AP Calculus AB Quiz: Integrating Using Substitution

What this quiz covers

This quiz focuses on Integrating Using Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $2\ln|3x-2|+C$ (correct answer)
2. $6\ln|3x-2|+C$
3. $\dfrac{1}{2}\ln|3x-2|+C$
4. $\dfrac{6}{3x-2}+C$
5. $2\ln|x-2|+C$

Explanation: The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 3x - 2, whose derivative 3 requires adjusting the 6 to 2. Set u = 3x - 2, so du = 3 dx and the integral becomes 2 ∫ du/u. Integrating gives 2 ln|u| + C, or 2 ln|3x - 2| + C. A tempting distractor like 6 ln|3x - 2| + C fails because it doesn't divide by the 3 from du, resulting in too large a coefficient. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.

Question 2

1. $-\dfrac{1}{2(x^2-x+6)^2}+C$ (correct answer)
2. $\dfrac{1}{2(x^2-x+6)^2}+C$
3. $-\dfrac{1}{(x^2-x+6)^2}+C$
4. $\dfrac{1}{(x^2-x+6)^2}+C$
5. $-\dfrac{1}{2}(x^2-x+6)^{-3}+C$

Explanation: This integral requires the skill of integration by substitution, often called u-substitution. The numerator (2x - 1) matches the derivative of the inner quadratic x² - x + 6 in the denominator raised to the third power. Setting u = x² - x + 6 gives du = (2x - 1) dx, simplifying to ∫ u^{-3} du. Integration results in -1/(2u²) + C, or -1/(2(x² - x + 6)²) + C. A tempting distractor is choice C, which omits the factor of 1/2, possibly from forgetting to divide by -2 when integrating u^{-3}. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

Question 3

1. $\sqrt{x^2+3x+1}+C$
2. $2\sqrt{x^2+3x+1}+C$ (correct answer)
3. $\dfrac{1}{2}\sqrt{x^2+3x+1}+C$
4. $2\ln|x^2+3x+1|+C$
5. $\sqrt{x^2+3x}+C$

Explanation: The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is x² + 3x + 1, whose derivative 2x + 3 exactly matches the numerator. Set u = x² + 3x + 1, so du = (2x + 3) dx, transforming the integral into ∫ du / √u. Integrating gives 2 √u + C, or 2 √(x² + 3x + 1) + C. A tempting distractor like √(x² + 3x + 1) + C fails because it misses the factor of 2 from integrating u^{-1/2}. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.

Question 4

1. $2\sqrt{x^3+x+5}+C$ (correct answer)
2. $\sqrt{x^3+x+5}+C$
3. $\dfrac{1}{2}\sqrt{x^3+x+5}+C$
4. $2\sqrt{x^3+x}+C$
5. $\ln|x^3+x+5|+C$

Explanation: The skill here is integrating using u-substitution. The denominator has a square root of a cubic, and the numerator 3x² + 1 is its derivative, so u = x³ + x + 5 with du = (3x² + 1) dx. This turns the integral into ∫ du / √u = 2√u + C. Back-substituting gives 2√(x³ + x + 5) + C. Choice C halves incorrectly, forgetting the 2 from integrating u^{-1/2}. Spot u as the inside of the root when the numerator is its derivative for these forms.

Question 5

1. $\dfrac{(4x^3-3x^2+8)^3}{3}+C$ (correct answer)
2. $(4x^3-3x^2+8)^2+C$
3. $\dfrac{(4x^3-3x^2+8)^3}{12}+C$
4. $\dfrac{(4x^3-3x^2)^3}{3}+C$
5. $\dfrac{(12x^2-6x)(4x^3-3x^2+8)^3}{3}+C$

Explanation: The skill here is integrating using u-substitution. The integrand has (4x³ - 3x² + 8)^2 multiplied by its derivative 12x² - 6x. Set u = 4x³ - 3x² + 8 with du = (12x² - 6x) dx, yielding ∫ u^2 du = u^3 / 3 + C. This is (4x³ - 3x² + 8)^3 / 3 + C. Choice C divides by 12 incorrectly, mishandling the power rule. Look for polynomials raised to powers multiplied by their derivatives for power-rule substitutions.

Question 6

1. $\sqrt{7x^2+3}+C$
2. $2\sqrt{7x^2+3}+C$ (correct answer)
3. $14\sqrt{7x^2+3}+C$
4. $\dfrac{1}{2}\sqrt{7x^2+3}+C$
5. $2\sqrt{7x+3}+C$

Explanation: The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 7x² + 3, whose derivative 14x exactly matches the numerator. Set u = 7x² + 3, so du = 14x dx, transforming the integral into ∫ du / √u. Integrating gives 2 √u + C, or 2 √(7x² + 3) + C. A tempting distractor like √(7x² + 3) + C fails because it misses the 2 from integrating u^{-1/2}. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.

Question 7

1. $\dfrac{1}{7}\sin(7x-3)+C$ (correct answer)
2. $\sin(7x-3)+C$
3. $-\dfrac{1}{7}\cos(7x-3)+C$
4. $\dfrac{1}{7}\cos(7x-3)+C$
5. $\dfrac{1}{7}\sin(7x)+3+C$

Explanation: This integral requires the skill of integration by substitution, often called u-substitution. The argument $7x - 3$ is the inner function of cosine, with derivative $7$. Setting $u = 7x - 3$ gives $du = 7 \, dx$, so $dx = \frac{du}{7}$, and the integral is $\frac{1}{7} \int \cos u \, du$. This integrates to $\frac{1}{7} \sin u + C$, or $\frac{1}{7} \sin(7x - 3) + C$. A tempting distractor is choice C, which uses cosine instead of sine, possibly from confusing the integral of cosine. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

Question 8

1. $-\dfrac{1}{(x^6+2)^3}+C$
2. $-\dfrac{1}{3(x^6+2)^3}+C$ (correct answer)
3. $\dfrac{1}{3(x^6+2)^3}+C$
4. $-\dfrac{6}{(x^6+2)^3}+C$
5. $-\dfrac{1}{(x+2)^3}+C$

Explanation: This integral requires the skill of integration by substitution, often called u-substitution. The numerator 6x^5 is the derivative of x^6 + 2 in the denominator raised to 4. Let $u = x^6 + 2$, $du = 6x^5 \, dx$, simplifying to $\int u^{-4} \, du$. Integration yields $-\frac{1}{3u^3} + C$, or $-\frac{1}{3(x^6 + 2)^3} + C$. A tempting distractor is choice A, omitting the 1/3, perhaps from forgetting to divide by -3 in the power rule. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

Question 9

1. $\dfrac{(2x+1)^9}{2}+C$ (correct answer)
2. $\dfrac{9(2x+1)^9}{9}+C$
3. $\dfrac{9(2x+1)^9}{2}+C$
4. $\dfrac{(2x+1)^9}{18}+C$
5. $(2x+1)^8+C$

Explanation: The skill here is integrating using u-substitution. The integrand is a constant times a power of a linear function, so set u = 2x + 1 with du = 2 dx. Rewrite the integral as (9/2) ∫ u^8 du, which integrates to (9/2)(u^9/9) = (1/2)u^9 + C. This simplifies to (2x + 1)^9 / 2 + C. Choice D divides by an extra factor, likely from mishandling the chain rule constant. Identify compositions where the outer function is a power and the inner's derivative appears as a factor.

Question 10

1. $\ln(x^2+1)+C$ (correct answer)
2. $\dfrac{1}{2}\ln(x^2+1)+C$
3. $\ln|x+1|+C$
4. $\ln|2x|+C$
5. $\dfrac{2x}{x^2+1}+C$

Explanation: The skill here is integrating using u-substitution. The denominator is x² + 1, with numerator 2x as its derivative. Set u = x² + 1 with du = 2x dx, yielding ∫ du/u = ln|u| + C. This is ln(x² + 1) + C. Choice B halves incorrectly, as the 2 is in du. Look for non-linear denominators with derivative numerators for logarithmic results.

Question 11

1. $\dfrac{(1-x^2)^{10}}{10}+C$
2. $-\dfrac{(1-x^2)^{10}}{10}+C$ (correct answer)
3. $-(1-x^2)^{10}+C$
4. $-\dfrac{2x(1-x^2)^{10}}{10}+C$
5. $-\dfrac{(1-x)^{10}}{10}+C$

Explanation: This integral requires the skill of integration by substitution, often called u-substitution. The factor 2x is the negative derivative of the inner 1 - x² in (1 - x²)^9. Let u = 1 - x², so du = -2x dx and 2x dx = -du, transforming to - ∫ u^9 du. Integration gives - (u^{10}/10) + C, or - (1 - x²)^{10}/10 + C. A tempting distractor is choice A, which has a positive sign, likely from missing the negative in du. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

Question 12

1. $\ln|5x+4|+C$ (correct answer)
2. $5\ln|5x+4|+C$
3. $\dfrac{1}{5}\ln|5x+4|+C$
4. $\ln|x+4|+C$
5. $\dfrac{5}{5x+4}+C$

Explanation: The skill here is integrating using u-substitution. The integrand is 5/(5x + 4), where the denominator's derivative is 5, matching the numerator. Set u = 5x + 4 with du = 5 dx, so ∫ du/u = ln|u| + C. This is ln|5x + 4| + C. Choice C divides by 5 extra, likely confusing the constant adjustment. Recognize linear denominators with matching derivative constants for logarithmic integrals.

Question 13

1. $\left.e^{x^3}\right|_0^1$ (correct answer)
2. $\left.3e^{x^3}\right|_0^1$
3. $\left.\dfrac{1}{3}e^{x^3}\right|_0^1$
4. $\left.x^2e^{x^3}\right|_0^1$
5. $\left.e^{x}\right|_0^1$

Explanation: The skill here is integrating using u-substitution for a definite integral. The integrand is $3x^2 e^{x^3}$, where $x^3$ is inside the exponential and $3x^2$ is its derivative. Set $u = x^3$ with $du = 3x^2 \, dx$, transforming to $\int e^u \, du = e^u$, evaluated from 0 to 1. This is $e^{x^3}$ from 0 to 1. Choice C divides by 3 incorrectly, as it's absorbed in du. Recognize exponentials with polynomial powers where the exponent's derivative appears.

Question 14

1. $\ln|7x^2+3|+C$ (correct answer)
2. $\dfrac{1}{7}\ln|7x^2+3|+C$
3. $14\ln|7x^2+3|+C$
4. $\ln|x^2+3|+C$
5. $\dfrac{7x^2}{7x^2+3}+C$

Explanation: The skill here is integrating using u-substitution. The denominator is $7x^2 + 3$, with numerator 14x being its derivative. Set $u = 7x^2 + 3$ with $du = 14x \, dx$, giving $\int \frac{du}{u} = \ln|u| + C$. This is $\ln|7x^2 + 3| + C$. Choice B divides by 7 unnecessarily, as the coefficient is in $du$. Watch for quadratic denominators with linear numerators matching derivatives for logs.

Question 15

1. $2\sqrt{\tfrac{1}{2}x^2+4x+7}+C$ (correct answer)
2. $\sqrt{\tfrac{1}{2}x^2+4x+7}+C$
3. $\dfrac{1}{2}\sqrt{\tfrac{1}{2}x^2+4x+7}+C$
4. $2\sqrt{\tfrac{1}{2}x^2+4x}+C$
5. $\ln\left|\tfrac{1}{2}x^2+4x+7\right|+C$

Explanation: The skill here is integrating using u-substitution. The square root holds (1/2)x² + 4x + 7, with numerator x + 4 as its derivative. Set u = (1/2)x² + 4x + 7 with du = (x + 4) dx, giving ∫ du/√u = 2√u + C. This is 2√((1/2)x² + 4x + 7) + C. Choice C halves wrongly, missing the 2 from integration. Note fractions in quadratics but ensure numerator matches derivative exactly.

Question 16

1. $-\cos(3t+\pi)+C$
2. $\dfrac{1}{3}\cos(3t+\pi)+C$
3. $-\dfrac{1}{3}\cos(3t+\pi)+C$ (correct answer)
4. $-\dfrac{1}{3}\sin(3t+\pi)+C$
5. $-\dfrac{1}{3}\cos(3t)+\pi+C$

Explanation: This integral requires the skill of integration by substitution, often called u-substitution. The integrand is sin(3t + π), where 3t + π is the inner function, and its derivative is 3, suggesting a need to account for the chain rule. Setting u = 3t + π gives du = 3 dt, so dt = du/3, transforming the integral to (1/3) ∫ sin u du. Integrating yields - (1/3) cos u + C, or - (1/3) cos(3t + π) + C. A tempting distractor is choice A, which omits the 1/3 factor, likely from forgetting to include the adjustment for du. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

Question 17

1. $-\dfrac{1}{6(x^3+2)^6}+C$ (correct answer)
2. $\dfrac{1}{6(x^3+2)^6}+C$
3. $-\dfrac{3x^2}{6(x^3+2)^6}+C$
4. $-\dfrac{1}{(x^3+2)^6}+C$
5. $\ln|x^3+2|+C$

Explanation: The skill here is integrating using u-substitution. The denominator is (x³ + 2)^7, with numerator 3x² matching its derivative. Set u = x³ + 2 with du = 3x² dx, so ∫ u^{-7} du = -1/(6u^6) + C. This is -1/[6(x³ + 2)^6] + C. Choice D omits the 6, forgetting to divide by -6. Spot high negative powers with polynomial numerators as derivatives.

Question 18

1. $\dfrac{(5x^3-1)^3}{3}+C$ (correct answer)
2. $(5x^3-1)^3+C$
3. $\dfrac{(5x^3-1)^3}{15}+C$
4. $\dfrac{15x^2(5x^3-1)^3}{3}+C$
5. $\dfrac{(x^3-1)^3}{3}+C$

Explanation: The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 5x³ - 1, whose derivative 15x² exactly matches the factor outside. Set u = 5x³ - 1, so du = 15x² dx, transforming the integral into ∫ u² du. Integrating gives (1/3) u³ + C, or (5x³ - 1)³ / 3 + C. A tempting distractor like (5x³ - 1)³ / 15 + C fails because it incorrectly divides by an extra 5, perhaps confusing the chain rule factor. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.

Question 19

1. $\tan(5x-2)+C$
2. $\tfrac{1}{5}\tan(5x-2)+C$ (correct answer)
3. $5\tan(5x-2)+C$
4. $\sec^2(5x-2)+C$
5. $\tfrac{1}{5}\tan x + C$

Explanation: This integral requires u-substitution because we have a composite function inside the secant squared. Let u = 5x - 2, so du = 5 dx, which means dx = ⅕ du. The integral becomes ∫ sec²(u) · ⅕ du = ⅕ ∫ sec²(u) du = ⅕ tan(u) + C = ⅕ tan(5x-2) + C. Choice A (tan(5x-2) + C) forgets the crucial factor of ⅕ that comes from the chain rule in reverse. When integrating trigonometric functions with linear arguments, always divide by the coefficient of x in the argument.

Question 20

1. $\ln|x^2+4x+1|+C$ (correct answer)
2. $\dfrac{1}{2}\ln|x^2+4x+1|+C$
3. $2\ln|x^2+4x+1|+C$
4. $\dfrac{2x+4}{x^2+4x+1}+C$
5. $\ln|2x+4|+C$

Explanation: This integral requires the skill of integration by substitution, often called u-substitution. The numerator 2x + 4 is the derivative of x² + 4x + 1 in the denominator. Set u = x² + 4x + 1, du = (2x + 4) dx, transforming to ∫ du/u. This integrates to ln|u| + C, or ln|x² + 4x + 1| + C. A tempting distractor is choice B, multiplying by 1/2, possibly from mistakenly halving the derivative. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.