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Antiderivatives & u-Substitution

The Fundamental Theorem made a promise: find an antiderivative and any definite integral falls. This page is the hunt — and the secret is that you already know every rule on it. They're your derivative rules, read right to left.
Antidifferentiation is pattern recognition, not new machinery. The basic rules are a deck to memorize cold; u-substitution is the chain rule run backward; and the "hard" integrals of 6.10 are easy ones wearing a disguise that algebra removes. One habit makes you bulletproof: check any answer by differentiating it — verification never requires integrating.

The reversible machine — the deck

Every derivative fact you know is an antiderivative fact read backward. Click any card to flip it; hit drill me to flip everything down and shuffle — then quiz yourself before peeking. These twelve are the entire vocabulary; everything else in the unit reduces to them.

What the +C really is

Every card ends in +C — here's the picture behind it. The integrand defines a slope field (Unit 7's language: dy/dx = f(x)), and an antiderivative is any curve that threads it. Slide C: the curve glides vertically, and it never stops fitting, because shifting a graph up or down changes nothing about its slopes. The antiderivative isn't one function — it's this whole family, and an initial condition is what pins down one member.

C = 0.40
slope at the dot: —
Here: f(x) = cos x
the familysin x + C
slope of every membercos x — identical
what picks one curvean initial condition

The teal tangent tick rides the dot as you slide C — and it never tilts. Same x, same slope, any C. That's why the indefinite integral must carry the +C: the derivative genuinely cannot tell the family members apart.

u-Substitution — the chain rule, rewound

Before any recipe, the idea. Every chain-rule derivative leaves a fingerprint: when you differentiate a composition, the inside function's derivative gets emitted as a factor. u-substitution is nothing but noticing that fingerprint in an integrand and running the tape backward.

Forward (the chain rule): differentiating the composition emits a copy of the inside's derivative. Concretely:
Backward (u-sub): so if an integrand contains a composition and that emitted factor, it MUST have come from differentiating F(inside) — and the integral is forced:
Naming the inside function "u" and its differential "du = u′ dx" is just bookkeeping for that reversal — everything in x converts, and a deck card finishes the job

That's the whole theory: u is the inside function, du is its fingerprint. If the fingerprint is missing a constant, you may adjust (constants commute with ∫); if it's missing an x, the substitution is dead. Now the recipe.

The full recipe, stepped

press step
The work, shown
Pick an integral

The third one is definite — watch what happens to the bounds. They live in x-world; once you switch to u, they must switch too.

Now train the eye — pick the u

You know the recipe; the exam-day skill is choosing u fast. Four integrals, four candidates each — pick the inside function whose fingerprint is standing in the integrand.

Which u collapses it?
What makes a good u
it's INSIDE somethingf( u )
its derivative is a FACTORdu = u′ dx
off by a constant?fine — adjust
off by a variable?dead — new u
The collapse
Pick a u to see where it leads.

Removing the disguise

Topic 6.10 isn't new integrals — it's old integrals in costume, and the costume is always a fraction. The trouble: integrals split over sums, never over quotients — you cannot integrate a fraction "top over bottom" piece by piece. So the whole game is algebra that converts the fraction into things the deck already handles. There are exactly two costumes, each with a tell.

Tell #1 — top-heavy (degree of numerator ≥ degree of denominator). An improper fraction hides a polynomial inside it, exactly like 17/5 hides "3 and a bit": 17/5 = 3 + 2/5. Long division extracts it — the polynomial part integrates by the power rule, and only a small proper remainder is left.
Check before anything else — is the numerator (a constant multiple of) the denominator's derivative? Then this isn't 6.10 at all: it's the ln pattern from u-substitution, finished in one line.
Tell #2 — a quadratic that won't factor (discriminant b² − 4ac < 0) under a plain numerator. Every quadratic is secretly a shifted square: x² + bx + c = (x + b/2)² + (the leftover). Completing the square exposes the shift, and the integral becomes the arctan card with u = the shifted x.
Either way, the algebra is the whole battle — the calculus at the end is one card from the deck
Completing the square = reading the vertex
the quadraticx² + 4x + 13
its vertex(−2, 9)
so it IS(x+2)² + 9

Every quadratic is the basic parabola, slid sideways and up. "Completing the square" isn't a ritual — it's locating that slide: left 2, up 9. The "+9" is the vertex height, and it becomes the a² = 9 in the arctan card. A quadratic that won't factor is one whose vertex sits above the axis — it never touches zero, which is exactly why arctan (and not ln of factors) is its destiny.

Now watch both costumes come off, step by step.

press step
The work, shown
Pick a disguise

The toolkit on one card

The u-sub template — u inside, du standing by, everything converts
Definite integrals: the bounds convert too — and then you never go back to x
Only constants slide through the integral sign — never an x

The ideas everything else builds on

+C is not decoration

Antiderivatives come in families — vertical shifts of one curve. Indefinite integral, write +C. Definite integral, it cancels and disappears.

Verify by differentiating

Integration is hard; differentiation is mechanical. Every answer you ever write can be checked in ten seconds — run it forward.

Choosing u

u is the inside function: under the power, inside the trig, up in the exponent, inside the log — whose derivative (up to a constant) is a factor in the integrand.

New world, new bounds

In a definite u-sub, convert the bounds with u(x) and finish entirely in u — or back-substitute and keep the x bounds. Never both.

Your turn

Five hunts. Try each before you reveal the worked answer.

Quick quiz

Eight fast checks across the whole lesson.

Question 1 of 8
Score: 0 / 0

Common mistakes & exam tips

Dropping the +C

On an indefinite integral it's a stated point on the rubric. Write it every time — and only collapse it after a definite integral's subtraction.

ln x instead of ln|x|

∫ dx/x = ln|x| + C. The absolute value isn't pedantry — without it the antiderivative doesn't exist for negative x, and the MC options will include both to catch you.

Sliding an x through the integral

∫ e^(x²) dx is NOT (1/2x)∫ 2x·e^(x²) dx. Only constants pass through the integral sign. If the missing factor has an x in it, u-sub is dead — pick a different u (or a different technique).

Forgetting du exists

∫ cos(5x) dx is sin(5x)/5, not sin(5x). Differentiate your answer: the chain rule coughs up a 5 that must be pre-cancelled. The constant adjustment IS the du bookkeeping.

Mixing worlds on a definite integral

Convert the bounds AND back-substitute, and you've transformed twice. Pick a lane: new bounds and finish in u, or old bounds after returning to x.

Losing the 1/a in arctan

∫ du/(u²+a²) = (1/a)·arctan(u/a) + C. After completing the square, that leading 1/a is the most-dropped factor in the unit.

On the AP exam

u-substitution undoes the chain rule — but products were never the chain rule's department. Next, the BC-only heavy machinery: Integration by Parts & Partial Fractions, where the product rule and rational functions get their reversals — including the tabular shortcut. Or head back to the Unit 6 Guide.