Antidifferentiation — finding a function from its derivative — is the algebraic backbone of every integral on the AP Calculus AB exam. Beyond the basic power rule, this unit tests a wide range of integration techniques: u-substitution for composite functions, trig identities that turn an unfamiliar integrand into a recognizable form, integrals involving logarithms and inverse trig functions, and initial value problems that use a known point to pin down the constant of integration. It also covers one of the most concrete applications of antidifferentiation: recovering velocity from acceleration and position from velocity in motion problems.
The single most useful habit for this unit is recognizing which substitution turns a hard integral into an easy one. An integrand built from a function and (something proportional to) its own derivative — like t·cos(4t²), or sin²θ·cosθ — is a signal to try u-substitution with that inner function. Trig integrals often require rewriting the integrand using a Pythagorean identity (like tan²θ = sec²θ−1) or a double-angle identity before any antiderivative rule applies directly. And motion problems are really just two antidifferentiation steps in a row: integrate acceleration once (using the initial velocity to solve for the constant) to get velocity, then integrate velocity again (using the initial position) to get a position function.
Below are 16 AP Calculus AB-style practice problems covering u-substitution, trig and logarithmic integrals, fractional-power integrands, initial value problems, and motion along a line. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus AB QBank.
∫ (6x1/2 − 10x3/2 − x−1/2) dx =
Apply the power rule to each term separately, remembering to increase each exponent by 1 and divide by the new exponent.
∫6x1/2dx = 6·(2/3)x3/2 = 4x3/2.
∫−10x3/2dx = −10·(2/5)x5/2 = −4x5/2.
∫−x−1/2dx = −1·2x1/2 = −2x1/2.
Combining all three: 4x3/2 − 4x5/2 − 2x1/2 + C. The correct answer is A.
∫ (x3+2x−1)/x2 dx =
Split the fraction into three separate terms before integrating: (x3+2x−1)/x2 = x + 2/x − 1/x2 = x + 2/x − x−2.
Integrate term by term: ∫x dx = x2/2; ∫2/x dx = 2ln|x|; ∫−x−2dx = x−1 = 1/x (the two negative signs cancel here).
Combining all three: x2/2 + 2ln|x| + 1/x + C. The correct answer is A.
∫ dy / [√y(1+√y)] =
Let u = √y, so y = u2 and dy = 2u du.
Substitute: ∫ (2u du) / [u(1+u)] = ∫ 2/(1+u) du (the u in the numerator and denominator cancels).
This is a standard logarithmic integral: 2∫1/(1+u) du = 2ln|1+u| + C.
Substitute back u = √y (and drop the absolute value, since 1+√y is always positive): 2ln(1+√y) + C. The correct answer is A.
Keep the momentum going.
Explore 1,500+ Free AP Calculus AB Problems∫ u du / √(9−4u2) =
Let w = 9−4u2, so dw = −8u du, meaning u du = −dw/8.
Substitute: ∫ (−dw/8) / √w = −(1/8)∫w−1/2dw = −(1/8)(2√w) = −√w/4.
Substitute back: −√(9−4u2)/4 + C.
The correct answer is A.
∫ cos2θ sinθ dθ =
Let u = cosθ, so du = −sinθ dθ, meaning sinθ dθ = −du.
Substitute: ∫u2(−du) = −∫u2du = −u3/3 + C.
Substitute back u=cosθ: −cos3θ/3 + C.
The correct answer is B.
∫ cos(√x) / √x dx =
Let u = √x, so du = dx/(2√x), meaning dx/√x = 2 du.
Substitute: ∫cos(u)·2 du = 2sin(u) + C.
Substitute back u=√x: 2sin(√x) + C.
The correct answer is A.
∫ t sin(4t2) dt =
Let u = 4t2, so du = 8t dt, meaning t dt = du/8.
Substitute: ∫sin(u)·(du/8) = (1/8)·(−cos(u)) + C = −cos(u)/8 + C.
Substitute back u = 4t2: −cos(4t2)/8 + C.
The correct answer is A.
∫ sin23x dx =
There's no direct antiderivative rule for sin2, so first rewrite it using the half-angle identity sin2u = (1−cos2u)/2, with u=3x: sin23x = (1−cos6x)/2.
Integrate term by term: ∫(1/2 − cos6x/2)dx = x/2 − (1/2)∫cos6x dx.
For the remaining integral, use u-substitution (u=6x, du=6dx): ∫cos6x dx = sin6x/6.
Combine: x/2 − (1/2)(sin6x/6) = x/2 − sin6x/12 + C. The correct answer is A.
8 down, 8 to go — keep practicing.
Explore 1,500+ Free AP Calculus AB Problems∫ cos3θ dθ =
The antiderivative of cos(u) is sin(u), but since the inner function is 3θ (derivative 3, not 1), a correction factor of 1/3 is needed.
∫cos3θ dθ = (1/3)sin3θ + C.
The correct answer is A. Forgetting the 1/3 correction factor (choice C, which instead multiplies by 3) is the most common mistake on this type of problem.
∫ dy / [y(4+(ln y)2)] =
Let u = ln y, so du = dy/y. The integral becomes ∫ du / (4+u2).
This matches the standard form ∫du/(a2+u2) = (1/a)tan−1(u/a) + C, with a=2: (1/2)tan−1(u/2) + C.
Substitute back u = ln|y|: (1/2)tan−1((ln|y|)/2) + C.
The correct answer is A.
∫ (cotθ+1)2 dθ =
Expand the square: (cotθ+1)2 = cot2θ + 2cotθ + 1. Use the Pythagorean identity cot2θ = csc2θ−1 to simplify: csc2θ−1+2cotθ+1 = csc2θ+2cotθ.
Integrate term by term: ∫csc2θ dθ = −cotθ, and ∫2cotθ dθ = 2ln|sinθ| (since ∫cotθdθ=ln|sinθ|+C).
Combine: −cotθ + 2ln|sinθ| + C.
The correct answer is A.
∫ dθ / (1−sinθ) =
Multiply the numerator and denominator by 1+sinθ (chosen so the denominator becomes a difference of squares): (1+sinθ) / [(1−sinθ)(1+sinθ)] = (1+sinθ)/(1−sin2θ) = (1+sinθ)/cos2θ.
Split the fraction: 1/cos2θ + sinθ/cos2θ = sec2θ + secθtanθ.
Integrate term by term: ∫sec2θ dθ = tanθ, and ∫secθtanθ dθ = secθ.
Combine: tanθ + secθ + C. The correct answer is A.
12 down, 4 to go.
Explore 1,500+ Free AP Calculus AB ProblemsA particle starting at rest at t = 0 moves along a line so that its acceleration at time t is 18t ft/sec2. How much distance does it cover during the first 3 seconds?
Integrate acceleration to get velocity: v(t) = ∫18t dt = 9t2 + C. Since the particle starts at rest, v(0)=0 → C=0, so v(t) = 9t2.
Since the particle starts at rest and the velocity never goes negative on this interval, distance covered equals the definite integral of velocity from 0 to 3: ∫03 9t2 dt = 3t3 |03.
= 3(27) − 0 = 81.
The correct answer is C.
The equation of the curve whose slope at point (x,y) is x2−4, and which contains the point (2, −5), is
The slope is dy/dx = x2−4, so integrate to find the general family of curves: y = ∫(x2−4) dx = x3/3 − 4x + C.
Use the given point (2, −5) to solve for C: −5 = (2)3/3 − 4(2) + C = 8/3 − 8 + C.
−5 = 8/3 − 24/3 + C = −16/3 + C → C = −5+16/3 = −15/3+16/3 = 1/3.
The equation is y = x3/3 − 4x + 1/3. The correct answer is A.
A particle moves along a line with acceleration 4+6t at time t. When t=0, its velocity equals 2 and it is at position s=1. When t=1, it is at position s =
Integrate acceleration to get velocity: v(t) = ∫(4+6t) dt = 4t+3t2+C1. Use v(0)=2: C1=2, so v(t) = 4t+3t2+2.
Integrate velocity to get position: s(t) = ∫(4t+3t2+2) dt = 2t2+t3+2t+C2. Use s(0)=1: C2=1, so s(t) = 2t2+t3+2t+1.
Evaluate at t=1: s(1) = 2(1)+1+2(1)+1 = 2+1+2+1.
= 6. The correct answer is C.
Find the acceleration (in ft/sec2) needed to bring a particle moving with a velocity of 100 ft/sec to a stop in 4 seconds.
With constant acceleration a, velocity follows v(t) = v0 + at, where v0 = 100 is the initial velocity.
"Comes to a stop in 4 seconds" means v(4) = 0: 0 = 100 + a(4).
4a = −100 → a = −25.
The correct answer is B.
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