here, not three.
The number of constants is set by the denominator's factorization alone, not the numerator.
Why
Each distinct linear factor contributes exactly one term with one unknown constant -- two factors, two constants.
A student decomposes \dfrac{1}{(x-1)^2} as \dfrac{A}{(x-1)^2} alone, with no second term. What's the flaw?
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A single term can't reproduce every possible numerator that division by (x-1)^2 could have left behind.
The original denominator's power must be preserved somewhere in the decomposition.
Partial fractions applies to any repeated linear factor, including (x-1)^2.
Why
A repeated factor (x-a)^k needs a separate term for each power from 1 to k -- (x-1)^2 needs both A/(x-1) and B/(x-1)^2.
A repeated factor (x-a)^2 needs one term for every power up to 2.
Decompose \dfrac{7x+3}{(x-1)(x+2)} into partial fractions of the form \dfrac{A}{x-1}+\dfrac{B}{x+2}. Enter the sum as a single expression in x.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
x=1 gives 10=3A, so A=10/3; x=-2 gives -11=-3B, so B=11/3.
Correct answer
Also accepted:
Why
Clear denominators to get 7x+3=A(x+2)+B(x-1), then substitute x=1 and x=-2.
For \dfrac{4x+1}{x^2}=\dfrac{A}{x}+\dfrac{B}{x^2}, what is A?
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Why
Clearing denominators gives 4x+1=Ax+B; matching coefficients of x gives A=4 (and B=1).
Correct answer
Also accepted:
Why
Clear denominators to get 4x+1=Ax+B, then match coefficients of x.