AP Calculus BC Practice Quiz: Integrating Functions Using Long Division and Completing the Square
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
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A) (1/2)x² + 2x - 4ln|x + 1| + C
B) x + 2 - 4/(x + 1)² + C
C) (1/2)x² + 2x + 4ln|x + 1| + C
D) x² + 2x - 4arctan(x) + C
Correct Answer: A
The degree of the numerator is greater than the degree of the denominator, so we use long division. Dividing (x² + 3x - 2) by (x + 1) gives a quotient of (x + 2) and a remainder of -4. The integral becomes ∫ (x + 2 - 4/(x + 1)) dx. Integrating term by term gives (1/2)x² + 2x - 4ln|x + 1| + C.
A) ln|x² + 6x + 10| + C
B) arctan(x + 3) + C
C) (1/2)arctan((x+3)/2) + C
D) -1 / (x² + 6x + 10) + C
Correct Answer: B
The denominator is an irreducible quadratic. We complete the square: x² + 6x + 10 = (x² + 6x + 9) + 1 = (x + 3)² + 1. The integral becomes ∫ 1 / ((x + 3)² + 1) dx. Using the substitution u = x + 3, du = dx, the integral is ∫ 1 / (u² + 1) du, which evaluates to arctan(u) + C. Substituting back gives arctan(x + 3) + C.
A) 4.5 + 2ln(2)
B) 2.5 + 2ln(2)
C) 6.5 - 2ln(2)
D) 4.5 - 2ln(2)
Correct Answer: A
First, rearrange the integrand by dividing each term in the numerator by x²: (3x³ / x²) + (2x / x²) - (4 / x²) = 3x + 2/x - 4x⁻². The antiderivative is (3/2)x² + 2ln|x| - 4(-1)x⁻¹ = (3/2)x² + 2ln|x| + 4/x. Now, evaluate from 1 to 2: [ (3/2)(2)² + 2ln(2) + 4/2 ] - [ (3/2)(1)² + 2ln(1) + 4/1 ] = [ 6 + 2ln(2) + 2 ] - [ 1.5 + 0 + 4 ] = (8 + 2ln(2)) - 5.5 = 2.5 + 2ln(2). Wait, I made a calculation error in my explanation. Let me re-calculate. [ (3/2)(4) + 2ln(2) + 2 ] - [ 3/2 + 0 + 4 ] = [6 + 2ln(2) + 2] - [5.5] = 8 + 2ln(2) - 5.5 = 2.5 + 2ln(2). Let me re-check the question and options. Ah, I see the error in my scratchpad. Let me re-do the problem from scratch. Integrand: 3x + 2/x - 4x⁻². Antiderivative: (3/2)x² + 2ln|x| + 4/x. Evaluate at 2: (3/2)(4) + 2ln(2) + 4/2 = 6 + 2ln(2) + 2 = 8 + 2ln(2). Evaluate at 1: (3/2)(1) + 2ln(1) + 4/1 = 1.5 + 0 + 4 = 5.5. Difference: (8 + 2ln(2)) - 5.5 = 2.5 + 2ln(2). My option B is correct. Let me re-evaluate the options to see if I made a typo. Let's assume the question was (3x^3 + 2x^2 - 4)/x^2. Then integrand is 3x + 2 - 4x^-2. Antiderivative is (3/2)x^2 + 2x + 4/x. At 2: (3/2)(4) + 2(2) + 4/2 = 6 + 4 + 2 = 12. At 1: (3/2)(1) + 2(1) + 4/1 = 1.5 + 2 + 4 = 7.5. Difference is 12 - 7.5 = 4.5. Let's use this version. The question is better. Changing question text to (3x³ + 2x² - 4) / x². New explanation: First, rearrange the integrand by dividing each term in the numerator by x²: (3x³ / x²) + (2x² / x²) - (4 / x²) = 3x + 2 - 4x⁻². The antiderivative is (3/2)x² + 2x + 4/x. Now, evaluate from 1 to 2: [ (3/2)(2)² + 2(2) + 4/2 ] - [ (3/2)(1)² + 2(1) + 4/1 ] = [ 6 + 4 + 2 ] - [ 1.5 + 2 + 4 ] = 12 - 7.5 = 4.5. This matches option A. Let's go with this. Final question text: "What is the value of the definite integral ∫ from 1 to 2 of (3x³ + 2x² - 4) / x² dx?" Correct Answer: A. Explanation: "First, rearrange the integrand by dividing each term in the numerator by x²: (3x³ / x²) + (2x² / x²) - (4 / x²) = 3x + 2 - 4x⁻². The antiderivative is (3/2)x² + 2x - 4(-1)x⁻¹ = (3/2)x² + 2x + 4/x. Now, evaluate from 1 to 2: [ (3/2)(2)² + 2(2) + 4/2 ] - [ (3/2)(1)² + 2(1) + 4/1 ] = [ 6 + 4 + 2 ] - [ 1.5 + 2 + 4 ] = 12 - 7.5 = 4.5."
A) ∫ (x³ + 1) / (x² - 4) dx
B) ∫ x / (x² - 4) dx
C) ∫ 1 / (x² - 4) dx
D) ∫ 1 / (x² - 4x + 5) dx
Correct Answer: D
Completing the square is used for irreducible quadratic expressions, typically in the denominator. Option A requires long division. Option B is best solved with a u-substitution. Option C uses partial fraction decomposition since x² - 4 factors into (x-2)(x+2). Option D has an irreducible quadratic denominator (x² - 4x + 5 = (x-2)² + 1), making completing the square the appropriate technique, which will lead to an arctangent form.
A) π/6
B) π/4
C) π/3
D) π/2
Correct Answer: C
The integral requires completing the square in the denominator inside the square root. 3 - 2x - x² = 3 - (x² + 2x) = 3 - (x² + 2x + 1 - 1) = 3 - ((x+1)² - 1) = 4 - (x+1)². The integral becomes ∫ from 0 to 1 of 1 / √(4 - (x+1)²) dx. This is in the form of an arcsin integral, ∫ 1/√(a²-u²) du, where a=2 and u=x+1. The antiderivative is arcsin((x+1)/2). Evaluating the definite integral: [arcsin((1+1)/2)] - [arcsin((0+1)/2)] = arcsin(1) - arcsin(1/2) = π/2 - π/6 = 2π/6 = π/3.
A) (1/2)x² - (1/2)ln(x² + 1) + C
B) (1/4)x⁴ / (x³/3 + x) + C
C) (1/2)x² - arctan(x) + C
D) x - ln(x² + 1) + C
Correct Answer: A
Since the degree of the numerator is greater than the degree of the denominator, perform long division. x³ divided by (x² + 1) gives a quotient of x and a remainder of -x. So, the integrand can be rewritten as ∫ (x - x / (x² + 1)) dx. This can be split into two integrals: ∫ x dx - ∫ x / (x² + 1) dx. The first integral is (1/2)x². The second integral can be solved with a u-substitution where u = x² + 1 and du = 2x dx. This gives (1/2) ∫ 1/u du = (1/2)ln|u| = (1/2)ln(x² + 1). Combining the parts, the result is (1/2)x² - (1/2)ln(x² + 1) + C.
A) (1/2)ln|x² + 2x + 5| + 4arctan(x+1) + C
B) ln|x² + 2x + 5| + arctan((x+1)/2) + C
C) (1/2)ln|x² + 2x + 5| + 2arctan((x+1)/2) + C
D) ln|x+1| + 2arctan((x+1)/2) + C
Correct Answer: C
The denominator is an irreducible quadratic, so we complete the square: x² + 2x + 5 = (x² + 2x + 1) + 4 = (x+1)² + 4. The derivative of the denominator is 2x + 2. We manipulate the numerator to contain this term: x + 5 = (1/2)(2x + 2) - 1 + 5 = (1/2)(2x + 2) + 4. Now, split the integral: ∫ [(1/2)(2x + 2) / (x² + 2x + 5)] dx + ∫ [4 / ((x+1)² + 4)] dx. The first part is a u-substitution resulting in (1/2)ln|x² + 2x + 5|. The second part is an arctan form: 4 ∫ 1/((x+1)² + 2²) dx = 4 * (1/2)arctan((x+1)/2) = 2arctan((x+1)/2). Combining them gives (1/2)ln|x² + 2x + 5| + 2arctan((x+1)/2) + C.