Calculus 2 Quiz: Choosing Integration Methods

What this quiz covers

This quiz focuses on Choosing Integration Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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. $\frac{A}{x-1} + \frac{B}{x^2+4}$
2. $\frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-2}$
3. $\frac{A}{x-1} + \frac{Bx+C}{x^2+4}$ (correct answer)
4. $\frac{Ax+B}{x-1} + \frac{C}{x^2+4}$

Question 2

1. $\frac{A}{x} + \frac{B}{x-1}$
2. $\frac{A}{x} + \frac{Bx+C}{(x-1)^2}$
3. $\frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$ (correct answer)
4. $\frac{A}{x} + \frac{B}{(x-1)^2}$

Question 3

1. Integration by parts, as it applies to any product of functions.
2. Trigonometric substitution, as it can handle any quadratic expression.
3. Partial fraction decomposition, as it is a systematic algorithm for all rational functions.
4. U-substitution, as it is the reverse of the chain rule and can simplify compositions of functions. (correct answer)

Question 4

1. $\int \frac{x+1}{(x-1)(x+2)} dx$ (correct answer)
2. $\int \frac{x^2+1}{x+1} dx$
3. $\int \frac{x+1}{x^2+1} dx$
4. $\int \frac{\sqrt{x}}{x-1} dx$

Question 5

1. Integration by parts with $u = x^2$ and $dv = e^{-x^2}dx$
2. Substitution $u = -x^2$ followed by integration by parts
3. Integration by parts with $u = x$ and $dv = xe^{-x^2}dx$ (correct answer)
4. Series expansion of $e^{-x^2}$ followed by term-by-term integration

Question 6

1. Substitution $u = \ln(x^2 + 1)$ followed by polynomial integration
2. Integration by parts with $u = \ln(x^2 + 1)$ and $dv = x^3 dx$ (correct answer)
3. Substitution $u = x^2 + 1$ to eliminate the logarithm
4. Partial fraction decomposition after expanding $\ln(x^2 + 1)$

Question 7

1. Integration by parts twice, first with $u = x^2$ and $dv = e^{x^3}\cos(e^{x^3})dx$
2. Substitution $u = x^3$, followed by integration by parts
3. Substitution $u = e^{x^3}$, then applying trigonometric integration (correct answer)
4. Direct application of the product rule for integration

Question 8

1. Use the power-reducing identities for $\sin^2(x)$ and $\cos^2(x)$ after rewriting the integrand using a double angle identity for sine. (correct answer)
2. Use the substitution $u = \sin(x)$, which transforms the integral to $\int u^2 \sqrt{1-u^2} du$ to be solved with trigonometric substitution.
3. Use integration by parts with $u = \sin^2(x)$ and $dv = \cos^2(x)dx$, which requires a second application of parts.
4. Rewrite the integrand as $\sin^2(x)(1-\sin^2(x))$ and distribute, then use power-reducing identities on $\sin^2(x)$ and $\sin^4(x)$.

Question 9

1. First, integration by parts with $u = e^{\sqrt{x}}$, then a u-substitution.
2. First, a u-substitution with $u = \sqrt{x}$, then integration by parts. (correct answer)
3. Only a single application of a u-substitution with $u = e^{\sqrt{x}}$.
4. Only a single application of integration by parts with $u=e^{\sqrt{x}}$ and $dv=dx$.

Question 10

1. Factor the denominator as $(x-1)(x+1)$ and apply partial fraction decomposition directly.
2. Perform polynomial long division because the degree of the numerator is greater than the degree of the denominator. (correct answer)
3. Use the substitution $u = x^2-1$ to simplify the denominator.
4. Apply trigonometric substitution by letting $x=\sec(\theta)$ since the denominator has the form $x^2-a^2$.

Question 11

1. $u = 1$ and $dv = \arctan(4x) dx$, because integrating $\arctan(4x)$ is the goal.
2. $u = dx$ and $dv = \arctan(4x)$, as this is the standard setup for inverse trigonometric functions.
3. $u = \arctan(4x)$ and $dv = dx$, because the derivative of $\arctan(4x)$ is a rational function. (correct answer)
4. This integral should be solved with trigonometric substitution, not integration by parts.

Question 12

1. Let $u = \sec(x)$, after reserving a $\sec(x)\tan(x)$ factor for $du$.
2. Let $u = \tan(x)$, after reserving a $\sec^2(x)$ factor for $du$. (correct answer)
3. Let $u = \sec^2(x)$, which simplifies the highest power term in the integrand.
4. Convert all terms to sines and cosines, resulting in $\int \frac{\sin^3(x)}{\cos^7(x)} dx$.

Question 13

1. Apply integration by parts three times, with $u$ as the polynomial term each time.
2. Recognize the integrand is an odd function and the interval of integration is symmetric about zero. (correct answer)
3. Use the substitution $u=x^3$ to simplify the polynomial part of the integrand.
4. Use a trigonometric identity to rewrite $\cos(x)$ before attempting integration by parts.

Question 14

1. $I$ is solved with u-substitution and $J$ is solved with integration by parts. (correct answer)
2. $I$ is solved with integration by parts and $J$ is solved with u-substitution.
3. Both integrals are most efficiently solved using integration by parts.
4. Both integrals are most efficiently solved using u-substitution.

Question 15

1. $\int x e^x dx$
2. $\int \ln(x) dx$
3. $\int x \sin(x^2) dx$ (correct answer)
4. $\int x^2 \cos(x) dx$

Question 16

1. Apply integration by parts a second time to the new integral.
2. Use the identity $\tan^2(x) = \sec^2(x) - 1$ on the new integral and solve for $\int \sec^3(x) dx$ algebraically. (correct answer)
3. Use the substitution $u=\tan(x)$ on the new integral.
4. Conclude that the method has failed because it is circular and restart with a different technique.

Question 17

1. Integration by parts with $u = \ln(x)$ and $dv = \frac{1}{x} dx$.
2. Integration by parts with $u = \frac{1}{\ln(x)}$ and $dv = \frac{1}{x} dx$.
3. Partial fraction decomposition.
4. U-substitution with $u = \ln(x)$. (correct answer)

Question 18

1. $\int \sec(\theta) d\theta$
2. $\int \sec^2(\theta) d\theta$
3. $\int \sec^3(\theta) d\theta$ (correct answer)
4. $\int \tan^2(\theta) \sec(\theta) d\theta$

Question 19

1. Integration by parts with $u = e^x$ and $dv = \frac{1}{e^{2x}+1} dx$.
2. Partial fraction decomposition by factoring the denominator.
3. Trigonometric substitution by letting $e^x = \tan(\theta)$.
4. U-substitution by letting $u = e^x$. (correct answer)

Question 20

1. Trigonometric substitution with $x = \tan(\theta)$, which transforms the integral into $\int \tan^3(\theta) \sec^3(\theta) d\theta$.
2. Integration by parts with $u = \sqrt{x^2+1}$ and $dv = x^3 dx$.
3. A u-substitution with $u = x^2+1$ after rewriting the integrand as $x^2 \sqrt{x^2+1} \cdot x$. (correct answer)
4. Integration by parts with $u = x^3$ and $dv = \sqrt{x^2+1} dx$.