U-substitution is an antidifferentiation technique in AP Calculus (Topic 6.9) where you replace part of the integrand with a new variable u, rewriting the integral in a simpler form. It works by reversing the chain rule, and for definite integrals you must also change the limits of integration.
U-substitution is the chain rule run backwards. When you differentiated a composite function in Unit 3, the chain rule spit out an "inside function" times the derivative of that inside. U-substitution spots that pattern inside an integral and undoes it. You set u equal to the inside function, compute du, and rewrite the entire integral in terms of u so it becomes one you actually know how to antidifferentiate.
The CED frames this under learning objective 6.9.A, which asks you to handle "integrands requiring substitution or rearrangements into equivalent forms" for both indefinite and definite integrals. The big definite-integral catch comes straight from the essential knowledge. When you substitute variables in a definite integral, you must also convert the limits of integration into u-values (or substitute back to x before plugging in). For example, in ∫₀² x√(3x² + 7) dx, setting u = 3x² + 7 means your new limits become u = 7 and u = 19, not 0 and 2.
U-substitution lives in Unit 6 (Integration and Accumulation of Change), specifically Topic 6.9, and it supports learning objective 6.9.A. It's the single most-used integration technique on the AP exam, and on the AB exam it's essentially THE technique beyond basic antiderivative rules. Topic 6.14 (Selecting Techniques for Antidifferentiation) then tests whether you can recognize when substitution is the right tool versus simple rewriting or completing the square. If you can't run u-sub cleanly, large chunks of Units 6 through 8 (area, volume, differential equations) become inaccessible, because those problems hand you integrals that need substitution before anything else can happen.
Keep studying AP Calculus Unit 6
Visual cheatsheet
view galleryChain Rule (Unit 3)
U-substitution is literally the chain rule in reverse. The chain rule produces f'(g(x)) · g'(x) when you differentiate; u-sub hunts for that exact "function and its derivative" pairing inside an integrand and unwinds it. If you can spot composite functions when differentiating, you already have the eye for choosing u.
Selecting Techniques for Antidifferentiation (Unit 6)
Topic 6.14 is the decision-making layer. Before computing anything, you ask whether the integral needs substitution, algebraic rearrangement, or just a basic rule. The telltale sign for u-sub is a function nested inside another with its derivative (up to a constant) sitting nearby as a factor.
Integration by Parts (Unit 6, BC only)
On the BC exam, u-sub competes with integration by parts for the same job. Substitution handles composite functions where the derivative of the inside appears; parts handles products of unrelated functions like x·eˣ. Try substitution first since it's faster when it works.
Constant of Integration (Unit 6)
Indefinite u-sub problems still end with + C, and you must substitute back to x for your final answer. Leaving an answer in terms of u, or dropping the + C, costs points even when every other step is right.
Multiple-choice questions test u-sub in two ways. Some ask you to recognize which integral requires substitution, like a stem listing four integrals where only one has the composite structure. Others make you execute it, such as rewriting ∫√(3x+7) dx with u = 3x + 7, or evaluating a definite integral like ∫₀² x√(3x² + 7) dx where the x out front signals that u = 3x² + 7 will work. The classic trap answers come from forgetting to change the limits of integration or from mishandling the constant when du = 6x dx but you only have x dx. On FRQs, u-substitution rarely gets named in the prompt, but it shows up as a required step inside area, volume, and accumulation problems. The rubric awards the antiderivative, so a botched substitution loses you that point and usually the answer point too.
Both swap an integral for an easier one, but they target different structures. U-substitution works on composite functions where the inside function's derivative appears as a factor, like x·cos(x²). Integration by parts (BC only) works on products of two unrelated functions, like x·cos(x), where no inside-derivative pairing exists. Quick check: if differentiating part of the integrand produces another part of it, that's u-sub territory.
U-substitution reverses the chain rule by setting u equal to an inside function whose derivative appears (up to a constant) elsewhere in the integrand.
For definite integrals, you must convert the limits of integration into u-values, or substitute back to x before evaluating. This is stated directly in the CED's essential knowledge for 6.9.A.
Choose u as the inside of the composite function. In ∫√(3x+7) dx, set u = 3x + 7, the expression under the radical.
If du is off by a constant factor, fix it with algebra (multiply and divide by the constant). If it's off by a variable factor, u-sub won't work and you need a different approach.
For indefinite integrals, always substitute back to the original variable and include + C in your final answer.
Topic 6.14 tests recognition, so before computing, scan the integrand for a function-and-its-derivative pair as your signal to use substitution.
U-substitution is an integration technique (Topic 6.9, learning objective 6.9.A) where you replace an inside function with a new variable u, rewrite the integral in terms of u and du, and antidifferentiate the simpler result. It's the reverse of the chain rule from Unit 3.
Yes, if you stay in terms of u. The CED's essential knowledge says substitution in a definite integral requires corresponding changes to the limits. For ∫₀² x√(3x² + 7) dx with u = 3x² + 7, the limits become 7 and 19. Alternatively, substitute back to x first and keep the original limits.
Look for a composite function whose inside derivative appears as a factor. ∫x·cos(x²) dx is u-sub because the x matches the derivative of x². ∫x·cos(x) dx has no such pairing, so it needs integration by parts (BC only). On the AB exam, u-sub is your main tool since parts isn't tested.
Set u equal to the inside function of the composition. Common picks are the expression under a radical, the exponent of e, the argument of a trig function, or the inside of a power. In ∫√(3x+7) dx, u = 3x + 7 because it sits under the square root.
Yes. It's Topic 6.9 in Unit 6 and tested on both AB and BC, in multiple choice and embedded in free-response problems. On AB it's the most advanced integration technique tested, while BC adds integration by parts and partial fractions on top of it.