AP Calculus BC expects fluency with a much wider integration toolkit than the AB exam: beyond u-substitution and basic trig integrals, BC regularly tests integration by parts, integrands that simplify dramatically once you recognize a hidden algebraic identity (like e2ln u = u2), rational functions that split apart via partial fractions, and integrands built from inverse trig or logarithmic pieces that don't fit any single standard formula on their own.
Integration by parts, ∫u dv = uv − ∫v du, is the single most important technique to have fully automatic for this unit — it's what makes integrals like ∫arctan(x)dx, ∫ln(x)dx, and ∫eθcos(θ)dθ tractable, and choosing which piece to call u versus dv is usually the whole battle. Trig integrands often need an identity first — expanding a squared binomial with a Pythagorean identity, or multiplying by a clever form of 1 (like (1−sinθ)/(1−sinθ)) to unlock a denominator — before any standard antiderivative rule applies. And whenever a denominator factors into simpler linear or quadratic pieces, partial fractions turn one hard integral into several easy ones.
Below are 23 AP Calculus BC-style practice problems covering integration by parts, substitution with exponential and logarithmic integrands, trigonometric integrals, partial fractions, inverse trig antiderivatives, and motion applications of antidifferentiation. 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 BC QBank.
The equation of the curve whose slope at point (x,y) is x2−5, and which contains the point (2,−1), is
Integrate the slope to find the family of curves: y = ∫(x2−5)dx = x3/3 − 5x + C.
Use the point (2,−1) to solve for C: −1 = 8/3 − 10 + C → C = −1−8/3+10 = 9−8/3 = 27/3−8/3 = 19/3.
The equation is y = x3/3 − 5x + 19/3. The correct answer is A.
A particle moves along a line with acceleration 4+12t at time t. When t=0, its velocity equals 1 and it is at position s=3. When t=2, it is at position s =
Integrate acceleration to get velocity: v(t) = ∫(4+12t)dt = 4t+6t2+C1. Use v(0)=1: C1=1, so v(t) = 4t+6t2+1.
Integrate velocity to get position: s(t) = ∫(4t+6t2+1)dt = 2t2+2t3+t+C2. Use s(0)=3: C2=3, so s(t) = 2t2+2t3+t+3.
Evaluate at t=2: s(2) = 2(4)+2(8)+2+3 = 8+16+2+3.
= 29. The correct answer is C.
Find the acceleration (in ft/sec2) needed to bring a particle moving with a velocity of 120 ft/sec to a stop in 4 seconds.
With constant acceleration a, v(t) = v0+at, where v0=120.
"Stops in 4 seconds" means v(4)=0: 0 = 120+4a → a = −120/4.
= −30. The correct answer is C.
∫ x2/(x2−4) dx =
Since the degree of the numerator equals the degree of the denominator, divide first: x2/(x2−4) = 1 + 4/(x2−4) (since x2 = (x2−4)+4).
Apply partial fractions to the remainder: 4/(x2−4) = 4/[(x−2)(x+2)] = 1/(x−2) − 1/(x+2) (solving for the constants gives A=1, B=−1).
Integrate term by term: ∫[1 + 1/(x−2) − 1/(x+2)]dx = x + ln|x−2| − ln|x+2| + C = x + ln|(x−2)/(x+2)| + C.
The correct answer is A.
Keep the momentum going.
Explore 1,500+ Free AP Calculus BC Problems∫ (3x−1)/√(6x−9x2) dx =
Notice that the derivative of the expression under the radical is proportional to the numerator: d/dx[6x−9x2] = 6−18x = −6(3x−1).
This means the integral has the form ∫ g′(x)/√g(x) dx = 2√g(x) + C, scaled by a constant. Let u=6x−9x2, du=−6(3x−1)dx, so (3x−1)dx = −du/6.
∫(−du/6)/√u = −(1/6)(2√u) = −√u/3.
Substituting back: −(1/3)√(6x−9x2) + C. The correct answer is D.
∫ e3x/(1+ex) dx =
Let u = ex, so du = exdx → dx = du/u. Since e3x = u3, the integral becomes ∫ u3/(1+u) · du/u = ∫ u2/(1+u) du.
Divide the polynomial: u2/(1+u) = u−1 + 1/(1+u) (since u2 = (u+1)(u−1)+1).
Integrate term by term: ∫[u−1+1/(1+u)]du = u2/2 − u + ln|1+u| + C.
Substitute back u=ex: e2x/2 − ex + ln(1+ex) + C. The correct answer is B.
∫ cosθ/(4+sin2θ) dθ =
Let u=sinθ, so du=cosθ dθ. 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=sinθ: (1/2)tan−1(sinθ/2) + C.
The correct answer is C.
7 down, 16 to go — keep practicing.
Explore 1,500+ Free AP Calculus BC Problems∫ arctan(2x) dx =
Use integration by parts with u=arctan(2x), dv=dx, so du=2/(1+4x2)dx and v=x.
∫u dv = uv−∫v du = x arctan(2x) − ∫2x/(1+4x2)dx.
For the remaining integral, let w=1+4x2, dw=8x dx, so 2x dx = dw/4: ∫(dw/4)/w = (1/4)ln|w| = (1/4)ln(1+4x2).
Combining: x arctan(2x) − (1/4)ln(1+4x2) + C. The correct answer is B.
∫ dx/(2−ex) =
Let w=ex, so dw=exdx = w dx → dx=dw/w. The integral becomes ∫dw/[w(2−w)].
Apply partial fractions: 1/[w(2−w)] = (1/2)[1/w + 1/(2−w)].
Integrate: (1/2)[ln|w| − ln|2−w|] + C. Substitute back w=ex: (1/2)[x − ln|2−ex|] + C.
This is x/2 − (1/2)ln|2−ex| + C. The correct answer is B.
∫ (3−y)2/(9√y) dy =
Expand the numerator: (3−y)2 = 9−6y+y2. Divide each term by 9√y = 9y1/2: y−1/2 − (2/3)y1/2 + (1/9)y3/2.
Integrate term by term: 2y1/2 − (2/3)(2/3)y3/2 + (1/9)(2/5)y5/2.
= 2√y − (4/9)y3/2 + (2/45)y5/2 + C.
The correct answer is D.
10 down, 13 to go.
Explore 1,500+ Free AP Calculus BC Problems∫ e3lnu du =
Simplify the integrand first using eln(k) = k: e3lnu = eln(u3) = u3.
The integral becomes the simple power rule: ∫u3du = u4/4 + C.
The correct answer is C. The other choices come from trying to integrate the original unsimplified exponential expression directly, which is a much harder (and unnecessary) path.
∫ dy/[y(4+(ln y)2)] =
Let u=ln y, so du=dy/y. The integral becomes ∫du/(4+u2).
Using 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 C.
∫ (cotθ+1)2 dθ =
Expand the square: (cotθ+1)2 = cot2θ+2cotθ+1. Use the 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 numerator and denominator by 1+sinθ (so the denominator becomes a difference of squares): (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.
14 down, 9 to go — over halfway there.
Explore 1,500+ Free AP Calculus BC 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 4 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 v(t) never goes negative on this interval, distance equals the definite integral of velocity: ∫049t2dt = 3t3|04 = 3(64)−0.
= 192. The correct answer is C.
∫ ln(x5) dx =
Simplify first using log rules: ln(x5) = 5 ln x, so the integral is 5∫ln x dx.
Use integration by parts on ∫ln x dx: let u=ln x, dv=dx, so du=dx/x, v=x. ∫ln x dx = x ln x − ∫x·(1/x)dx = x ln x − x.
Multiply by the constant 5: 5(x ln x−x) = 5x(ln x−1)+C.
The correct answer is B.
∫ ln y/y3 dy =
Use integration by parts: let u=ln y, dv=y−3dy, so du=dy/y, and v=−1/(2y2).
∫u dv = uv−∫v du = −ln y/(2y2) − ∫−1/(2y2)·(1/y)dy = −ln y/(2y2) + (1/2)∫y−3dy.
= −ln y/(2y2) + (1/2)(−1/(2y2)) = −ln y/(2y2) − 1/(4y2).
Combining over a common denominator: −(2ln y+1)/(4y2) + C. The correct answer is B.
17 down, 6 to go.
Explore 1,500+ Free AP Calculus BC Problems∫ √x(√x−2) dx =
Distribute first: √x(√x−2) = x−2√x = x−2x1/2.
Integrate term by term: ∫(x−2x1/2)dx = x2/2 − 2(2/3)x3/2+C.
= x2/2 − (4/3)x3/2 + C.
The correct answer is D.
∫ eθsinθ dθ =
Apply integration by parts twice (with u=eθ both times), which eventually returns the original integral, denoted I: I = eθsinθ − eθcosθ − I (after two applications).
Solve for I algebraically: 2I = eθ(sinθ−cosθ) → I = (1/2)eθ(sinθ−cosθ).
The correct answer is D. (This "solve for the original integral algebraically" trick is the standard approach whenever integration by parts on eθ times a trig function loops back on itself.)
∫ (2−ln t)2/t dt =
Let u = 2−ln t, so du = −dt/t, meaning dt/t = −du.
Substitute: ∫u2(−du) = −u3/3+C.
Substitute back u=2−ln t: −(2−ln t)3/3+C.
The correct answer is D.
20 down, 3 to go — almost there.
Explore 1,500+ Free AP Calculus BC Problems∫ u·csc2(u) du =
Use integration by parts, treating u itself as the "u" part: let dv=csc2(u)du, so v=−cot u.
∫u dv = uv−∫v du = −u cot u − ∫−cot u du = −u cot u + ∫cot u du.
Since ∫cot u du = ln|sin u|: −u cot u + ln|sin u| + C.
The correct answer is C.
∫ (2x+3)/(9+x2) dx =
Split into two integrals that each match a standard form: ∫2x/(9+x2)dx + ∫3/(9+x2)dx.
For the first (u=9+x2, du=2x dx): ∫du/u = ln(9+x2).
For the second, use ∫dx/(a2+x2) = (1/a)tan−1(x/a) with a=3: 3·(1/3)tan−1(x/3) = tan−1(x/3) (the factor of 3 from the numerator exactly cancels the 1/a=1/3 from the formula).
Combining: ln(9+x2) + tan−1(x/3) + C. The correct answer is B.
∫ (2−x)/√(4−x2) dx =
Split into two integrals: ∫2/√(4−x2)dx − ∫x/√(4−x2)dx.
For the first, use ∫dx/√(a2−x2) = sin−1(x/a) with a=2: 2· sin−1(x/2).
For the second (u=4−x2, du=−2x dx, so x dx=−du/2): −∫(−du/2)/√u = (1/2)(2√u) = √(4−x2) (this piece is subtracted, but the two negative signs cancel to give a positive result).
Combining: 2sin−1(x/2) + √(4−x2) + C. The correct answer is C.
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