4.2.10 · D5Calculus II — Integration
Question bank — Partial fractions — linear, repeated, irreducible quadratic factors
The four sections below cover: what's true (and why), where people slip, the deep "why" behind each rule, and the strange boundary cases.
True or false — justify
The fraction can be partial-fractioned as it stands.
False — it is improper (), so no sum of proper pieces can equal it; you must do Polynomial long division first, then decompose the remainder.
Every rational function of real polynomials can be split into linear and quadratic pieces.
True — over the reals any polynomial factors into linear and irreducible quadratic factors (Factoring polynomials over the reals), so those are the only two block types the templates ever need.
integrates to a pure arctan.
False — only the constant part gives arctan; the part is an exact-derivative log , so in general you get both a log and an arctan.
A repeated factor contributes exactly three unknown constants.
True — you need one term for each power , and the total unknown count across all blocks always equals .
integrates to a logarithm.
False — the power rule gives ; you only get a log when the exponent is exactly , which a squared factor is not.
If two proper fractions are equal for infinitely many , their partial-fraction constants must match.
True — the decomposition is unique, so equal rational functions force identical numerators and hence identical constants; this is why coefficient-matching is valid.
needs a block.
False — its discriminant is , so it factors as : a repeated linear factor, needing , not a quadratic block.
Spot the error
A student writes and nothing else — what's missing?
The term is missing; a lone constant top can only reach constant numerators, so it cannot span the linear freedom a repeated factor genuinely allows (see the parent's "why repeated" box).
A student sets treating the bottom as an irreducible quadratic.
Error — is reducible, so it should be two linear blocks ; always test factorability before assuming "quadratic top".
A student covers up in and sets to find the constant.
Error — cover-up on the highest power gives the constant cleanly, but the lower constant has no simple cover-up and needs coefficient-matching.
Someone integrates by completing the square first and forgets the term.
Error — completing the square only helps the arctan half; you must first split off the exact-derivative piece (a log), then complete the square on what remains.
A student factors as irreducible because "it's a quadratic".
Error — discriminant , so it factors ; "being a quadratic" is not the test, having no real roots is.
Someone writes the improper directly.
Error — so it is improper; long division gives first, and only the remainder gets decomposed.
Why questions
Why must the numerator over a quadratic be and not just a constant?
The numerator degree must be one less than the denominator's; a degree-2 bottom needs a degree-1 top to represent every possible fraction, including those whose true numerator carries an term.
Why does the cover-up trick kill every term except the one you want?
Multiplying through by leaves that term's constant bare while every other term still carries a factor , which becomes when you substitute — Integration by substitution-style limit reasoning.
Why does give a logarithm at all?
Because ; the shape is precisely the derivative of a log (Standard integrals — log and arctan), which is why linear pieces are so easy to integrate.
Why does completing the square turn a quadratic denominator into an arctan?
Completing the square rewrites as , matching the standard form whose antiderivative is .
Why does the total number of unknown constants always equal ?
Each factor contributes exactly as many constants as its degree (a gives , a quadratic gives ), and the factor degrees sum to — this is your built-in bookkeeping check.
Why is partial fractions even reversible — how do we know the pieces exist?
Adding fractions over a common denominator is a reversible algebraic operation; since factors into those exact pieces, the original fraction must have been assembled from fractions with those denominators.
Why is completing the square unnecessary for a linear factor?
A linear denominator is already in the simplest shape whose integral is a log directly; there is no square to complete because there's no quadratic curvature to unwind.
Why does the same partial-fraction machinery power inverse Laplace transforms?
Laplace transforms — inverse via partial fractions splits a rational transform into standard pieces whose inverses are known table entries, exactly mirroring how we split into log/arctan-integrable pieces here.
Edge cases
What if exactly (not strictly greater)?
Still improper — long division yields a constant plus a proper remainder; equality of degrees already violates the requirement.
What happens if you accidentally include a block when the bottom was actually -reducible over reals as... it isn't. Is ever reducible over the reals?
No — has discriminant and no real roots, so it is irreducible over the reals and correctly takes a block; it only "factors" over complex numbers, which we do not use here.
What is the decomposition when the numerator is ?
Every constant comes out , so the whole fraction is ; the machinery still runs consistently and returns the trivial answer.
For a repeated quadratic , how many constants appear?
Four — a block, matching contribution of .
What if a "quadratic" denominator is a perfect square like — quadratic or repeated-linear?
Repeated linear — it has a real double root, so use ; the template is reserved for quadratics with no real roots.
Can cover-up ever give the constant over an irreducible quadratic block?
No — cover-up needs a real root to substitute, and irreducible quadratics have none; those constants must be found by clearing denominators and matching coefficients.
What if the same linear factor appears in both numerator and denominator, e.g. ?
Cancel it first to get ; failing to cancel invents a spurious term that will simply come out but wastes work and can mask a removable point.
Recall One-line self-test
If someone hands you , what are the first two checks before writing any template? First check ::: Is it proper? If not, long-divide. Second check ::: Is every quadratic factor genuinely irreducible (discriminant )? Factor any that aren't.