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Parts & Partial FractionsBC ONLY

Two integrals u-substitution can't touch: products, and rational functions whose denominators factor. u-sub undoes the chain rule — but products were never the chain rule's department. This is the heavy machinery.
Two new reversals. Integration by parts runs the product rule backward — it doesn't solve an integral so much as trade it for a different one, and the skill is trading down. Partial fractions un-adds a rational function — a fraction whose denominator factors is secretly a sum of deck-card fractions, and algebra recovers the pieces. Both end, as always, on cards you already own.

Where parts comes from — the product rule, rewound

Same move as last lesson, different rule. Differentiate a product and two terms come out; integrate that equation and rearrange, and you get a machine for trading one integral for another.

Integrate both sides — the left side collapses by the Fundamental Theorem, and the two right-hand pieces appear:
Rearrange — solve for the piece you were stuck on. This is the formula, and it's a TRADE: you pay ∫u dv and receive uv − ∫v du:
The trade is only worth making if the NEW integral ∫v du is easier — which is the whole question of choosing u and dv

Choosing u — LIATE, and why it works

The trade improves when u gets simpler under differentiation (its job is to die) and dv is something you can actually integrate. Rank function types by how eagerly they simplify when differentiated, and you get the classic priority list:

Logs · Inverse trig · Algebraic · Trig · Exponential
u = the earliest type on the list. Logs and inverse trig differentiate into algebra (ln x → 1/x) — they simplify dramatically, but are painful to integrate. Perfect u material.u dies ↓
dv = the latest type. Exponentials and trig integrate without complaint, forever (eˣ → eˣ). Perfect dv material — and remember dv must swallow the dx.dv recycles ∞
It's a principle, not a spell: u should simplify when differentiated, dv should be integrable. LIATE is just that principle pre-sorted.why > what

Feel the trade — cast it yourself

The formula doesn't care which factor you call u — but the integral does. Pick a casting and the page runs that round of parts honestly. Try both for each integral: one trade improves things, the other makes them visibly worse.

That round of parts, run honestly
Pick a casting to see where the trade lands.

The recipe, stepped

press step
The work, shown
Pick an integral

Watch the second one — LIATE puts the log ahead of the polynomial, which surprises most students. And the third has no visible product at all… until you invent one.

The tabular method — repeated parts, compressed

When u is a polynomial (it differentiates to zero in a few steps) and dv integrates forever (eˣ, sin, cos), parts will need several rounds — and every round is the same bookkeeping. The tabular method runs all the rounds at once: differentiate down the left column until you hit 0, integrate down the right, attach alternating signs, and multiply along the diagonals. Build it live:

press step
What each step is doing
When tabular applies
u columnpolynomial → dies at 0
dv columneˣ / sin / cos → forever
signs+ − + − …
read the answeralong the diagonals ↘

Now you build one — ∫ x² sin x dx

Same machine, your hands. Each press asks for the next cell — and the antiderivative column of sin x is exactly where sign errors breed: sin → −cos → −sin → cos. The signs in the left column come free; the cells don't.

The cyclic case — when the integral comes back

∫eˣ sin x dx breaks the LIATE worldview: neither factor ever dies. Run parts twice and something strange happens — the original integral reappears on the right side. That's not failure; it's the mechanism. Name it and solve for it like any unknown.

press step
The work, shown

Partial fractions — un-adding a fraction

Adding fractions is something you've done since grade school: 1/(x+1) + 2/(x+3) combines into one big fraction. Partial fractions runs that backward: a rational function whose denominator factors into distinct linear pieces is secretly such a sum, every piece is the deck's ∫du/u card, and algebra recovers them. This also completes Lesson 3's decision tree — the "it factors" branch finally has its answer.

The template: one constant over each distinct linear factor. (BC scope: nonrepeating linear factors — that's all the exam asks.)
The cover-up shortcut: to find A, cover its factor in the original and evaluate what's left at that factor's root. Each constant in one line, no systems of equations.
The completed decision tree for any rational integrand:

The cover-up, performed

The trick is named after a physical gesture — so perform it. Click a factor in the denominator: it gets covered, and the survivors are evaluated at that factor's root. Each constant falls out in one line.

Click (x + 1) or (x + 3) to cover it.
0 of 2 constants found
What the cover-up computes
Cover a factor to see its constant fall out.
press step
The work, shown
Pick a fraction

The second one is the full exam gauntlet: top-heavy AND factorable — division first, then partial fractions on the remainder. Two costumes, removed in order.

The machinery on one card

Tabular schema — differentiate down, integrate down, alternate signs, multiply diagonals
Cyclic integrals — when the original returns, name it I and solve algebraically
Partial fractions template for distinct linear factors

The ideas everything else builds on

Parts is a trade, not a solution

∫u dv buys you uv − ∫v du. The trade is good only if the new integral is easier — if it got worse, your u and dv are backwards.

u dies, dv recycles

Pick u to simplify under differentiation and dv to be integrable. LIATE is this principle alphabetized — and dv always includes the dx.

Tabular = parts on repeat

Nothing new — just every round of parts at once. Valid exactly when the u-column reaches zero.

Factor first, always

Rational integrand? Check the denominator: factors → partial fractions; won't factor → complete the square. And if the numerator is the denominator's derivative, it was a one-line u-sub all along.

Your turn

Five problems in the exam's voice — including one trap. Try each before revealing.

Quick quiz

Eight fast checks across the whole lesson.

Question 1 of 8
Score: 0 / 0

Common mistakes & exam tips

u and dv backwards

Pick u = eˣ and the new integral is WORSE than the old one. If ∫v du looks harder than what you started with, stop and swap — the formula isn't wrong, the casting is.

dv forgot its dx

dv is a chunk of the integrand INCLUDING dx. Splitting ∫x eˣ dx as u = x, dv = eˣ (no dx) makes v meaningless. Write dv = eˣ dx every time.

Sign slips in the tabular

The signs alternate starting from + and attach to the DIAGONAL products. Writing them next to the derivatives and then multiplying straight across is the classic way to lose every other sign.

Stopping the table early

The tabular column must run until the derivative is exactly 0. Stopping at the constant (instead of differentiating it once more) silently drops the last term.

Partial fractions on the wrong fraction

PF needs a FACTORABLE denominator and a top-light fraction. Top-heavy? Divide first. Won't factor? That's completing the square. Check both before writing A and B.

Panicking in the cycle

When ∫eˣ sin x reappears, students think they went in a circle and start over with different choices — forever. The reappearance IS the solution: name it I, collect, divide.

On the AP exam

The toolkit is complete — every kind of integrand the BC exam can throw now has a technique. One frontier remains: integrals that run to infinity, or across a point where the function blows up. Improper Integrals closes the unit — and opens the door to Unit 10's series. Or head back to the Unit 6 Guide.