AP Calculus BC 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 BC.

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

Explanation: This integral requires u-substitution for a rational function with a quadratic denominator raised to a power. Let u = x² + 1, so du = 2x dx, which means 10x dx = 5 du. The integral becomes ∫(5/u²) du = 5∫u^(-2) du = 5(-1/u) = -5/u = -5/(x² + 1). The antiderivative of u^(-2) is -1/u. Choice A has the wrong sign, while choice E represents the derivative rather than the antiderivative. When the numerator is a multiple of the derivative of the base of a power in the denominator, use u-substitution.

Question 2

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

Explanation: This integral uses u-substitution for a rational function with a linear denominator. Let u = 3x + 2, so du = 3 dx, which means 6 dx = 2 du. The integral becomes ∫(2/u) du = 2 ln|u| + C = 2 ln|3x + 2| + C. The coefficient 6 divided by the derivative factor 3 gives 2. Choice A incorrectly multiplies by 6, while choice C omits the adjustment factor. When integrating a constant over a linear expression ax + b, divide the constant by a.

Question 3

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

Explanation: This integral uses u-substitution for a rational function where the numerator is a multiple of the derivative of the denominator. Let u = x² + 1, so du = 2x dx, which means 5(2x) dx = 10x dx becomes ∫(5/u) du = 5 ln|u| + C = 5 ln(x² + 1) + C. Since x² + 1 > 0, we can omit absolute value bars. Choice B uses 5/2 incorrectly, while choice D would result from an arctangent substitution. When the numerator is a multiple of the derivative of the denominator, use logarithmic integration.

Question 4

1. $-2\ln|\cos(3x)|+C$ (correct answer)
2. $2\ln|\cos(3x)|+C$
3. $-6\ln|\cos(3x)|+C$
4. $2\ln|\sin(3x)|+C$
5. $2\tan(3x)+C$

Explanation: This integral requires u-substitution for a rational trigonometric function. The integrand 6sin(3x)/cos(3x) = 6tan(3x) can be rewritten, but it's better to use u = cos(3x), so du = -3sin(3x) dx. This gives us ∫6sin(3x)/cos(3x) dx = ∫(-2/u) du = -2ln|u| = -2ln|cos(3x)|. The factor 6 divided by -3 gives -2. Choice B has the wrong sign, while choice E represents the derivative form. When integrating f'(x)/f(x), use u-substitution to get ln|f(x)| + C.

Question 5

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

Explanation: This integral requires u-substitution to handle the rational function with a quadratic denominator. Let u = t² + 4, so du = 2t dt, which means 6t dt = 3 du. The integral becomes ∫(3/u) du = 3 ln|u| + C. Substituting back gives 3 ln(t² + 4) + C (absolute value bars can be dropped since t² + 4 > 0). Choice D might tempt students who think of arctangent formulas, but that applies to ∫1/(t² + 4) dt, not our integral with t in the numerator. When the numerator contains the derivative of the denominator (up to a constant), use u-substitution with the denominator as u.

Question 6

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

Explanation: This integral uses u-substitution for an exponential function with a negative linear exponent. Let u = -6t, so du = -6 dt, which means dt = -(1/6) du. The integral becomes ∫12e^u(-1/6) du = -2∫e^u du = -2e^u = -2e^(-6t). The factor 12 combined with -1/6 gives -2. Choice B has the wrong sign, while choice C incorrectly multiplies by -12. For exponential functions with negative exponents, the antiderivative will be negative times the exponential divided by the coefficient.

Question 7

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

Explanation: This integral uses u-substitution for a power function with a quadratic inner expression. Let u = x² + 4x + 5, so du = (2x + 4) dx, making the integral ∫u³ du = u⁴/4 = (1/4)(x² + 4x + 5)⁴. The derivative of the inner function 2x + 4 exactly matches the linear factor in the integrand. Choice B omits the factor of 1/4, while choice C has an incorrect denominator. When the linear factor exactly matches the derivative of a quadratic expression, use u-substitution with the quadratic as u.

Question 8

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

Explanation: This integral uses u-substitution for a function involving a square root with a linear expression inside. Let $u = 1 - 4x$, so $du = -4 \, dx$, which means $dx = -\dfrac{1}{4} \, du$. The integral becomes $\int 2 u^{-1/2} (-1/4) \, du = -\dfrac{1}{2} \int u^{-1/2} \, du = -\dfrac{1}{2} (2 \sqrt{u}) = -\sqrt{u} = -\sqrt{1-4x}$. The antiderivative of $u^{-1/2}$ is $2 u^{1/2} = 2 \sqrt{u}$, and we include the negative sign from the substitution. Choice C has an extra factor of 1/2, while choice D would result from an inverse trigonometric substitution. For integrals of the form $\dfrac{f'(x)}{\sqrt{f(x)}}$, use u-substitution with $u = f(x)$.

Question 9

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

Explanation: This integral uses u-substitution for a rational function where the numerator is a multiple of the derivative of the denominator. Let u = x⁴ + 2, so du = 4x³ dx, which means 10x³ dx = (10/4) du = (5/2) du. The integral becomes ∫(5/2)(1/u) du = (5/2) ln|u| + C = (5/2) ln|x⁴ + 2| + C. Choice A incorrectly uses 10 instead of 5/2, while choice C uses 10/4 = 5/2 but writes it as a fraction. When the numerator is a multiple of the derivative of the denominator, divide by the appropriate factor.

Question 10

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

Explanation: This integral uses u-substitution for a rational function with a linear denominator. Let $u = 1 + 2x$, so $du = 2 \, dx$, which means $3 \, dx = \frac{3}{2} du$. The integral becomes $\int \frac{3}{2} \frac{1}{u} \, du = \frac{3}{2} \ln|u| + C = \frac{3}{2} \ln|1 + 2x| + C$. The coefficient 3 divided by the derivative factor 2 gives 3/2. Choice B incorrectly multiplies by 3 instead of dividing by 2, while choice C omits the adjustment factor. When integrating over a linear expression $ax + b$, divide the coefficient by a.

Question 11

1. $\dfrac{(3x+1)^8}{8}+C$
2. $\dfrac{(3x+1)^8}{24}+C$ (correct answer)
3. $\dfrac{(3x+1)^8}{3}+C$
4. $\dfrac{(3x+1)^7}{7}+C$
5. $\dfrac{(3x+1)^8}{9}+C$

Explanation: This integral requires u-substitution for a power function with a linear inner expression. Let u = 3x + 1, so du = 3 dx, which means dx = $\frac{1}{3}$ du. The integral becomes $\int u^7 \cdot \frac{1}{3} \, du = \frac{1}{3} \left( \frac{u^8}{8} \right) = \frac{1}{24} u^8 = \frac{1}{24} (3x+1)^8$. The chain rule adjustment divides by 3, and the power rule gives u^8/8, resulting in the coefficient 1/24. Choice A omits the chain rule factor, while choice C uses the wrong denominator. For power functions with linear arguments, substitute the linear expression and adjust for its derivative.

Question 12

1. $-14\sqrt{9-x}+C$ (correct answer)
2. $14\sqrt{9-x}+C$
3. $-7\sqrt{9-x}+C$
4. $-\dfrac{7}{2}\sqrt{9-x}+C$
5. $7\arcsin\left(\dfrac{x}{9}\right)+C$

Explanation: This integral uses u-substitution for a function with a square root and linear inner expression. Let u = 9 - x, so du = -dx, which means 7 dx = -7 du. The integral becomes ∫(7/√u)(-1) du = -7∫u^(-1/2) du = -7(2√u) = -14√u = -14√(9 - x). The antiderivative of u^(-1/2) is 2u^(1/2), and we include the negative from the substitution. Choice B omits the negative sign that comes from the chain rule. When the inner function has a negative derivative, include the negative sign in the result.

Question 13

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

Explanation: This integral uses u-substitution for a simple rational function with a linear denominator. Let u = x - 4, so du = dx, making the integral ∫(6/u) du = 6 ln|u| + C = 6 ln|x - 4| + C. The constant 6 remains as a coefficient in front of the natural logarithm. Choice A omits the coefficient 6, while choice C incorrectly divides by 6. When integrating a constant over a linear expression of the form x - a, the result is the constant times ln|x - a| + C.

Question 14

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

Explanation: This integral requires u-substitution for a rational function where the numerator is the derivative of the denominator. Let $u = x^2 + x + 3$, so $du = (2x + 1) \, dx$, making the integral $\int \frac{1}{u} \, du = \ln|u| + C = \ln(x^2 + x + 3) + C$. Since $x^2 + x + 3 = (x + \frac{1}{2})^2 + \frac{11}{4} > 0$, we can omit absolute value bars. Choice B incorrectly includes a factor of 1/2, while choice D loses the quadratic terms. When the numerator exactly equals the derivative of the denominator, use logarithmic integration.

Question 15

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

Explanation: This integral uses u-substitution for a trigonometric function where we can identify the inner function. Let u = 1 + sin x, so du = cos x dx, making the integral ∫u du = u²/2 = (1/2)(1 + sin x)². This expands to (1/2)(1 + 2sin x + sin²x), which includes both linear and quadratic sine terms. Choice B omits the factor of 1/2 from integration, while choice C incorrectly substitutes cosine. When integrating a function times the derivative of an inner function, substitute the inner function.

Question 16

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

Explanation: This integral requires u-substitution to handle the compound logarithmic function. Let u = ln x, so du = (1/x) dx, making the integral ∫(4/u) du = 4 ln|u| + C. Substituting back gives 4 ln|ln x| + C, which correctly handles the natural logarithm of a logarithm. The absolute value is necessary since ln x can be negative for 0 < x < 1. Choice B omits the constant 4, while choice C would result from incorrectly substituting just x instead of ln x. When the integrand has the form f'(x)/f(x), use u-substitution with u = f(x) to get ln|f(x)| + C.

Question 17

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

Explanation: This integral uses u-substitution for an exponential function with a linear exponent. Let u = 2x - 1, so du = 2 dx, which means 8 dx = 4 du. The integral becomes ∫4eᵘ du = 4eᵘ = 4e^(2x-1). The chain rule adjustment divides 8 by 2 to give the coefficient 4. Choice B incorrectly keeps the factor 8, while choice C has the wrong coefficient. For exponential functions with linear exponents ax + b, substitute u = ax + b and divide by a.

Question 18

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

Explanation: This integral uses u-substitution for a function with a linear expression raised to a negative power. Let $u = 2x + 1$, so $du = 2 \, dx$, which means $4 \, dx = 2 \, du$. The integral becomes $\int \frac{2}{u^3} \, du = 2 \int u^{-3} \, du = 2 \left( -\frac{1}{2} u^{-2} \right) = - \frac{1}{u^2} = -\frac{1}{(2x + 1)^2}$. The antiderivative of $u^{-3}$ is $-\frac{1}{2} u^{-2}$. Choice A doubles the coefficient incorrectly. When integrating powers of linear expressions, substitute and use the power rule with chain rule adjustment.

Question 19

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

Explanation: This integral requires u-substitution for a trigonometric function with a linear argument. Let u = 5x, so du = 5 dx, making the integral ∫sin(u) du = -cos(u) + C = -cos(5x) + C. The antiderivative of sin(u) is -cos(u), and we don't need to adjust for the chain rule here since we already have the factor 5 in front. Choice B has the wrong sign, while choice C incorrectly keeps the factor 5 with the cosine. When integrating sin(ax), the result is -(1/a)cos(ax) + C, but here we have the full derivative factor already present.

Question 20

1. $4\ln\left(x^2+9\right)+C$ (correct answer)
2. $8\ln\left(x^2+9\right)+C$
3. $4\arctan\left(\dfrac{x}{3}\right)+C$
4. $\dfrac{4}{x^2+9}+C$
5. $4\ln|x|+C$

Explanation: This integral uses u-substitution to handle a rational function where the numerator is the derivative of the denominator. Let u = x² + 9, so du = 2x dx, which means 8x dx = 4 du. The integral becomes ∫(4/u) du = 4 ln|u| + C = 4 ln(x² + 9) + C. Since x² + 9 > 0 for all real x, we can drop the absolute value bars. Choice C might tempt students thinking of arctangent formulas, but those apply when the numerator is constant, not when it contains x. When the numerator is a constant multiple of the derivative of the denominator, use ln of the denominator.