Calculus 2 Quiz: Integration By Substitution
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Integration By SubstitutionQuestion 1 of 20

Evaluate ∫1xln⁡(x3)dx\int \frac{1}{x \ln(x^3)} dx.

3ln⁡∣ln⁡x∣+C3 \ln|\ln x| + C
ln⁡∣ln⁡(x3)∣+C\ln|\ln(x^3)| + C
−12(xln⁡(x3))2+C\frac{-1}{2(x \ln(x^3))^2} + C
13ln⁡∣ln⁡x∣+C\frac{1}{3} \ln|\ln x| + C
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Calculus 2 Quiz

Calculus 2 Quiz: Integration By Substitution

Practice Integration By Substitution 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 Integration By Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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

Evaluate ∫1xln⁡(x3)dx\int \frac{1}{x \ln(x^3)} dx.

  1. 3ln⁡∣ln⁡x∣+C3 \ln|\ln x| + C
  2. ln⁡∣ln⁡(x3)∣+C\ln|\ln(x^3)| + C
  3. −12(xln⁡(x3))2+C\frac{-1}{2(x \ln(x^3))^2} + C
  4. 13ln⁡∣ln⁡x∣+C\frac{1}{3} \ln|\ln x| + C (correct answer)
Explanation: When you encounter an integral with logarithmic functions in both the numerator and denominator, substitution is typically your best approach. The key insight here is recognizing that the derivative of ln⁡(x3)\ln(x^3) appears (almost) in the numerator. Let's use the substitution u=ln⁡(x3)u = \ln(x^3). First, simplify using logarithm properties: u=ln⁡(x3)=3ln⁡(x)u = \ln(x^3) = 3\ln(x), so du=3⋅1xdxdu = 3 \cdot \frac{1}{x} dx, which means 1xdx=13du\frac{1}{x} dx = \frac{1}{3} du. Substituting into our integral: ∫1xln⁡(x3)dx=∫1ln⁡(x3)⋅1xdx=∫1u⋅13du=13∫1udu=13ln⁡∣u∣+C\int \frac{1}{x \ln(x^3)} dx = \int \frac{1}{\ln(x^3)} \cdot \frac{1}{x} dx = \int \frac{1}{u} \cdot \frac{1}{3} du = \frac{1}{3} \int \frac{1}{u} du = \frac{1}{3} \ln|u| + C Substituting back: 13ln⁡∣ln⁡(x3)∣+C=13ln⁡∣3ln⁡(x)∣+C=13ln⁡∣3∣+13ln⁡∣ln⁡(x)∣+C\frac{1}{3} \ln|\ln(x^3)| + C = \frac{1}{3} \ln|3\ln(x)| + C = \frac{1}{3} \ln|3| + \frac{1}{3} \ln|\ln(x)| + C Since 13ln⁡∣3∣\frac{1}{3} \ln|3| is just a constant, we can absorb it into CC, giving us 13ln⁡∣ln⁡x∣+C\frac{1}{3} \ln|\ln x| + C. Choice A misses the factor of 13\frac{1}{3} and incorrectly has coefficient 3. Choice B fails to simplify ln⁡(x3)=3ln⁡(x)\ln(x^3) = 3\ln(x) and doesn't account for the substitution factor. Choice C incorrectly attempts integration by parts or treats this as a power function rather than recognizing the logarithmic structure. Strategy tip: When you see nested functions like f(g(x))f(g(x)), always check if the derivative of the inner function g′(x)g'(x) appears elsewhere in the integrand—this signals u-substitution.

Question 2

After applying the substitution u=exu = e^x, which integral is equivalent to ∫ex1+e2xdx\int \frac{e^x}{1+e^{2x}} dx?

  1. ∫11+u2du\int \frac{1}{1+u^2} du (correct answer)
  2. ∫u1+u2du\int \frac{u}{1+u^2} du
  3. ∫11+udu\int \frac{1}{1+u} du
  4. ∫1u(1+u2)du\int \frac{1}{u(1+u^2)} du
Explanation: Let u=exu = e^x. Then du=exdxdu = e^x dx. Also, note that e2x=(ex)2=u2e^{2x} = (e^x)^2 = u^2. Substituting these into the original integral gives ∫11+u2du\int \frac{1}{1+u^2} du.

Question 3

Evaluate ∫x+1x2+1dx\int \frac{x+1}{x^2+1} dx.

  1. ln⁡(x2+1)+arctan⁡(x)+C\ln(x^2+1) + \arctan(x) + C
  2. 12ln⁡(x2+1)+arctan⁡(x)+C\frac{1}{2}\ln(x^2+1) + \arctan(x) + C (correct answer)
  3. arctan⁡(x)+C\arctan(x) + C
  4. 12ln⁡(x2+1)+C\frac{1}{2}\ln(x^2+1) + C
Explanation: This integral should be split into two parts: ∫xx2+1dx+∫1x2+1dx\int \frac{x}{x^2+1} dx + \int \frac{1}{x^2+1} dx. The second integral is the standard form for arctan⁡(x)\arctan(x). For the first integral, use the substitution u=x2+1u=x^2+1, so du=2xdxdu=2xdx or 12du=xdx\frac{1}{2}du=xdx. This integral becomes 12∫duu=12ln⁡∣u∣=12ln⁡(x2+1)\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+1). Combining the two parts gives 12ln⁡(x2+1)+arctan⁡(x)+C\frac{1}{2}\ln(x^2+1) + \arctan(x) + C.

Question 4

The substitution u=x−3u=\sqrt{x}-3 transforms a definite integral ∫abf(x)dx\int_a^b f(x) dx into ∫−2−12(u+3)eudu\int_{-2}^{-1} 2(u+3)e^u du. What was the original integral?

  1. ∫14ex−3dx\int_1^4 e^{\sqrt{x}-3} dx (correct answer)
  2. ∫12eudu\int_1^2 e^u du
  3. ∫14(x−3)ex−3dx\int_1^4 (\sqrt{x}-3)e^{\sqrt{x}-3} dx
  4. ∫14ex−32xdx\int_1^4 \frac{e^{\sqrt{x}-3}}{2\sqrt{x}} dx
Explanation: From u=x−3u = \sqrt{x}-3, we find x=u+3\sqrt{x} = u+3 and x=(u+3)2x=(u+3)^2. Then dx=2(u+3)dudx = 2(u+3)du. The transformed integral is ∫−2−1eu⋅2(u+3)du\int_{-2}^{-1} e^u \cdot 2(u+3)du. The part 2(u+3)du2(u+3)du corresponds to dxdx, so the original function f(x)f(x) must correspond to eue^u. Substituting u=x−3u=\sqrt{x}-3 gives f(x)=ex−3f(x) = e^{\sqrt{x}-3}. For the limits: if u=−2u=-2, then −2=a−3  ⟹  a=1  ⟹  a=1-2=\sqrt{a}-3 \implies \sqrt{a}=1 \implies a=1. If u=−1u=-1, then −1=b−3  ⟹  b=2  ⟹  b=4-1=\sqrt{b}-3 \implies \sqrt{b}=2 \implies b=4. Thus, the original integral was ∫14ex−3dx\int_1^4 e^{\sqrt{x}-3} dx.

Question 5

Find ∫sec⁡2(x)tan⁡2(x)+4tan⁡(x)+3 dx\int \frac{\sec^2(x)}{\tan^2(x) + 4\tan(x) + 3} \, dx

  1. ln⁡∣tan⁡(x)+1∣−ln⁡∣tan⁡(x)+3∣+C\ln|\tan(x) + 1| - \ln|\tan(x) + 3| + C
  2. 12ln⁡∣tan⁡(x)+1tan⁡(x)+3∣+C\frac{1}{2}\ln|\frac{\tan(x) + 1}{\tan(x) + 3}| + C (correct answer)
  3. ln⁡∣tan⁡(x)+1tan⁡(x)+3∣+C\ln|\frac{\tan(x) + 1}{\tan(x) + 3}| + C
  4. 12ln⁡∣tan⁡(x)+1∣−12ln⁡∣tan⁡(x)+3∣+C\frac{1}{2}\ln|\tan(x) + 1| - \frac{1}{2}\ln|\tan(x) + 3| + C
Explanation: Let u=tan⁡(x)u = \tan(x), so du=sec⁡2(x)dxdu = \sec^2(x) dx. The integral becomes ∫1u2+4u+3du\int \frac{1}{u^2 + 4u + 3} du. Factor the denominator: u2+4u+3=(u+1)(u+3)u^2 + 4u + 3 = (u + 1)(u + 3). Using partial fractions: 1(u+1)(u+3)=Au+1+Bu+3\frac{1}{(u+1)(u+3)} = \frac{A}{u+1} + \frac{B}{u+3}. Solving: 1=A(u+3)+B(u+1)1 = A(u+3) + B(u+1). Setting u=−1u = -1: 1=2A1 = 2A, so A=12A = \frac{1}{2}. Setting u=−3u = -3: 1=−2B1 = -2B, so B=−12B = -\frac{1}{2}. Therefore: ∫(1/2u+1−1/2u+3)du=12ln⁡∣u+1∣−12ln⁡∣u+3∣+C=12ln⁡∣u+1u+3∣+C=12ln⁡∣tan⁡(x)+1tan⁡(x)+3∣+C\int (\frac{1/2}{u+1} - \frac{1/2}{u+3}) du = \frac{1}{2}\ln|u+1| - \frac{1}{2}\ln|u+3| + C = \frac{1}{2}\ln|\frac{u+1}{u+3}| + C = \frac{1}{2}\ln|\frac{\tan(x)+1}{\tan(x)+3}| + C. Choice A omits the 12\frac{1}{2} coefficient. Choice C has coefficient 1 instead of 12\frac{1}{2}. Choice D shows the individual logarithm terms but is equivalent to choice B.

Question 6

Evaluate ∫x3x2+4 dx\int \frac{x^3}{\sqrt{x^2 + 4}} \, dx

  1. 13(x2+4)3/2−4x2+4+C\frac{1}{3}(x^2 + 4)^{3/2} - 4\sqrt{x^2 + 4} + C (correct answer)
  2. 13(x2+4)3/2−2x2+4+C\frac{1}{3}(x^2 + 4)^{3/2} - 2\sqrt{x^2 + 4} + C
  3. 23(x2+4)3/2−4x2+4+C\frac{2}{3}(x^2 + 4)^{3/2} - 4\sqrt{x^2 + 4} + C
  4. 12(x2+4)3/2−4x2+4+C\frac{1}{2}(x^2 + 4)^{3/2} - 4\sqrt{x^2 + 4} + C
Explanation: Using substitution u=x2+4u = x^2 + 4, so du=2x dxdu = 2x \, dx and x2=u−4x^2 = u - 4. The integral becomes ∫x2⋅xx2+4 dx=12∫u−4u du=12∫(u1/2−4u−1/2) du=12[23u3/2−8u1/2]+C=13(x2+4)3/2−4x2+4+C\int \frac{x^2 \cdot x}{\sqrt{x^2 + 4}} \, dx = \frac{1}{2} \int \frac{u - 4}{\sqrt{u}} \, du = \frac{1}{2} \int (u^{1/2} - 4u^{-1/2}) \, du = \frac{1}{2}[\frac{2}{3}u^{3/2} - 8u^{1/2}] + C = \frac{1}{3}(x^2 + 4)^{3/2} - 4\sqrt{x^2 + 4} + C. Choice B has an incorrect coefficient (-2 instead of -4). Choice C has an incorrect first coefficient (2/3 instead of 1/3). Choice D has an incorrect first coefficient (1/2 instead of 1/3).

Question 7

Find the indefinite integral ∫sin⁡(x)cos⁡(x)dx\int \sin(x) \cos(x) dx. Which of the following is NOT a correct representation of the antiderivative?

  1. 12sin⁡2(x)+C\frac{1}{2}\sin^2(x) + C
  2. −12cos⁡2(x)+C-\frac{1}{2}\cos^2(x) + C
  3. −14cos⁡(2x)+C-\frac{1}{4}\cos(2x) + C
  4. sin⁡2(x)+C\sin^2(x) + C (correct answer)
Explanation: This integral can be solved in multiple ways. 1) Let u=sin⁡(x)u=\sin(x), du=cos⁡(x)dxdu=\cos(x)dx, giving ∫udu=u22+C=12sin⁡2(x)+C\int u du = \frac{u^2}{2}+C = \frac{1}{2}\sin^2(x) + C. 2) Let u=cos⁡(x)u=\cos(x), du=−sin⁡(x)dxdu=-\sin(x)dx, giving ∫−udu=−u22+C=−12cos⁡2(x)+C\int -u du = -\frac{u^2}{2}+C = -\frac{1}{2}\cos^2(x) + C. 3) Use the identity sin⁡(x)cos⁡(x)=12sin⁡(2x)\sin(x)\cos(x) = \frac{1}{2}\sin(2x). Then ∫12sin⁡(2x)dx=12⋅−cos⁡(2x)2+C=−14cos⁡(2x)+C\int \frac{1}{2}\sin(2x) dx = \frac{1}{2} \cdot \frac{-\cos(2x)}{2} + C = -\frac{1}{4}\cos(2x) + C. All three results (A, B, C) are correct antiderivatives (they differ by a constant). Choice D is incorrect because it is missing the factor of 12\frac{1}{2}.

Question 8

Evaluate the integral ∫x2x3+83dx\int \frac{x^2}{\sqrt[3]{x^3+8}} dx.

  1. 12(x3+8)2/3+C\frac{1}{2}(x^3+8)^{2/3} + C (correct answer)
  2. (x3+8)2/3+C(x^3+8)^{2/3} + C
  3. 13ln⁡∣x3+83∣+C\frac{1}{3}\ln|\sqrt[3]{x^3+8}| + C
  4. 12(x3+8)−2/3+C\frac{1}{2}(x^3+8)^{-2/3} + C
Explanation: Let u=x3+8u = x^3+8. Then du=3x2dxdu = 3x^2 dx, so 13du=x2dx\frac{1}{3}du = x^2 dx. The integral becomes ∫1u3⋅13du=13∫u−1/3du\int \frac{1}{\sqrt[3]{u}} \cdot \frac{1}{3}du = \frac{1}{3} \int u^{-1/3} du. Using the power rule for integration, this is 13u2/32/3+C=13⋅32u2/3+C=12u2/3+C\frac{1}{3} \frac{u^{2/3}}{2/3} + C = \frac{1}{3} \cdot \frac{3}{2} u^{2/3} + C = \frac{1}{2}u^{2/3} + C. Substituting back gives 12(x3+8)2/3+C\frac{1}{2}(x^3+8)^{2/3} + C.

Question 9

Evaluate the definite integral ∫02x9−x2dx\int_0^2 \frac{x}{\sqrt{9-x^2}} dx.

  1. arcsin⁡(2/3)\arcsin(2/3)
  2. 3−53 - \sqrt{5} (correct answer)
  3. 5−3\sqrt{5} - 3
  4. 12(5−3)\frac{1}{2}(\sqrt{5}-3)
Explanation: Let u=9−x2u = 9-x^2. Then du=−2xdxdu = -2x dx, so −12du=xdx-\frac{1}{2}du = x dx. The limits of integration change: when x=0x=0, u=9−02=9u=9-0^2=9; when x=2x=2, u=9−22=5u=9-2^2=5. The integral becomes ∫951u(−12du)=12∫59u−1/2du\int_9^5 \frac{1}{\sqrt{u}} (-\frac{1}{2}du) = \frac{1}{2} \int_5^9 u^{-1/2} du. Integrating gives 12[2u1/2]59=[u]59=9−5=3−5\frac{1}{2}[2u^{1/2}]_5^9 = [\sqrt{u}]_5^9 = \sqrt{9}-\sqrt{5} = 3-\sqrt{5}.

Question 10

Evaluate the definite integral ∫0π/4sec⁡2(x)etan⁡(x)dx.\int_0^{\pi/4} \sec^2(x) e^{\tan(x)} dx.

  1. ee
  2. e−1e-1 (correct answer)
  3. 11
  4. eπ/4−1e^{\pi/4}-1
Explanation: Let u=tan⁡(x)u = \tan(x). Then du=sec⁡2(x)dxdu = \sec^2(x) dx. We must also change the limits of integration. When x=0x=0, u=tan⁡(0)=0u = \tan(0) = 0. When x=π/4x=\pi/4, u=tan⁡(π/4)=1u = \tan(\pi/4) = 1. The integral becomes ∫01eudu\int_0^1 e^u du, which evaluates to [eu]01=e1−e0=e−1[e^u]_0^1 = e^1 - e^0 = e - 1.

Question 11

Find the indefinite integral ∫cos⁡(ln⁡x)xdx\int \frac{\cos(\ln x)}{x} dx.

  1. −sin⁡(ln⁡x)+C-\sin(\ln x) + C
  2. sin⁡(ln⁡x)+C\sin(\ln x) + C (correct answer)
  3. cos⁡(ln⁡x)ln⁡∣x∣+C\cos(\ln x) \ln|x| + C
  4. −sin⁡(ln⁡x)x2+C-\frac{\sin(\ln x)}{x^2} + C
Explanation: Use the substitution u=ln⁡xu = \ln x, for which du=1xdxdu = \frac{1}{x} dx. The integral transforms to ∫cos⁡(u)du\int \cos(u) du. The antiderivative of cos⁡(u)\cos(u) is sin⁡(u)\sin(u). Substituting back for uu gives sin⁡(ln⁡x)+C\sin(\ln x) + C.

Question 12

Which of the following is an antiderivative of tan⁡(x)\tan(x)?

  1. sec⁡2(x)+C\sec^2(x) + C
  2. ln⁡∣sec⁡(x)∣+C\ln|\sec(x)| + C (correct answer)
  3. ln⁡∣cos⁡(x)∣+C\ln|\cos(x)| + C
  4. tan⁡2(x)2+C\frac{\tan^2(x)}{2} + C
Explanation: To integrate tan⁡(x)\tan(x), rewrite it as ∫sin⁡(x)cos⁡(x)dx\int \frac{\sin(x)}{\cos(x)} dx. Let u=cos⁡(x)u = \cos(x), so du=−sin⁡(x)dxdu = -\sin(x) dx. The integral becomes ∫−duu=−ln⁡∣u∣+C=−ln⁡∣cos⁡(x)∣+C\int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos(x)| + C. Using logarithm properties, this is equivalent to ln⁡∣(cos⁡(x))−1∣+C=ln⁡∣sec⁡(x)∣+C\ln|(\cos(x))^{-1}| + C = \ln|\sec(x)| + C.

Question 13

A particle moves along a line with velocity v(t)=t1+t2v(t) = \frac{t}{1+t^2} m/s. What is the net displacement of the particle from t=0t=0 to t=3t=3 seconds?

  1. 12arctan⁡(3)\frac{1}{2} \arctan(3) m
  2. ln⁡(10)\ln(10) m
  3. 12ln⁡(10)\frac{1}{2} \ln(10) m (correct answer)
  4. arctan⁡(3)\arctan(3) m
Explanation: Net displacement is the definite integral of velocity: ∫03t1+t2dt\int_0^3 \frac{t}{1+t^2} dt. Let u=1+t2u = 1+t^2. Then du=2tdtdu = 2t dt, so 12du=tdt\frac{1}{2}du = t dt. The limits change: when t=0t=0, u=1+02=1u=1+0^2=1; when t=3t=3, u=1+32=10u=1+3^2=10. The integral is ∫1101u⋅12du=12[ln⁡∣u∣]110=12(ln⁡(10)−ln⁡(1))=12ln⁡(10)\int_1^{10} \frac{1}{u} \cdot \frac{1}{2}du = \frac{1}{2}[\ln|u|]_1^{10} = \frac{1}{2}(\ln(10) - \ln(1)) = \frac{1}{2}\ln(10) meters.

Question 14

Which expression represents the integral ∫x1−x4dx\int \frac{x}{\sqrt{1-x^4}} dx?

  1. 12arcsin⁡(x2)+C\frac{1}{2}\arcsin(x^2) + C (correct answer)
  2. arcsin⁡(x2)+C\arcsin(x^2) + C
  3. −12(1−x4)1/2+C-\frac{1}{2}(1-x^4)^{1/2} + C
  4. 12ln⁡∣1−x4∣+C\frac{1}{2}\ln|\sqrt{1-x^4}| + C
Explanation: The denominator suggests an arcsin form. Let u=x2u = x^2. Then du=2xdxdu = 2x dx, so 12du=xdx\frac{1}{2}du = x dx. The integral becomes ∫11−u2⋅12du=12∫11−u2du\int \frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{2}du = \frac{1}{2} \int \frac{1}{\sqrt{1-u^2}} du. The antiderivative is 12arcsin⁡(u)+C\frac{1}{2} \arcsin(u) + C. Substituting back for uu yields 12arcsin⁡(x2)+C\frac{1}{2}\arcsin(x^2) + C.

Question 15

Which substitution would be most effective for evaluating the integral ∫sin⁡(x)xcos⁡3(x)dx\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \cos^3(\sqrt{x}) dx?

  1. u=xu = \sqrt{x}
  2. u=sin⁡(x)u = \sin(\sqrt{x})
  3. u=cos⁡(x)u = \cos(\sqrt{x}) (correct answer)
  4. u=xu = x
Explanation: Let u=cos⁡(x)u = \cos(\sqrt{x}). Its derivative is du=−sin⁡(x)⋅12xdxdu = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} dx. The integral contains the term sin⁡(x)xdx\frac{\sin(\sqrt{x})}{\sqrt{x}} dx, which is equal to −2du-2du. The integral transforms to ∫u3(−2du)=−2∫u3du\int u^3 (-2du) = -2 \int u^3 du, which is easily solvable. The substitution u=xu=\sqrt{x} would simplify the argument but would still leave a trigonometric integral ∫2sin⁡(u)cos⁡3(u)du\int 2\sin(u)\cos^3(u) du, requiring another substitution.

Question 16

Find the antiderivative of f(x)=x2sin⁡(x3+1)f(x) = x^2 \sin(x^3+1).

  1. −13cos⁡(x3+1)+C-\frac{1}{3}\cos(x^3+1) + C (correct answer)
  2. 13cos⁡(x3+1)+C\frac{1}{3}\cos(x^3+1) + C
  3. −3cos⁡(x3+1)+C-3\cos(x^3+1) + C
  4. 2xcos⁡(x3+1)+C2x\cos(x^3+1) + C
Explanation: To evaluate ∫x2sin⁡(x3+1)dx\int x^2 \sin(x^3+1) dx, let u=x3+1u = x^3+1. Then du=3x2dxdu = 3x^2 dx, which means 13du=x2dx\frac{1}{3}du = x^2 dx. The integral becomes ∫sin⁡(u)⋅13du=13∫sin⁡(u)du\int \sin(u) \cdot \frac{1}{3} du = \frac{1}{3} \int \sin(u) du. This evaluates to 13(−cos⁡(u))+C=−13cos⁡(x3+1)+C\frac{1}{3}(-\cos(u)) + C = -\frac{1}{3}\cos(x^3+1) + C.

Question 17

To evaluate the integral ∫x+2x+1dx\int \frac{x+2}{\sqrt{x+1}} dx, a substitution is used. Which of the following is the correct antiderivative?

  1. 23(x+1)3/2+4(x+1)1/2+C\frac{2}{3}(x+1)^{3/2} + 4(x+1)^{1/2} + C
  2. 23(x+1)3/2−2(x+1)1/2+C\frac{2}{3}(x+1)^{3/2} - 2(x+1)^{1/2} + C
  3. 23(x+1)3/2+2(x+1)1/2+C\frac{2}{3}(x+1)^{3/2} + 2(x+1)^{1/2} + C (correct answer)
  4. (x+2)ln⁡∣x+1∣+C(x+2) \ln|\sqrt{x+1}| + C
Explanation: Let u=x+1u = x+1. This implies du=dxdu = dx and x=u−1x = u-1. The numerator becomes x+2=(u−1)+2=u+1x+2 = (u-1)+2 = u+1. The integral transforms to ∫u+1udu=∫(u1/2+u−1/2)du\int \frac{u+1}{\sqrt{u}} du = \int (u^{1/2} + u^{-1/2}) du. Integrating term-by-term gives u3/23/2+u1/21/2+C=23u3/2+2u1/2+C\frac{u^{3/2}}{3/2} + \frac{u^{1/2}}{1/2} + C = \frac{2}{3}u^{3/2} + 2u^{1/2} + C. Substituting back u=x+1u=x+1 yields the correct answer.

Question 18

Evaluate ∫e3x1+e3xdx\int e^{3x} \sqrt{1+e^{3x}} dx.

  1. 29(1+e3x)3/2+C\frac{2}{9}(1+e^{3x})^{3/2} + C (correct answer)
  2. 23(1+e3x)3/2+C\frac{2}{3}(1+e^{3x})^{3/2} + C
  3. 13(1+e3x)3/2+C\frac{1}{3}(1+e^{3x})^{3/2} + C
  4. 12(1+e3x)1/2+C\frac{1}{2}(1+e^{3x})^{1/2} + C
Explanation: Let u=1+e3xu = 1+e^{3x}. Then du=3e3xdxdu = 3e^{3x} dx, which means 13du=e3xdx\frac{1}{3}du = e^{3x} dx. The integral becomes ∫u⋅13du=13∫u1/2du\int \sqrt{u} \cdot \frac{1}{3}du = \frac{1}{3} \int u^{1/2} du. Integrating gives 13u3/23/2+C=13⋅23u3/2+C=29u3/2+C\frac{1}{3} \frac{u^{3/2}}{3/2} + C = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{9}u^{3/2} + C. Substituting back results in 29(1+e3x)3/2+C\frac{2}{9}(1+e^{3x})^{3/2} + C.

Question 19

If the substitution u=5x+2u = 5x+2 is used to evaluate the definite integral ∫015x+2 dx\int_0^1 \sqrt{5x+2} \, dx, what are the resulting limits of integration?

  1. From 0 to 1
  2. From 2 to 7 (correct answer)
  3. From 0 to 5
  4. From 5/2 to 7/2
Explanation: When using u-substitution for definite integrals, the limits of integration must be converted to u-values. The original limits are for xx. The lower limit is x=0x=0, so the new lower limit is u=5(0)+2=2u = 5(0) + 2 = 2. The upper limit is x=1x=1, so the new upper limit is u=5(1)+2=7u = 5(1) + 2 = 7.

Question 20

Evaluate the integral ∫sin⁡(2x)1+cos⁡2(x)dx\int \frac{\sin(2x)}{1+\cos^2(x)} dx.

  1. ln⁡(1+cos⁡2(x))+C\ln(1+\cos^2(x)) + C
  2. arctan⁡(cos⁡(x))+C\arctan(\cos(x)) + C
  3. −ln⁡(1+cos⁡2(x))+C-\ln(1+\cos^2(x)) + C (correct answer)
  4. −2ln⁡(1+cos⁡2(x))+C-2\ln(1+\cos^2(x)) + C
Explanation: First, use the identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x). The integral becomes ∫2sin⁡(x)cos⁡(x)1+cos⁡2(x)dx\int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)} dx. Let u=1+cos⁡2(x)u = 1+\cos^2(x). Then du=2cos⁡(x)(−sin⁡(x))dx=−2sin⁡(x)cos⁡(x)dxdu = 2\cos(x)(-\sin(x)) dx = -2\sin(x)\cos(x) dx. Thus, −du=2sin⁡(x)cos⁡(x)dx-du = 2\sin(x)\cos(x) dx. The integral transforms to ∫−duu=−ln⁡∣u∣+C\int \frac{-du}{u} = -\ln|u| + C. Substituting back gives −ln⁡(1+cos⁡2(x))+C-\ln(1+\cos^2(x)) + C. The absolute value is not needed as 1+cos⁡2(x)1+\cos^2(x) is always positive.