AP Calculus BC Quiz: Selecting Techniques For Antidifferentiation

What this quiz covers

This quiz focuses on Selecting Techniques For Antidifferentiation, 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. Partial fraction decomposition
2. Substitution using $u=\sin(2x)$
3. Repeated integration by parts (correct answer)
4. Trigonometric substitution
5. Long division then integrate

Explanation: Selecting an appropriate integration technique requires recognizing products of exponential and trigonometric functions. The integrand $e^{2x}\sin(2x)$ is a product where both functions have the same argument coefficient (2). This structure requires repeated integration by parts, typically applied twice to obtain a system that can be solved for the original integral. Substitution using $u = \sin(2x)$ wouldn't work effectively because the exponential function doesn't integrate to something involving sine. When you have products of exponential and trigonometric functions with matching coefficients, expect to use integration by parts repeatedly until a pattern emerges that allows solving for the original integral.

Question 2

1. Integration by parts (correct answer)
2. u‑substitution
3. Partial fraction decomposition
4. Trigonometric substitution
5. Use a trig identity to rewrite the integrand

Explanation: Selecting the appropriate technique for antidifferentiation is crucial for evaluating integrals efficiently. For the integral ∫ x e^{3x} dx from 0 to 2, integration by parts is most appropriate because the integrand is a product of a polynomial and an exponential function. Set u = x and dv = e^{3x} dx, so du = dx and v = (1/3) e^{3x}, leading to uv - ∫ v du, which simplifies easily. This method reduces the power of x, making the remaining integral straightforward. A tempting distractor like u-substitution might be considered if one mistakes the exponential for a composite function, but it fails because there's no clear inner function whose derivative matches the rest of the integrand. Recognize products of polynomials and exponentials or trig functions as prime candidates for integration by parts.

Question 3

1. Partial fractions
2. Substitution using $u=x^2+x$ (correct answer)
3. Integration by parts
4. Trigonometric substitution
5. Use a Taylor series

Explanation: Selecting an appropriate integration technique requires identifying substitution patterns in radical expressions. The integrand $\frac{2x+1}{\sqrt{x^2+x}}$ suggests substitution because the numerator $2x+1$ is the derivative of the expression under the radical (since $\frac{d}{dx}[x^2+x] = 2x+1$). Using $u = x^2+x$ gives $du = (2x+1)dx$, transforming the integral to $\int u^{-1/2}\,du$. Partial fractions isn't applicable to radical expressions, and trigonometric substitution would be more complex. When the numerator is the derivative of the expression under a radical, substitution using that expression provides the most direct solution path.

Question 4

1. Partial fraction decomposition
2. Substitution using $u=9+x^2$
3. Trigonometric substitution (correct answer)
4. Integration by parts
5. Use a Taylor polynomial

Explanation: Selecting an appropriate integration technique requires recognizing radical expressions with quadratic forms. The integrand $\sqrt{9+x^2}$ has the form $\sqrt{a^2+x^2}$ where $a = 3$. This structure specifically calls for trigonometric substitution using $x = 3\tan\theta$, which transforms the radical into $3\sec\theta$. Substitution using $u = 9+x^2$ wouldn't work because we don't have $2x$ as a factor in the integrand. Integration by parts isn't suitable since we don't have a clear product structure. When encountering $\sqrt{a^2+x^2}$, $\sqrt{a^2-x^2}$, or $\sqrt{x^2-a^2}$, trigonometric substitution is the standard technique for eliminating the radical.

Question 5

1. Integration by parts
2. Substitution using $u=x^2+1$ (correct answer)
3. Partial fractions
4. Trigonometric substitution
5. Use a power series

Explanation: Selecting the right integration technique requires identifying substitution opportunities. The integrand $\frac{4x}{(x^2+1)^3}$ contains $4x$ in the numerator, which is twice the derivative of $x^2+1$ (since $\frac{d}{dx}[x^2+1] = 2x$). Using substitution $u = x^2+1$ gives $du = 2x\,dx$, so $4x\,dx = 2du$. The integral becomes $2\int u^{-3}\,du$, which is straightforward. Trigonometric substitution would be much more complex and unnecessary given this clean substitution. When the numerator is a constant multiple of the derivative of an expression in the denominator, substitution using that expression is the most efficient approach.

Question 6

1. Integration by parts
2. Trigonometric substitution
3. Polynomial long division then integrate (correct answer)
4. Partial fraction decomposition
5. Substitution using $u=x^2+1$

Explanation: Selecting the right integration technique involves recognizing improper rational functions. The integrand $\frac{x^3}{x^2+1}$ has a numerator degree greater than the denominator degree, requiring polynomial long division first. Dividing gives $\frac{x^3}{x^2+1} = x - \frac{x}{x^2+1}$, creating $\int x\,dx - \int \frac{x}{x^2+1}\,dx$ where the second integral uses substitution. Integration by parts would be unnecessarily complex for this rational function structure. When dealing with improper rational functions (numerator degree ≥ denominator degree), always perform polynomial long division before applying other integration techniques.

Question 7

1. Integration by parts (correct answer)
2. Partial fractions
3. Trigonometric substitution
4. Substitution using $u=\arctan x$
5. Long division then integrate

Explanation: Selecting an appropriate integration technique involves recognizing inverse trigonometric integration by parts. The integrand $\arctan x$ is an inverse trigonometric function alone, which requires integration by parts with $u = \arctan x$ and $dv = dx$. This gives $du = \frac{1}{1+x^2}dx$ and $v = x$, leading to $x\arctan x - \int \frac{x}{1+x^2}dx$. The remaining integral is manageable with substitution. Substitution using $u = \arctan x$ alone wouldn't work since we don't have $\frac{1}{1+x^2}$ as a factor. For integrals of inverse trigonometric functions by themselves, integration by parts is the standard approach.

Question 8

1. Trigonometric substitution
2. Integration by parts
3. Partial fraction decomposition (correct answer)
4. Substitution using $u=x^2-9$
5. Use numerical integration

Explanation: Selecting an appropriate integration technique involves recognizing difference of squares patterns. The expression $\frac{1}{x^2-9}$ has a denominator that factors as $(x-3)(x+3)$, indicating partial fraction decomposition is needed. The partial fraction form is $\frac{A}{x-3} + \frac{B}{x+3}$, which leads to logarithmic terms after integration. Trigonometric substitution isn't the most direct approach here since the quadratic is factorable over the reals. When you have a rational function with a denominator that's a difference of squares (or any factorable quadratic), partial fractions typically provides a more straightforward path than trigonometric substitution.

Question 9

1. Substitution using $u=\sqrt{x}$ (correct answer)
2. Integration by parts
3. Partial fraction decomposition
4. Trigonometric substitution
5. Long division then integrate

Explanation: Selecting an appropriate integration technique requires recognizing substitution with radicals involving square roots. The integrand $\frac{\sqrt{x}}{1+x}$ suggests substitution $u = \sqrt{x}$, which gives $x = u^2$ and $dx = 2u\,du$. The integral transforms to $\int \frac{u}{1+u^2} \cdot 2u\,du = 2\int \frac{u^2}{1+u^2}\,du$, which can be handled with polynomial division. Integration by parts would be complex due to the mixed radical and algebraic structure. When dealing with integrands involving square roots in both numerator and a function of the variable, substitution using the square root often simplifies the expression significantly.

Question 10

1. Partial fraction decomposition
2. Integration by parts
3. Trigonometric substitution (correct answer)
4. Substitution using $u=1+4x^2$
5. Use Simpson’s Rule

Explanation: Selecting an appropriate integration technique requires recognizing trigonometric substitution with radicals. The integrand $\sqrt{1+4x^2}$ can be rewritten as $\sqrt{1+(2x)^2}$, which has the form $\sqrt{a^2+u^2}$ where $a = 1$ and $u = 2x$. This suggests trigonometric substitution using $2x = \tan\theta$, which transforms the radical into $\sec\theta$. Partial fractions doesn't apply to radical expressions, and simple substitution wouldn't eliminate the radical effectively. When encountering $\sqrt{a^2+u^2}$ forms, trigonometric substitution with $u = a\tan\theta$ is the standard technique for handling the radical expression.

Question 11

1. Integration by parts
2. Substitution using $u=\tan x$ (correct answer)
3. Trigonometric substitution
4. Partial fractions
5. Use a series expansion

Explanation: Selecting an appropriate integration technique requires identifying substitution opportunities with trigonometric functions. The integrand $\frac{\sec^2 x}{\tan x}$ can be rewritten as $\frac{\sec^2 x}{\tan x}$, where the numerator $\sec^2 x$ is the derivative of $\tan x$. Using substitution $u = \tan x$ gives $du = \sec^2 x\,dx$, transforming the integral to $\int \frac{1}{u}\,du = \ln|u| + C$. Trigonometric substitution isn't needed when a simpler algebraic substitution works. When the numerator is the derivative of the denominator (or a constant multiple), substitution using the denominator expression is the most efficient approach.

Question 12

1. Integration by parts
2. Trigonometric substitution
3. Simple $u$-substitution (correct answer)
4. Partial fraction decomposition
5. Trigonometric identities (power-reduction)

Explanation: This question tests the skill of selecting the appropriate technique for antidifferentiation. The integral $\int t e^{t^2} \, dt$ features a product where t is the derivative of $t^2$, which appears in the exponent, making it ideal for substitution. By letting $u = t^2$, then $du = 2t \, dt$, so $t \, dt = \frac{du}{2}$, transforming the integral into $\frac{1}{2} \int e^u \, du$, which is straightforward. This simple u-substitution directly matches the structure where the integrand is a composite function with its derivative present. A tempting distractor like integration by parts might seem applicable due to the product, but it would complicate things unnecessarily since no reduction formula is needed here. Always look for substitution opportunities when the integrand includes a function and its derivative multiplied together.

Question 13

1. Integration by parts (correct answer)
2. Simple $u$-substitution
3. Partial fraction decomposition
4. Trigonometric substitution
5. Improper integral comparison test

Explanation: This question tests the skill of selecting the appropriate technique for antidifferentiation. The integral ∫ x ln(x) dx is a product of x and ln(x), where ln(x) is not easily substituted but can be handled by assigning u = ln(x) and dv = x dx in integration by parts. This yields du = (1/x) dx and v = (1/2)x², leading to (1/2)x² ln(x) - ∫ (1/2)x dx, which simplifies nicely. Integration by parts is fitting because it reduces the logarithmic term effectively when paired with a polynomial. Simple u-substitution might tempt if one lets u = ln(x), but then du = (1/x) dx lacks the extra x factor, making it ineffective. Recognize products involving logs, inverse trig, or exponentials with polynomials as prime candidates for integration by parts.

Question 14

1. Partial fraction decomposition
2. Use trigonometric identities and substitution for odd/even powers (correct answer)
3. Integration by parts
4. Trigonometric substitution
5. Simple $u$-substitution only (no identities needed)

Explanation: This question tests the skill of selecting the appropriate technique for antidifferentiation. The integral ∫ sin^5(x) cos^2(x) dx has odd power on sin(x) and even on cos(x), so save one sin(x) for du and use the identity sin²(x) = 1 - cos²(x) for the rest. Letting u = cos(x), du = -sin(x) dx transforms it into a polynomial in u, integrable directly. This combines trigonometric identities for power reduction with substitution, matching the structure of odd-even trig power products. Integration by parts might seem appealing for repeated application, but it's more cumbersome than the identity-substitution method here. When integrating powers of sine and cosine, check for odd powers to guide the use of identities and substitution.

Question 15

1. Integration by parts
2. Trigonometric substitution
3. Simple $u$-substitution
4. Multiply by a conjugate and use algebraic simplification (correct answer)
5. Partial fraction decomposition

Explanation: This question tests the skill of selecting the appropriate technique for antidifferentiation. The integral ∫ 1/(1 + sin(x)) dx can be simplified by multiplying numerator and denominator by the conjugate 1 - sin(x), yielding (1 - sin(x))/cos²(x) = sec²(x) - sec(x) tan(x). This algebraic manipulation allows direct integration to tan(x) - sec(x) + C. The technique fits because it rationalizes the trigonometric denominator without substitution. Simple u-substitution might tempt with u = sin(x), but du = cos(x) dx doesn't match the denominator directly, leaving it unresolved. When facing trig functions in denominators, consider multiplying by conjugates or using Weierstrass substitution as algebraic strategies before other methods.

Question 16

1. Partial fraction decomposition
2. Trigonometric substitution (correct answer)
3. Integration by parts
4. u-substitution with $u=1-x^2$
5. Long division then integrate

Explanation: This problem requires selecting the technique for ∫₀¹ 1/√(1-x²) dx, which involves recognizing a specific antidifferentiation pattern. The integrand 1/√(1-x²) is the derivative of arcsin(x), making this integral evaluate directly to arcsin(x) from 0 to 1, which equals π/2. However, if we didn't recognize this standard form, trigonometric substitution with x = sin(θ) would transform √(1-x²) into √(1-sin²θ) = cos(θ), converting the integral to ∫dθ = θ + C = arcsin(x) + C. The substitution u = 1-x² (option D) would fail because du = -2x dx, and we don't have an x in the numerator to absorb. The recognition strategy is that expressions of the form √(a²-x²), √(x²+a²), or √(x²-a²) typically call for trigonometric substitution, with the specific substitution determined by the sign pattern.

Question 17

1. Integration by parts (correct answer)
2. u-substitution
3. Trigonometric substitution
4. Partial fraction decomposition
5. Use a trig identity then integrate directly

Explanation: This problem asks for the best technique to evaluate ∫(x²+1)eˣ dx, which requires selecting an appropriate antidifferentiation method. The integrand is a product of a polynomial (x²+1) and an exponential function eˣ, which is the classic structure for integration by parts. Using the LIATE hierarchy (Logarithmic, Inverse trig, Algebraic, Trig, Exponential), we choose u = x²+1 (algebraic) and dv = eˣ dx (exponential). This requires applying integration by parts twice: first to reduce x²+1 to 2x, then again to reduce 2x to 2, ultimately yielding (x²-2x+3)eˣ + C. U-substitution (option B) fails here because there's no composite function structure where one part is the derivative of another. The recognition principle is that products of different function types, especially polynomial times exponential or trigonometric, typically require integration by parts.

Question 18

1. Use a trigonometric identity, then integrate directly
2. Integration by parts
3. Trigonometric substitution
4. Partial fraction decomposition (correct answer)
5. Reverse chain rule (u-substitution)

Explanation: This problem requires choosing a technique for a rational function. The integrand $\frac{2x^2+3x-1}{x^3-x}$ is a proper rational function since the degree of the numerator is less than the degree of the denominator. The denominator factors as $x(x^2-1) = x(x-1)(x+1)$, giving us three distinct linear factors. This structure is ideal for partial fraction decomposition, where we express the fraction as $\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$ and solve for the constants. Trigonometric identities are irrelevant here since we have no trigonometric functions present. When dealing with rational functions where the denominator can be factored, partial fraction decomposition is the standard approach.

Question 19

1. Integration by parts (correct answer)
2. Reverse chain rule (u-substitution)
3. Partial fraction decomposition
4. Trigonometric substitution
5. Recognize as an improper integral and use limits only

Explanation: This problem asks for the best technique to evaluate an integral with a product structure. The integrand $tln(t)$ consists of a polynomial factor $t$ and a logarithmic factor $ln(t)$, which is a classic setup for integration by parts. Using the LIATE hierarchy (Logarithmic, Inverse trig, Algebraic, Trig, Exponential), we choose $u = ln(t)$ and $dv = t,dt$, which leads to a simpler integral after applying the parts formula. U-substitution fails here because neither factor is the derivative of the other, making it impossible to find a suitable substitution. When you encounter products involving logarithms multiplied by polynomials, integration by parts is typically the most effective approach.

Question 20

1. Partial fraction decomposition
2. Integration by parts (correct answer)
3. Trigonometric substitution
4. Use a trigonometric identity, then integrate directly
5. Recognize as a geometric series and integrate term-by-term

Explanation: This problem asks for the appropriate technique for integrating a quotient involving a logarithm. The integrand $\frac{\ln x}{x^2}$ can be rewritten as $x^{-2}\ln x$, which is a product of a power function and a logarithmic function. Following the LIATE hierarchy for integration by parts, we set $u = \ln x$ (giving $du = \frac{1}{x}dx$) and $dv = x^{-2}dx$ (giving $v = -x^{-1}$). Partial fraction decomposition is not applicable here since we don't have a rational function with a factorable denominator. When integrating products involving logarithms and powers of x, integration by parts is the standard technique.