Calculus 2 Quiz: Selecting Integration Techniques

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Selecting Integration TechniquesQuestion 1 of 20

Which trigonometric substitution is appropriate for the integral $\int \frac{dx}{(9-4x^2)^{3/2}}$ ?

2x = 3\sin\theta

x = 3\sin\theta

2x = 3\tan\theta

4x^2 = 9\sec\theta

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Calculus 2 Quiz

Calculus 2 Quiz: Selecting Integration Techniques

Practice Selecting Integration Techniques in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Selecting Integration Techniques, 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.

All questions

Question 1

Which trigonometric substitution is appropriate for the integral $\int \frac{dx}{(9-4x^2)^{3/2}}$ ?

$2x = 3\sin\theta$ (correct answer)
$x = 3\sin\theta$
$2x = 3\tan\theta$
$4x^2 = 9\sec\theta$

Explanation: The expression

9-4x^2

is of the form

a^2-u^2

, where

a^2=9

(so

a=3

) and

u^2=4x^2

(so

u=2x

). The correct trigonometric substitution for this form is

u = a\sin\theta

, which translates to

2x = 3\sin\theta

Question 2

What is the correct form of the partial fraction decomposition for the expression $\frac{3x^2 - 5x + 1}{(x-1)^2(x^2+4)}$ ?

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

Explanation: The denominator has a repeated linear factor

(x-1)^2

and an irreducible quadratic factor

(x^2+4)

. The repeated linear factor requires a term for each power,

\frac{A}{x-1}

and

\frac{B}{(x-1)^2}

. The irreducible quadratic factor requires a linear numerator,

\frac{Cx+D}{x^2+4}

. Combining these gives the correct form.

Question 3

Consider the integral $\int \frac{dx}{\sqrt{x}(x+1)}$ . Which substitution is most effective for evaluating this integral?

A rationalizing substitution $u = \sqrt{x}$ , which transforms the integrand into a rational function of $u$ . (correct answer)
Integration by parts with $u = \frac{1}{\sqrt{x}}$ and $dv = \frac{1}{x+1} dx$ .
A trigonometric substitution $x = \tan^2\theta$ , motivated by the $x+1$ term.
Partial fraction decomposition applied directly to the original integrand.

Explanation: Letting

u = \sqrt{x}

means

u^2 = x

and

2u du = dx

. Substituting these into the integral yields

\int \frac{2u du}{u(u^2+1)} = \int \frac{2}{u^2+1} du

. This is a standard arctangent integral. This substitution is far more direct than the others.

Question 4

A student needs to evaluate $\int x^2 \sqrt{9-x^2} dx$ . After trying integration by parts with $u = x^2$ and $dv = \sqrt{9-x^2} dx$ , they realize this approach leads to a more complex integral. What is the most efficient alternative strategy?

Use trigonometric substitution $x = 3\sin\theta$ , then apply integration by parts (correct answer)
Use the substitution $u = 9-x^2$ , then integrate by parts
Split the integral as $\int x \cdot x\sqrt{9-x^2} dx$ and use integration by parts
Use trigonometric substitution $x = 3\tan\theta$ , then simplify

Explanation: The presence of

\sqrt{9-x^2}

suggests trigonometric substitution with

x = 3\sin\theta

, which transforms the square root to

3\cos\theta

. After substitution, the integral becomes

\int (3\sin\theta)^2 \cdot 3\cos\theta \cdot 3\cos\theta d\theta = 81\int \sin^2\theta \cos^2\theta d\theta

. This still requires integration by parts or trigonometric identities. Choice B with

u = 9-x^2

doesn't handle the

x^2

factor effectively. Choice C doesn't fundamentally change the complexity. Choice D uses the wrong trigonometric substitution for

\sqrt{9-x^2}

Question 5

For the integral $\int \frac{dx}{x \ln(x) \ln(\ln x)}$ , which technique should be applied?

A single u-substitution with $u = \ln(\ln x)$ . (correct answer)
Integration by parts with $u = \frac{1}{\ln(\ln x)}$ and $dv = \frac{dx}{x \ln x}$ .
Partial fraction decomposition after a substitution.
Repeated u-substitutions, starting with $u = x \ln(x)$ .

Explanation: This integral requires recognizing a chain of derivatives. Let

u = \ln(\ln x)

. Using the chain rule,

du = \frac{1}{\ln x} \cdot \frac{1}{x} dx

. The integral perfectly transforms into

\int \frac{du}{u}

, which is

\ln|u| + C

. This single substitution resolves the entire integral.

Question 6

An attempt to integrate $\int \frac{e^x}{e^{2x}-e^x-2} dx$ begins with a substitution. Which substitution most effectively simplifies the integral into a form that can be solved with partial fractions?

$u = e^{2x}$
$u = e^x - 2$
$u = e^x$ (correct answer)
$u = e^{2x}-e^x-2$

Explanation: Let

u = e^x

. Then

du = e^x dx

and

u^2 = e^{2x}

. The integral transforms into

\int \frac{du}{u^2-u-2}

. The denominator factors as

(u-2)(u+1)

, making it a standard partial fraction decomposition problem. The other substitutions are not as effective.

Question 7

Which of the following integrals is most appropriately solved using integration by parts, as opposed to a u-substitution?

$\int x \sin(x^2) dx$
$\int x^2 \sin(x) dx$ (correct answer)
$\int \frac{\sin(\sqrt{x})}{\sqrt{x}} dx$
$\int \sin(x) \cos^3(x) dx$

Explanation: The integral

\int x^2 \sin(x) dx

requires integration by parts (specifically, two applications). The other integrals are all solvable with a direct u-substitution: (A) let

u=x^2

; (C) let

u=\sqrt{x}

; (D) let

u=\cos(x)

Question 8

For the integral $\int x^2 \sin(x^3+1) dx$ , a student correctly identifies that substitution should be used, but then struggles with the setup. What is the key insight for choosing the substitution variable?

Let $u = x^3$ because it's the highest degree term and simplifies to $du = 3x^2 dx$
Let $u = x^2$ because it's the polynomial factor not inside the trigonometric function
Let $u = \sin(x^3+1)$ to eliminate the trigonometric function directly
Let $u = x^3+1$ because $du = 3x^2 dx$ , and $x^2 dx$ appears in the integrand (correct answer)

Explanation: When you encounter an integral with a composition of functions like this, the key to u-substitution is finding a substitution where the derivative of your u-variable already appears (or can easily be made to appear) elsewhere in the integrand. Looking at

\int x^2 \sin(x^3+1) dx

, you should scan for patterns where the "inside" of a composite function has its derivative present. Here,

x^3+1

is inside the sine function, and its derivative is

3x^2

. Since

x^2

appears as a factor in the integrand, this creates a perfect match—you just need to account for the constant factor of 3. Setting

u = x^3+1

gives

du = 3x^2 dx

, which means

x^2 dx = \frac{1}{3}du

. The integral becomes

\frac{1}{3}\int \sin(u) du = -\frac{1}{3}\cos(u) + C = -\frac{1}{3}\cos(x^3+1) + C

. Answer A chooses

u = x^3

, missing the "+1" and creating unnecessary complications when substituting back into the sine function. Answer B selects

u = x^2

, which doesn't help simplify the

\sin(x^3+1)

portion at all. Answer C attempts

u = \sin(x^3+1)

, but this creates

du = \cos(x^3+1) \cdot 3x^2 dx

, introducing cosine where you don't want it. Strategy tip: In u-substitution, always look for the "inside function" of a composition, then check if its derivative appears elsewhere in the integrand. This pattern—derivative of the inside function present as a factor—signals the right substitution choice.

Question 9

A student needs to integrate $\int \frac{e^x}{e^{2x} + e^x + 1} dx$ and considers several approaches. Which technique would lead to the most straightforward evaluation?

Factor out $e^x$ from both numerator and denominator, then use partial fractions
Use the substitution $u = e^x$ , converting to a rational function in $u$ (correct answer)
Multiply numerator and denominator by $e^{-x}$ , then integrate the resulting expression
Use integration by parts with $u = \frac{1}{e^{2x} + e^x + 1}$ and $dv = e^x dx$

Explanation: With

u = e^x

, we have

du = e^x dx

, so the integral becomes

\int \frac{1}{u^2 + u + 1} du

. The denominator

u^2 + u + 1

can be completed to

(u + \frac{1}{2})^2 + \frac{3}{4}

, leading to an arctangent form:

\frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right) + C = \frac{2}{\sqrt{3}}\arctan\left(\frac{2e^x+1}{\sqrt{3}}\right) + C

. Choice A doesn't work because we can't factor

e^x

from the constant term 1. Choice C gives

\int \frac{1}{e^x + 1 + e^{-x}} dx

, which is more complex. Choice D makes

dv

integration much harder than the original problem.

Question 10

Consider $\int \frac{x^3}{\sqrt{x^2-4}} dx$ where $x > 2$ . A student attempts trigonometric substitution but realizes that the $x^3$ factor complicates the approach. What strategy would be most efficient?

Use trigonometric substitution $x = 2\sec\theta$ , then handle the resulting $\sec^3\theta$ integral
Rewrite as $\int \frac{x^2 \cdot x}{\sqrt{x^2-4}} dx$ and use integration by parts with $u = x^2$
Split as $\int x \cdot \frac{x^2}{\sqrt{x^2-4}} dx$ and use substitution $u = x^2-4$ after algebraic manipulation (correct answer)
Factor out $x^2$ to get $\int x^2 \cdot \frac{x}{\sqrt{x^2-4}} dx$ , then use substitution $u = \sqrt{x^2-4}$

Explanation: The most efficient approach is to rewrite

x^2 = (x^2-4) + 4

, so

\frac{x^3}{\sqrt{x^2-4}} = \frac{x(x^2-4)}{\sqrt{x^2-4}} + \frac{4x}{\sqrt{x^2-4}} = x\sqrt{x^2-4} + \frac{4x}{\sqrt{x^2-4}}

. The first term integrates using

u = x^2-4

, and the second is a standard form. Choice A works but leads to a complex

\sec^3\theta

integral requiring reduction formulas. Choice B with integration by parts creates a more complex integral. Choice D's substitution

u = \sqrt{x^2-4}

gives

du = \frac{x dx}{\sqrt{x^2-4}}

, but doesn't handle the full

x^3

efficiently.

Question 11

Consider the integral $\int e^{2x} \cos(3x) dx$ . If a student applies integration by parts twice and returns to the original integral with coefficient $k$ , what technique should be used to complete the evaluation?

Set up an algebraic equation using the coefficient $k$ and solve for the original integral (correct answer)
Apply integration by parts a third time with a different choice of $u$ and $dv$
Use the substitution $u = 2x$ to simplify before continuing with integration by parts
Convert to complex exponentials using Euler's formula, then integrate directly

Explanation: When integrating

\int e^{ax}\cos(bx)dx

by parts twice, you return to the original integral with coefficient

-(a^2+b^2)

. Here, after two applications, you get

I = \text{(expression)} - 13I

where

I

is the original integral. Solving algebraically:

I + 13I = \text{(expression)}

, so

14I = \text{(expression)}

and

I = \frac{\text{(expression)}}{14}

. Choice B leads to infinite recursion. Choice C doesn't address the fundamental structure. Choice D works but is unnecessarily complex.

Question 12

For $\int \frac{x}{\sqrt{x-1}} dx$ where $x > 1$ , two students propose different substitution strategies. Student A uses $u = \sqrt{x-1}$ while Student B uses $u = x-1$ . Which analysis of their approaches is correct?

Both substitutions work equally well and lead to the same computational complexity
Student A's substitution is more efficient because it directly eliminates the square root (correct answer)
Student B's substitution is more efficient because it avoids fractional exponents in the result
Student A's substitution works but Student B's requires additional manipulation to handle the square root

Explanation: Student A:

u = \sqrt{x-1}

gives

x = u^2 + 1

and

dx = 2u du

. The integral becomes

\int \frac{u^2+1}{u} \cdot 2u du = \int 2(u^2+1) du = \frac{2u^3}{3} + 2u + C

. Student B:

u = x-1

gives

x = u+1

and

dx = du

. The integral becomes

\int \frac{u+1}{\sqrt{u}} du = \int (u^{1/2} + u^{-1/2}) du = \frac{2u^{3/2}}{3} + 2u^{1/2} + C

. While both work, Student A's approach avoids fractional exponents throughout the integration process and is more direct. Student B's method requires integrating fractional powers, which is slightly more prone to errors.

Question 13

To evaluate $\int \frac{x^3 + 2x^2 + 3x + 1}{x^2(x+1)^2} dx$ , which sequence of integration techniques should be applied?

Partial fraction decomposition, then integration by parts for one specific term (correct answer)
Partial fraction decomposition, then basic antiderivative rules only
Integration by parts first, then partial fraction decomposition afterwards
Trigonometric substitution, then partial fraction decomposition methods

Explanation: Since the degree of the numerator equals the degree of the denominator, we first perform polynomial long division, obtaining quotient 1 and remainder

x+1

. Then we apply partial fraction decomposition to

\frac{x+1}{x^2(x+1)^2} = \frac{1}{x^2(x+1)}

. The decomposition yields terms

\frac{A}{x}

\frac{B}{x^2}

, and

\frac{C}{x+1}

. While most terms integrate directly using basic rules, careful handling of the rational function structure may require recognizing certain derivative forms, making this more complex than pure basic integration.

Question 14

To evaluate $\int \frac{\sqrt{x}}{\sqrt{x}+\sqrt[3]{x}} dx$ , which substitution would be most effective in simplifying the integrand?

$u = \sqrt{x}$ , which eliminates the square root but complicates the cube root
$u = \sqrt[3]{x}$ , which eliminates the cube root but complicates the square root
$u = \sqrt[6]{x}$ , which handles both roots simultaneously using a common denominator (correct answer)
$u = \sqrt{x} + \sqrt[3]{x}$ , which directly substitutes the denominator expression

Explanation: The key insight is that

\sqrt{x} = x^{1/2} = (x^{1/6})^3 = u^3

and

\sqrt[3]{x} = x^{1/3} = (x^{1/6})^2 = u^2

when

u = x^{1/6}

. This transforms the integral to

\int \frac{u^3}{u^3+u^2} \cdot 6u^5 du = \int \frac{6u^8}{u^2(u+1)} du = \int \frac{6u^6}{u+1} du

, which can be handled by polynomial long division. Choice A gives

\sqrt[3]{x} = u^{2/3}

, creating fractional exponents. Choice B gives

\sqrt{x} = u^{3/2}

, also fractional. Choice D makes

du

very complicated to express in terms of

x

Question 15

The integral $\int x^5 \sqrt{1+x^3} dx$ is best solved using a substitution that simplifies the radical. Which approach is most effective?

A u-substitution with $u=1+x^3$ , after rewriting $x^5$ as $x^2 \cdot x^3$ . (correct answer)
Integration by parts with $u=x^5$ and $dv = \sqrt{1+x^3} dx$ .
A trigonometric substitution after recognizing a sum of squares.
A u-substitution with $u=x^3$ , which simplifies the integrand but leaves a more complex radical.

Explanation: Let

u = 1+x^3

. Then

du = 3x^2 dx

, or

\frac{1}{3}du = x^2 dx

. We also have

x^3 = u-1

. We can rewrite the integral as

\int x^3 \sqrt{1+x^3} \cdot x^2 dx

. Substituting gives

\frac{1}{3} \int (u-1)\sqrt{u} du

, which can be easily solved by distributing

\sqrt{u}

Question 16

Which of the following describes the most significant challenge when selecting a technique for the improper integral $\int_1^\infty \frac{\ln x}{x^2} dx$ ?

Determining the correct limit expression $\lim_{b \to \infty}$ to define the improper integral.
Finding the antiderivative of $\frac{\ln x}{x^2}$ , which requires integration by parts. (correct answer)
Evaluating the antiderivative at the lower bound of integration, $x=1$ .
Deciding whether to use the comparison test versus direct evaluation to determine convergence.

Explanation: While setting up the limit (A) is the definition of an improper integral, the core calculus-2 skill tested is finding the antiderivative. The function

\frac{\ln x}{x^2}

is not a basic integral and requires integration by parts (with

u=\ln x

and

dv=x^{-2}dx

), which is the main computational step. The other steps are relatively straightforward once the antiderivative is known.

Question 17

Which of the following integrals is a candidate for partial fraction decomposition without requiring an initial step of polynomial long division?

$\int \frac{x^3}{x^2-1} dx$
$\int \frac{2x^2+x-1}{x^2+1} dx$
$\int \frac{x-5}{x^2+x-6} dx$ (correct answer)
$\int \frac{x^2}{(x-1)^2} dx$

Explanation: Partial fraction decomposition can be applied directly only when the degree of the numerator is strictly less than the degree of the denominator. In choice (C), the numerator has degree 1 and the denominator has degree 2, so it is a proper rational function. Choices (A), (B), and (D) are all improper rational functions where the degree of the numerator is greater than or equal to the degree of the denominator, so they require long division first.

Question 18

The integral $\int \frac{1}{x^2-4x+3} dx$ and the integral $\int \frac{1}{x^2-4x+5} dx$ require different integration techniques. What is the key difference?

The first requires partial fractions because its denominator factors, while the second requires completing the square for an arctangent form. (correct answer)
The first requires completing the square for an inverse hyperbolic function, while the second requires partial fractions.
Both integrals can be solved using the same u-substitution, but with different results.
The first requires integration by parts, while the second requires a trigonometric substitution.

Explanation: The key is the nature of the quadratic denominator. For

\int \frac{1}{x^2-4x+3} dx

, the denominator factors into

(x-1)(x-3)

, making it suitable for partial fraction decomposition. For

\int \frac{1}{x^2-4x+5} dx

, the discriminant is

(-4)^2 - 4(1)(5) = -4 < 0

, so it is an irreducible quadratic. The correct technique is to complete the square,

(x-2)^2 + 1

, which leads to an arctangent integral.

Question 19

To evaluate $\int \frac{\ln(x+1)}{x^2} dx$ , what is the most promising first step?

A u-substitution with $u=x+1$ , which simplifies the argument of the logarithm.
Integration by parts with $u = \ln(x+1)$ and $dv = \frac{1}{x^2} dx$ . (correct answer)
Partial fraction decomposition after attempting to rewrite the integrand.
Integration by parts with $u = \frac{1}{x^2}$ and $dv = \ln(x+1) dx$ .

Explanation: This integral is a classic case for integration by parts. The LIATE (Logarithmic, Inverse Trig, Algebraic, Trig, Exponential) mnemonic suggests choosing the logarithmic function as 'u'. So, let

u = \ln(x+1)

and

dv = x^{-2} dx

. Then

du = \frac{1}{x+1} dx

and

v = -x^{-1}

. The new integral,

\int v du

, becomes

\int -\frac{1}{x} \frac{1}{x+1} dx

, which can be solved with partial fractions. Choosing

dv = \ln(x+1) dx

(Choice D) is not practical as its integral is not immediately known.

Question 20

For the integral $\int \tan^4(x) dx$ , which of the following strategies is the correct first move?

Use integration by parts with $u = \tan^4(x)$ and $dv = dx$ .
Rewrite the integrand as $\int \frac{\sin^4(x)}{\cos^4(x)} dx$ and use substitutions for sine and cosine.
Split the integrand into $\tan^2(x) \cdot \tan^2(x)$ and use the identity $\tan^2(x) = \sec^2(x) - 1$ . (correct answer)
Apply the substitution $u = \tan(x)$ , which transforms the integral into $\int u^4 \frac{du}{1+u^2}$ .

Explanation: The standard technique for integrating even powers of tangent is to split off a

\tan^2(x)

and replace it with

\sec^2(x) - 1

. This gives

\int \tan^2(x)(\sec^2(x) - 1) dx = \int \tan^2(x)\sec^2(x) dx - \int \tan^2(x) dx

. The first integral can be solved with

u=\tan(x)

, and the process is repeated on the second integral.