Calculus 1 Quiz: Integration By Substitution

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Integration By SubstitutionQuestion 1 of 20

The substitution $u = 3x+1$ is used to find $\int \frac{x}{(3x+1)^3} \,dx$ . Which of the following is the correct antiderivative?

\frac{x^2/2}{(3x+1)^4/4} + C

-\frac{1}{3(3x+1)} + \frac{1}{6(3x+1)^2} + C

\frac{1}{9} \ln|3x+1| + \frac{1}{18(3x+1)^2} + C

-\frac{1}{9(3x+1)} + \frac{1}{18(3x+1)^2} + C

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Calculus 1 Quiz

Calculus 1 Quiz: Integration By Substitution

Practice Integration By Substitution in Calculus 1 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 Integration By Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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.

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Question 1

The substitution $u = 3x+1$ is used to find $\int \frac{x}{(3x+1)^3} \,dx$ . Which of the following is the correct antiderivative?

$\frac{x^2/2}{(3x+1)^4/4} + C$
$-\frac{1}{3(3x+1)} + \frac{1}{6(3x+1)^2} + C$
$\frac{1}{9} \ln|3x+1| + \frac{1}{18(3x+1)^2} + C$
$-\frac{1}{9(3x+1)} + \frac{1}{18(3x+1)^2} + C$ (correct answer)

Explanation: When you encounter an integral with a polynomial in the numerator and a more complex expression in the denominator, substitution often helps simplify the problem. The key insight here is that after substitution, you'll need to express everything in terms of the new variable. Let's work through this substitution systematically. With

u = 3x + 1

, we have

du = 3dx

, so

dx = \frac{du}{3}

. Also,

x = \frac{u-1}{3}

. Substituting into our integral:

\int \frac{x}{(3x+1)^3} dx = \int \frac{\frac{u-1}{3}}{u^3} \cdot \frac{du}{3} = \frac{1}{9} \int \frac{u-1}{u^3} du

This splits into:

\frac{1}{9} \int (u^{-2} - u^{-3}) du = \frac{1}{9}(-u^{-1} + \frac{u^{-2}}{2}) + C

Converting back to

x

-\frac{1}{9(3x+1)} + \frac{1}{18(3x+1)^2} + C

Choice A incorrectly applies integration rules without proper substitution. Choice B has the wrong coefficients—the denominators don't match our calculation. Choice C mistakenly introduces a logarithm, which would only appear if we had

u^{-1}

in our integral, but we have

u^{-2}

and

u^{-3}

terms. Choice D matches our result exactly. Study tip: When using substitution with rational functions, always split complex fractions into simpler terms before integrating. Double-check your coefficients by differentiating your final answer—it should give you the original integrand.

Question 2

Evaluate the indefinite integral $\int \frac{x+1}{x^2+1} dx$ .

$\ln(x^2+1) + \arctan(x) + C$
$\frac{1}{2}\ln(x^2+1) + \arctan(x) + C$ (correct answer)
$2\ln(x^2+1) + \arctan(x) + C$
$\frac{1}{2}\ln(x^2+1) - \frac{1}{x^2+1} + C$

Explanation: When you encounter a rational function where the numerator has multiple terms, the key strategy is to split the integral into separate, more manageable pieces. This allows you to use different integration techniques for each part. Split the fraction:

\int \frac{x+1}{x^2+1} dx = \int \frac{x}{x^2+1} dx + \int \frac{1}{x^2+1} dx

For the first integral

\int \frac{x}{x^2+1} dx

, use substitution with

u = x^2+1

, so

du = 2x dx

, which means

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

. This gives you

\int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+1)

. For the second integral

\int \frac{1}{x^2+1} dx

, this is a standard form that equals

\arctan(x)

. Combining both results:

\frac{1}{2}\ln(x^2+1) + \arctan(x) + C

, which is answer B. Answer A is incorrect because it's missing the factor of

\frac{1}{2}

in front of the logarithm—a common error when students forget to account for the substitution factor. Answer C has the wrong coefficient (2 instead of

\frac{1}{2}

), likely from incorrectly handling the substitution. Answer D incorrectly integrates the second term as

-\frac{1}{x^2+1}

instead of recognizing the standard

\arctan(x)

form. Remember: when splitting rational functions, always double-check your substitution factors, and memorize that

\int \frac{1}{x^2+1} dx = \arctan(x) + C

Question 3

Evaluate $\int_0^{\ln(2)} \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$ .

$\ln\left(\frac{3}{4}\right)$
$\ln\left(\frac{3}{2}\right)$
$\ln\left(\frac{5}{2}\right)$
$\ln\left(\frac{5}{4}\right)$ (correct answer)

Explanation: When you encounter an integral with exponential expressions like

\frac{e^x - e^{-x}}{e^x + e^{-x}}

, look for substitution opportunities. This particular form should remind you of hyperbolic functions or suggest a u-substitution. Let's use substitution with

u = e^x + e^{-x}

. Then

du = (e^x - e^{-x})dx

, which is exactly our numerator! This transforms our integral into

\int \frac{1}{u} du = \ln|u| + C

. Substituting back:

\int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx = \ln|e^x + e^{-x}| + C

Now evaluate the definite integral from 0 to

\ln(2)

: At

x = \ln(2)

e^{\ln(2)} + e^{-\ln(2)} = 2 + \frac{1}{2} = \frac{5}{2}

x = 0

e^0 + e^0 = 1 + 1 = 2

Therefore:

\ln\left(\frac{5}{2}\right) - \ln(2) = \ln\left(\frac{5/2}{2}\right) = \ln\left(\frac{5}{4}\right)

Choice A gives

\ln\left(\frac{3}{4}\right)

, likely from incorrectly calculating the bounds. Choice B gives

\ln\left(\frac{3}{2}\right)

, possibly from an error in the upper bound calculation. Choice C gives

\ln\left(\frac{5}{2}\right)

, which occurs if you forget to subtract the lower bound evaluation—a common mistake in definite integrals. The answer is D. Key strategy: When you see rational expressions involving

e^x

and

e^{-x}

, check if the numerator is the derivative of the denominator. This often leads to a clean logarithmic antiderivative.

Question 4

Evaluate the indefinite integral $\int \frac{\sin(\ln x)}{x \cos^2(\ln x)} dx$ .

$\ln|\cos(\ln x)| + C$
$\tan(\ln x) + C$
$-\sec(\ln x) + C$
$\sec(\ln x) + C$ (correct answer)

Explanation: When you encounter an integral with

\ln x

inside trigonometric functions, substitution is your key strategy. The presence of

\frac{1}{x}

in the integrand is a strong hint that

u = \ln x

will work perfectly. Let

u = \ln x

, so

du = \frac{1}{x}dx

. This transforms our integral into:

\int \frac{\sin u}{\cos^2 u} du

Now you can recognize this as

\int \sin u \cdot \sec^2 u \, du

. To evaluate this, use integration by parts or recognize that this equals

\int \tan u \sec u \, du

, which has the antiderivative

\sec u + C

. Substituting back:

\sec(\ln x) + C

, which is answer D. Let's examine why the other options are wrong. Answer A,

\ln|\cos(\ln x)| + C

, would result from integrating

\int \frac{\sin(\ln x)}{x\cos(\ln x)}dx

(notice the missing squared term in the denominator). Answer B,

\tan(\ln x) + C

, comes from incorrectly thinking that

\int \frac{\sin u}{\cos^2 u}du = \tan u

, but the derivative of

\tan u

\sec^2 u

, not

\frac{\sin u}{\cos^2 u}

. Answer C has the wrong sign—you might get this if you made an error with the integration by parts or misremembered the antiderivative. Remember: when you see

\ln x

in the integrand along with

\frac{1}{x}

, immediately consider

u = \ln x

substitution. This pattern appears frequently and dramatically simplifies these problems.

Question 5

Evaluate the indefinite integral $\int \frac{5^{x}}{1+5^{x}} \,dx$

$\frac{5^x}{\ln 5} + x + C$
$\ln(1+5^x) + C$
$x - \frac{\ln(1+5^x)}{\ln 5} + C$
$\frac{1}{\ln 5} \ln(1+5^x) + C$ (correct answer)

Explanation: When you encounter an integral with exponential functions in both numerator and denominator, look for substitution opportunities that can simplify the expression. For

\int \frac{5^{x}}{1+5^{x}} \,dx

, let's use the substitution

u = 5^x

. Then

du = 5^x \ln 5 \, dx

, which means

dx = \frac{du}{5^x \ln 5} = \frac{du}{u \ln 5}

. Substituting into our integral:

\int \frac{5^{x}}{1+5^{x}} \,dx = \int \frac{u}{1+u} \cdot \frac{du}{u \ln 5} = \frac{1}{\ln 5} \int \frac{1}{1+u} \,du

This simplifies to

\frac{1}{\ln 5} \ln(1+u) + C = \frac{1}{\ln 5} \ln(1+5^x) + C

, confirming answer D. Let's examine why the other choices are incorrect: Choice A,

\frac{5^x}{\ln 5} + x + C

, can be eliminated by differentiation - this would give

5^x + 1

, not our original integrand. Choice B,

\ln(1+5^x) + C

, is missing the crucial factor of

\frac{1}{\ln 5}

that comes from the substitution process. Choice C,

x - \frac{\ln(1+5^x)}{\ln 5} + C

, incorrectly includes an

x

term and has the wrong sign on the logarithmic term. Study tip: When integrating rational functions involving exponentials, try substituting

u

equal to the exponential expression. The derivative will often provide exactly the factor needed to simplify the integral, but don't forget to account for constants like

\ln 5

that appear in exponential derivatives.

Question 6

Find the indefinite integral $\int \frac{dx}{\sqrt{x}(1+\sqrt{x})^2}$

$\frac{2}{1+\sqrt{x}} + C$
$-\frac{2}{1+\sqrt{x}} + C$ (correct answer)
$\ln((1+\sqrt{x})^2) + C$
$-\frac{1}{2(1+\sqrt{x})} + C$

Explanation: Let

u = 1 + \sqrt{x}

. Then

du = \frac{1}{2\sqrt{x}} \,dx

, which means

2 \,du = \frac{1}{\sqrt{x}} \,dx

. Substituting into the integral:

\int \frac{1}{(1+\sqrt{x})^2} \cdot \frac{1}{\sqrt{x}} \,dx = \int \frac{1}{u^2} \cdot 2 \,du = 2 \int u^{-2} \,du

Integrating with respect to

u

gives:

2 \left( \frac{u^{-1}}{-1} \right) + C = -2u^{-1} + C = -\frac{2}{u} + C

Substituting back

u = 1 + \sqrt{x}

yields:

-\frac{2}{1+\sqrt{x}} + C

A: This answer has a sign error in the final result, likely from forgetting the negative in the power rule for

u^{-2}

. C: This answer results from incorrectly applying a logarithm integration rule. D: This answer omits the factor of 2 that arises from the differential

du

Question 7

Evaluate the definite integral $\int_{0}^{\ln(\sqrt{3})} \frac{e^x}{1+e^{2x}} \,dx$

$\frac{\pi}{12}$ (correct answer)
$\frac{\pi}{6}$
$\frac{1}{2}\ln(2)$
$\frac{\pi}{3}$

Explanation: Use the substitution

u = e^x

. Then

du = e^x \,dx

. We change the limits of integration: when

x=0

u=e^0=1

. When

x=\ln(\sqrt{3})

u=e^{\ln(\sqrt{3})}=\sqrt{3}

. The integral becomes:

\int_{1}^{\sqrt{3}} \frac{1}{1+u^2} \,du = [\arctan(u)]_{1}^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(1) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}

B: This answer corresponds to

\arctan(\sqrt{3}) - \arctan(0)

, which would arise if the lower limit

u=1

was incorrectly evaluated as

u=0

. C: This answer results from incorrectly integrating

1/(1+u^2)

\ln(1+u^2)

, a common error confusing it with

u/(1+u^2)

. D: This answer corresponds to just

\arctan(\sqrt{3})

, forgetting to subtract the value at the lower limit.

Question 8

Evaluate $\int \sec^3(x) \tan(x) \,dx$

$\frac{1}{3}\tan^3(x) + C$
$\frac{1}{4}\sec^4(x) + C$
$\frac{1}{3}\sec^3(x) + C$ (correct answer)
$\frac{1}{4}\tan^4(x) + C$

Explanation: This integral can be solved with a u-substitution. Let

u = \sec(x)

. The derivative is

du = \sec(x)\tan(x) \,dx

. We can rewrite the integrand to make this substitution clear:

\int \sec^2(x) (\sec(x)\tan(x)) \,dx

Now substitute

u = \sec(x)

and

du = \sec(x)\tan(x) \,dx

\int u^2 \,du = \frac{u^3}{3} + C

Substituting back for

u

gives:

\frac{\sec^3(x)}{3} + C

A: This answer results from an incorrect substitution, likely letting

u=\tan(x)

and making an error with the remaining

\sec^2(x)

. B: This answer results from an incorrect power in the final result. It would be the result of integrating

\int u^3 du

, which would correspond to an integrand of

\sec^4(x)\tan(x)

. D: This answer results from an incorrect substitution, perhaps

u = \tan(x)

, and mismanaging the powers of secant.

Question 9

Evaluate the indefinite integral $\int \frac{\arcsin(x)}{\sqrt{1-x^2}} \,dx$

$2\sqrt{\arcsin(x)} + C$
$\ln|\arcsin(x)| + C$
$\frac{(\arcsin x)^2}{2} + C$ (correct answer)
$\arctan(x) + C$

Explanation: This integral is set up for a direct u-substitution. Let

u = \arcsin(x)

. The derivative is

du = \frac{1}{\sqrt{1-x^2}} \,dx

. Substituting these into the integral gives:

\int u \,du

The integral is

\frac{u^2}{2} + C

. Substituting back for

u

yields:

\frac{(\arcsin x)^2}{2} + C

A: This answer would result if the integrand were

\frac{1}{\sqrt{1-x^2}\sqrt{\arcsin(x)}}

. B: This answer would result if the integrand were

\frac{1}{\arcsin(x)\sqrt{1-x^2}}

. D: This answer confuses the arcsin function with the arctan function and the structure of the integral.

Question 10

The integral $\int \frac{x+2}{x^2+4} \,dx$ can be split into two separate integrals. What is the resulting antiderivative?

$\frac{1}{2}\ln(x^2+4) + \arctan(\frac{x}{2}) + C$ (correct answer)
$\ln(x^2+4) + 2\arctan(x) + C$
$\frac{1}{2}\ln(x^2+4) + \arctan(x) + C$
$\frac{1}{2}\ln(x^2+4) + 2\arctan(\frac{x}{2}) + C$

Explanation: First, split the integral into two parts:

\int \frac{x+2}{x^2+4} \,dx = \int \frac{x}{x^2+4} \,dx + \int \frac{2}{x^2+4} \,dx

For the first integral, let

u = x^2+4

, so

du = 2x \,dx

and

\frac{1}{2}du = x \,dx

\int \frac{x}{x^2+4} \,dx = \frac{1}{2} \int \frac{1}{u} \,du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+4)

For the second integral, we use the arctangent formula

\int \frac{1}{a^2+x^2}dx = \frac{1}{a}\arctan(\frac{x}{a})

. Here

a=2

\int \frac{2}{x^2+4} \,dx = 2 \int \frac{1}{x^2+2^2} \,dx = 2 \left(\frac{1}{2}\arctan\left(\frac{x}{2}\right)\right) = \arctan\left(\frac{x}{2}\right)

Combining the results gives:

\frac{1}{2}\ln(x^2+4) + \arctan(\frac{x}{2}) + C

. B: This answer misses the

1/2

factor on the logarithm and uses the incorrect form for the arctangent. C: This answer incorrectly integrates the second part as if it were

\int \frac{1}{x^2+1}dx

. D: This answer incorrectly handles the constant in the second integral, resulting in

2\arctan(x/2)

instead of

\arctan(x/2)

Question 11

Which of the following integrals results from making the substitution $u = x^2$ in $\int_{1}^{2} x e^{x^2} \,dx$ ?

$\frac{1}{2} \int_{1}^{2} e^u \,du$
$\int_{1}^{4} e^u \,du$
$2 \int_{1}^{2} e^u \,du$
$\frac{1}{2} \int_{1}^{4} e^u \,du$ (correct answer)

Explanation: When you encounter a substitution problem, you need to systematically replace three things: the variable, the differential, and the limits of integration. Starting with

u = x^2

, you first find the differential by taking the derivative:

\frac{du}{dx} = 2x

, so

du = 2x \, dx

. This means

x \, dx = \frac{1}{2} du

. Next, update the limits of integration. When

x = 1

u = 1^2 = 1

. When

x = 2

u = 2^2 = 4

. So your new limits are from 1 to 4. Now substitute everything into the original integral:

\int_{1}^{2} x e^{x^2} \,dx = \int_{1}^{4} e^u \cdot \frac{1}{2} \,du = \frac{1}{2} \int_{1}^{4} e^u \,du

This matches answer choice D. Looking at the wrong answers: A keeps the original limits (1 to 2) instead of updating them to reflect the substitution. B has the correct new limits but missing the

\frac{1}{2}

factor that comes from

x \, dx = \frac{1}{2} du

. C has the wrong coefficient (2 instead of

\frac{1}{2}

) and keeps the original limits. Study tip: For u-substitution, always follow the three-step checklist: substitute the function, substitute the differential (don't forget coefficients!), and update the limits using your substitution equation. Missing any of these steps leads to the common wrong answers you see here.

Question 12

Which of the following is the indefinite integral of $\int x(x+1)^5 \,dx$ ?

$\frac{(x+1)^7}{7} - \frac{(x+1)^6}{6} + C$ (correct answer)
$\frac{x^2}{2} \frac{(x+1)^6}{6} + C$
$\frac{x(x+1)^6}{6} + C$
$\frac{(x+1)^7}{7} + \frac{(x+1)^6}{6} + C$

Explanation: To solve

\int x(x+1)^5 \,dx

, let

u = x+1

. This implies

x = u-1

and

du = dx

. Substituting these into the integral gives:

\int (u-1)u^5 \,du = \int (u^6 - u^5) \,du

Now, integrate term by term:

\frac{u^7}{7} - \frac{u^6}{6} + C

Finally, substitute back

u = x+1

\frac{(x+1)^7}{7} - \frac{(x+1)^6}{6} + C

B: This answer incorrectly attempts to integrate the factors separately, which is not a valid integration rule. C: This is the result of applying integration by parts incorrectly. D: This answer has a sign error in the second term, resulting from an error in distributing

(u-1)u^5

Question 13

Evaluate the definite integral $\int_{0}^{\pi/2} \sin(x) e^{\cos(x)} \,dx$

1 - e
e - 1 (correct answer)
e
1

Explanation: To evaluate this integral, we use u-substitution. Let

u = \cos(x)

. Then

du = -\sin(x) \,dx

, which means

-du = \sin(x) \,dx

. We must also change the limits of integration. When

x = 0

u = \cos(0) = 1

. When

x = \pi/2

u = \cos(\pi/2) = 0

. Substituting these into the integral gives:

\int_{1}^{0} e^u (-du) = -\int_{1}^{0} e^u \,du = \int_{0}^{1} e^u \,du

Evaluating this integral gives

[e^u]_0^1 = e^1 - e^0 = e - 1

. A: This answer results from a sign error, likely from forgetting the negative sign in

du = -\sin(x) dx

and evaluating

-\int_0^1 e^u \,du

. C: This answer results from incorrectly evaluating

e^0

as 0 instead of 1. D: This answer results from incorrectly evaluating the bounds, perhaps calculating

e^1 - e^1

e^0 - e^0

Question 14

To evaluate the integral $\int \frac{e^{3x}}{e^{2x}+1} dx$ , a student uses the substitution $u = e^x$ . Which of the following integrals results from this substitution?

$\int \frac{u^2}{u^2+1} du$ (correct answer)
$\int \frac{u^3}{u^2+1} du$
$\int \frac{u}{u^2+1} du$
$\int \left(u - \frac{u}{u^2+1}\right) du$

Explanation: Given the substitution

u = e^x

, we find the differential

du = e^x dx

. We can rewrite the integrand in terms of

u

e^{3x} = (e^x)^3 = u^3

and

e^{2x} = (e^x)^2 = u^2

. The integral can be written as

\int \frac{e^{2x} \cdot e^x}{e^{2x}+1} dx

. Substituting

u

for

e^x

and

du

for

e^x dx

yields:

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

Distractor B arises from substituting for

x

but not for

dx

. Distractor C results from an incorrect substitution for the numerator. Distractor D would be the next step in integrating distractor B after performing polynomial long division, representing a student who starts with the wrong substitution and continues down that incorrect path.

Question 15

The rate of consumption of a resource is given by $R(t) = \frac{t}{(t^2+4)^2}$ units per year, where $t$ is the number of years from the start of a project. Find the total amount of the resource consumed from $t=0$ to $t=2$ years.

$\frac{1}{8}$
$\frac{1}{16}$ (correct answer)
$\ln(2)$
$\frac{1}{32}$

Explanation: When you see a rate function and need to find the total amount consumed over time, you're looking at a fundamental theorem of calculus problem. The rate

R(t) = \frac{t}{(t^2+4)^2}

tells you how fast the resource is being consumed at any moment, so to find the total consumption, you need to integrate this rate from

t=0

t=2

. To evaluate

\int_0^2 \frac{t}{(t^2+4)^2} dt

, use substitution. Let

u = t^2 + 4

, so

du = 2t \, dt

, which means

t \, dt = \frac{1}{2} du

. When

t=0

u=4

; when

t=2

u=8

. The integral becomes:

\int_4^8 \frac{1}{2u^2} du = \frac{1}{2} \int_4^8 u^{-2} du = \frac{1}{2} \left[-u^{-1}\right]_4^8 = \frac{1}{2}\left(-\frac{1}{8} + \frac{1}{4}\right) = \frac{1}{2} \cdot \frac{1}{8} = \frac{1}{16}

Answer B is correct. Answer A (

\frac{1}{8}

) occurs if you forget the factor of

\frac{1}{2}

from the substitution. Answer C (

\ln(2)

) suggests incorrectly treating this as

\int \frac{1}{u} du

instead of

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

. Answer D (

\frac{1}{32}

) results from making an error in the arithmetic, possibly doubling the denominator incorrectly. Remember: when you see a rate function, always integrate to find the total quantity. Look for substitution opportunities when you have a function and its derivative in the same integral.

Question 16

If $\int_{0}^{k} \frac{x}{x^2+1} \,dx = \ln(3)$ , what is the value of $k$ ? Assume $k>0$ .

8
3
$\sqrt{8}$ (correct answer)
$\sqrt{3}-1$

Explanation: First, evaluate the integral. Let

u = x^2+1

, so

du = 2x \,dx

x \,dx = \frac{1}{2}du

. Change the limits: when

x=0, u=1

; when

x=k, u=k^2+1

\int_{1}^{k^2+1} \frac{1}{u} \cdot \frac{1}{2}du = \frac{1}{2} [\ln|u|]_1^{k^2+1} = \frac{1}{2}(\ln(k^2+1) - \ln(1)) = \frac{1}{2}\ln(k^2+1)

We are given this equals

\ln(3)

\frac{1}{2}\ln(k^2+1) = \ln(3) \implies \ln(k^2+1) = 2\ln(3) \implies \ln(k^2+1) = \ln(3^2) = \ln(9)

Therefore,

k^2+1 = 9

, which means

k^2 = 8

. Since

k>0

k = \sqrt{8} = 2\sqrt{2}

. A: This answer results from setting

k^2=8

but then forgetting the square root, or from setting

k+1=9

. B: This answer incorrectly equates

\frac{1}{2}\ln(k^2+1)

with

\ln(3)

by simply setting

k^2+1=3

and ignoring the

1/2

coefficient and solving

k^2=2

. D: This answer might arise from incorrectly solving the equation

k^2+1=3

Question 17

To evaluate $\int \frac{1}{e^x + e^{-x}} \,dx$ , a common first step is to multiply the numerator and denominator by $e^x$ . After this step and a subsequent u-substitution, what is the resulting antiderivative?

$\ln(e^x + e^{-x}) + C$
$\arctan(e^x) + C$ (correct answer)
$\frac{1}{2}x - \frac{1}{4}e^{-2x} + C$
$-\frac{1}{e^x + e^{-x}} + C$

Explanation: First, multiply the numerator and denominator by

e^x

\int \frac{1}{e^x + e^{-x}} \cdot \frac{e^x}{e^x} \,dx = \int \frac{e^x}{e^{2x} + 1} \,dx

Now, perform a u-substitution. Let

u = e^x

. Then

du = e^x \,dx

. The integral becomes:

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

This is the standard integral for the inverse tangent function:

\arctan(u) + C

Substituting back

u = e^x

gives the final answer:

\arctan(e^x) + C

. This function is also

\tanh(x)

, whose integral is not a standard Calc 1 result, but the method described leads to the correct answer. A: This would be the result if the numerator were

e^x - e^{-x}

. C: This comes from the integral of

\cosh^2(x)

, suggesting a misremembered hyperbolic identity. D: This is a plausible but incorrect result from misapplying the power rule.

Question 18

Evaluate the definite integral $\int_0^1 x^3 \sqrt{1-x^2} \, dx$ .

$\frac{2}{15}$ (correct answer)
$-\frac{2}{15}$
$\frac{8}{15}$
$\frac{1}{12}$

Explanation: To evaluate this integral, use the substitution

u = 1-x^2

. This implies

du = -2x \, dx

, so

x \, dx = -\frac{1}{2} du

. We also need to express

x^2

in terms of

u

, which is

x^2 = 1-u

. The limits of integration must be changed: when

x=0

u=1-0^2=1

, and when

x=1

u=1-1^2=0

. The integral can be rewritten as

\int_0^1 x^2 \sqrt{1-x^2} (x \, dx)

. Substituting gives:

\int_1^0 (1-u)\sqrt{u} \left(-\frac{1}{2} du\right) = \frac{1}{2} \int_0^1 (1-u)u^{1/2} du = \frac{1}{2} \int_0^1 (u^{1/2} - u^{3/2}) du

Evaluating the integral:

\frac{1}{2} \left[ \frac{2}{3}u^{3/2} - \frac{2}{5}u^{5/2} \right]_0^1 = \frac{1}{2} \left( (\frac{2}{3} - \frac{2}{5}) - 0 \right) = \frac{1}{2} \left( \frac{10-6}{15} \right) = \frac{1}{2} \left( \frac{4}{15} \right) = \frac{2}{15}

Question 19

Let $f$ be a continuous function such that $\int_0^9 f(x) \, dx = 12$ . What is the value of $\int_0^3 x f(x^2) \, dx$ ?

24
12
6 (correct answer)
36

Explanation: This problem tests your ability to use substitution in definite integrals, specifically recognizing when a substitution will transform one integral into another. To evaluate

\int_0^3 x f(x^2) \, dx

, use the substitution

u = x^2

. Then

du = 2x \, dx

, which means

x \, dx = \frac{1}{2} du

. When

x = 0

, we have

u = 0^2 = 0

. When

x = 3

, we have

u = 3^2 = 9

. Substituting:

\int_0^3 x f(x^2) \, dx = \int_0^9 f(u) \cdot \frac{1}{2} \, du = \frac{1}{2} \int_0^9 f(u) \, du

Since the variable of integration is just a dummy variable,

\int_0^9 f(u) \, du = \int_0^9 f(x) \, dx = 12

. Therefore:

\int_0^3 x f(x^2) \, dx = \frac{1}{2} \cdot 12 = 6

Looking at the wrong answers: Choice (A) gives 24, which you'd get if you mistakenly multiplied by 2 instead of

\frac{1}{2}

when handling the substitution. Choice (B) gives 12, which ignores the factor of

\frac{1}{2}

entirely—this happens when students forget that

x \, dx = \frac{1}{2} du

, not just

du

. Choice (D) gives 36, likely from multiplying the given integral value by 3 (the upper limit) without proper substitution work. Strategy tip: When you see

x f(x^2)

or similar compositions, immediately think substitution with

u = x^2

. The key is carefully tracking the factor that comes from

du = 2x \, dx

Question 20

Evaluate the definite integral $\int_{e}^{e^4} \frac{dx}{x\sqrt{\ln x}}$ .

$4$
$1$
$2$ (correct answer)
$\ln(4)$

Explanation: When you encounter an integral with a logarithmic expression under a square root, substitution is typically your best approach. The key insight here is recognizing that the derivative of

\ln x

\frac{1}{x}

, which appears as a factor in our integrand. Let's use the substitution

u = \ln x

, so

du = \frac{1}{x}dx

. We need to change our limits of integration: when

x = e

, we have

u = \ln e = 1

, and when

x = e^4

, we have

u = \ln(e^4) = 4

. Our integral becomes:

\int_{1}^{4} \frac{du}{\sqrt{u}} = \int_{1}^{4} u^{-1/2} du

Using the power rule for integration:

\int u^{-1/2} du = \frac{u^{1/2}}{1/2} = 2u^{1/2} = 2\sqrt{u}

Evaluating from 1 to 4:

2\sqrt{u}\Big|_{1}^{4} = 2\sqrt{4} - 2\sqrt{1} = 2(2) - 2(1) = 4 - 2 = 2

This confirms answer (C) 2. Looking at the wrong answers: (A) 4 would result from forgetting to subtract the lower limit evaluation. (B) 1 might come from incorrectly handling the coefficient when applying the power rule. (D) $\ln(4)$ could result from confusion about which substitution to use or incorrectly thinking the antiderivative involves logarithms. Strategy tip: When you see

\frac{1}{x}

paired with a function of

\ln x

, immediately consider the substitution

u = \ln x

. This pattern appears frequently and transforms complex-looking integrals into simple power functions.