Calculus 1 Quiz: Antiderivatives And Indefinite Integrals
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Antiderivatives And Indefinite IntegralsQuestion 1 of 20

If f(x)=x2cos⁡(x)f(x) = x^2 \cos(x), which of the following is ∫f′(x)dx\int f'(x) dx?

2xcos⁡(x)−x2sin⁡(x)+C2x\cos(x) - x^2\sin(x) + C
x2cos⁡(x)+Cx^2 \cos(x) + C
x33sin⁡(x)+C\frac{x^3}{3} \sin(x) + C
f′′(x)+Cf''(x) + C
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Calculus 1 Quiz

Calculus 1 Quiz: Antiderivatives And Indefinite Integrals

Practice Antiderivatives And Indefinite Integrals 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 Antiderivatives And Indefinite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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

If f(x)=x2cos⁡(x)f(x) = x^2 \cos(x), which of the following is ∫f′(x)dx\int f'(x) dx?

  1. 2xcos⁡(x)−x2sin⁡(x)+C2x\cos(x) - x^2\sin(x) + C
  2. x2cos⁡(x)+Cx^2 \cos(x) + C (correct answer)
  3. x33sin⁡(x)+C\frac{x^3}{3} \sin(x) + C
  4. f′′(x)+Cf''(x) + C
Explanation: This question tests the fundamental relationship between differentiation and integration. The indefinite integral of the derivative of a function, ∫f′(x)dx\int f'(x) dx, is the original function f(x)f(x) plus a constant of integration, CC. This is a direct consequence of the Fundamental Theorem of Calculus. Therefore, ∫f′(x)dx=f(x)+C\int f'(x) dx = f(x) + C. Given f(x)=x2cos⁡(x)f(x) = x^2 \cos(x), the result is x2cos⁡(x)+Cx^2 \cos(x) + C.

Question 2

Let g(x)=x44−cos⁡(x)g(x) = \frac{x^4}{4} - \cos(x). Which of the following represents ∫g′(x)dx\int g'(x) dx?

  1. x3−sin⁡(x)+Cx^3 - \sin(x) + C
  2. x520+sin⁡(x)+C\frac{x^5}{20} + \sin(x) + C
  3. x44−cos⁡(x)+C\frac{x^4}{4} - \cos(x) + C (correct answer)
  4. x3+sin⁡(x)+Cx^3 + \sin(x) + C
Explanation: This question tests the relationship between a function, its derivative, and its antiderivative. The expression ∫g′(x)dx\int g'(x) dx asks for the antiderivative of the derivative of g(x)g(x). According to the Fundamental Theorem of Calculus, integrating the derivative of a function returns the original function, plus an arbitrary constant of integration. Therefore, ∫g′(x)dx=g(x)+C\int g'(x) dx = g(x) + C. Substituting the given expression for g(x)g(x) yields x44−cos⁡(x)+C\frac{x^4}{4} - \cos(x) + C.

Question 3

Evaluate ∫(3et−sec⁡2(t))dt\int (3e^t - \sec^2(t)) dt.

  1. 3et−tan⁡(t)+C3e^t - \tan(t) + C (correct answer)
  2. 3et−sec⁡3(t)3+C3e^t - \frac{\sec^3(t)}{3} + C
  3. 3tet−1−2sec⁡2(t)tan⁡(t)+C3te^{t-1} - 2\sec^2(t)\tan(t) + C
  4. 3et−cot⁡(t)+C3e^t - \cot(t) + C
Explanation: We integrate term by term using standard integration rules. ∫(3et−sec⁡2(t))dt=∫3etdt−∫sec⁡2(t)dt\int (3e^t - \sec^2(t)) dt = \int 3e^t dt - \int \sec^2(t) dt For the first term, the integral of ete^t is ete^t. For the second term, the integral of sec⁡2(t)\sec^2(t) is tan⁡(t)\tan(t). 3∫etdt−∫sec⁡2(t)dt=3et−tan⁡(t)+C3\int e^t dt - \int \sec^2(t) dt = 3e^t - \tan(t) + C

Question 4

Evaluate the indefinite integral ∫(x−1xx)dx\int \left(\sqrt{x} - \frac{1}{x\sqrt{x}}\right) dx.

  1. 23x3/2−2x−1/2+C\frac{2}{3}x^{3/2} - 2x^{-1/2} + C
  2. 12x−1/2+32x−5/2+C\frac{1}{2}x^{-1/2} + \frac{3}{2}x^{-5/2} + C
  3. 23x3/2−2x1/2+C\frac{2}{3}x^{3/2} - 2x^{1/2} + C
  4. 23x3/2+2x−1/2+C\frac{2}{3}x^{3/2} + 2x^{-1/2} + C (correct answer)
Explanation: First, rewrite the integrand using rational exponents: x=x1/2\sqrt{x} = x^{1/2} and xx=x1⋅x1/2=x3/2x\sqrt{x} = x^1 \cdot x^{1/2} = x^{3/2}. So, 1xx=x−3/2\frac{1}{x\sqrt{x}} = x^{-3/2}. The integral becomes: ∫(x1/2−x−3/2)dx\int (x^{1/2} - x^{-3/2}) dx Now, apply the power rule for integration, ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C: x1/2+11/2+1−x−3/2+1−3/2+1+C=x3/23/2−x−1/2−1/2+C=23x3/2+2x−1/2+C\frac{x^{1/2+1}}{1/2+1} - \frac{x^{-3/2+1}}{-3/2+1} + C = \frac{x^{3/2}}{3/2} - \frac{x^{-1/2}}{-1/2} + C = \frac{2}{3}x^{3/2} + 2x^{-1/2} + C

Question 5

Find the family of functions whose derivative is g′(t)=1t23+t2g'(t) = \frac{1}{\sqrt[3]{t^2}} + t^{\sqrt{2}}.

  1. −t−5/3+2t2−1+C-t^{-5/3} + \sqrt{2}t^{\sqrt{2}-1} + C
  2. 3t1/3+t2+12+1+C3t^{1/3} + \frac{t^{\sqrt{2}+1}}{\sqrt{2}+1} + C (correct answer)
  3. 35t5/3+t2+1ln⁡(2)+C\frac{3}{5}t^{5/3} + \frac{t^{\sqrt{2}+1}}{\ln(\sqrt{2})} + C
  4. ln⁡(t23)+t2+12+1+C\ln(\sqrt[3]{t^2}) + \frac{t^{\sqrt{2}+1}}{\sqrt{2}+1} + C
Explanation: First, rewrite the function g′(t)g'(t) using rational exponents: g′(t)=t−2/3+t2g'(t) = t^{-2/3} + t^{\sqrt{2}}. Now, integrate term by term using the power rule ∫tndt=tn+1n+1+C\int t^n dt = \frac{t^{n+1}}{n+1} + C. For the first term, n=−2/3n = -2/3: ∫t−2/3dt=t−2/3+1−2/3+1=t1/31/3=3t1/3\int t^{-2/3} dt = \frac{t^{-2/3 + 1}}{-2/3 + 1} = \frac{t^{1/3}}{1/3} = 3t^{1/3} For the second term, n=2n = \sqrt{2}: ∫t2dt=t2+12+1\int t^{\sqrt{2}} dt = \frac{t^{\sqrt{2}+1}}{\sqrt{2}+1} Combining these results gives the general antiderivative: g(t)=3t1/3+t2+12+1+Cg(t) = 3t^{1/3} + \frac{t^{\sqrt{2}+1}}{\sqrt{2}+1} + C

Question 6

Find the family of functions whose derivative is h′(x)=x3−5x3h'(x) = \frac{x^3 - 5}{x^3}.

  1. ln⁡∣x3−5∣−3ln⁡∣x3∣+C\ln|x^3 - 5| - 3\ln|x^3| + C
  2. 14x4−5x14x4+C\frac{\frac{1}{4}x^4 - 5x}{\frac{1}{4}x^4} + C
  3. x+52x−2+Cx + \frac{5}{2}x^{-2} + C (correct answer)
  4. x−54x−4+Cx - \frac{5}{4}x^{-4} + C
Explanation: To find the family of functions h(x)h(x), we need to compute the indefinite integral of h′(x)h'(x). First, simplify the integrand: ∫x3−5x3dx=∫(x3x3−5x3)dx=∫(1−5x−3)dx\int \frac{x^3 - 5}{x^3} dx = \int \left(\frac{x^3}{x^3} - \frac{5}{x^3}\right) dx = \int (1 - 5x^{-3}) dx Now, integrate term-by-term using the power rule: ∫1dx−5∫x−3dx=x−5(x−3+1−3+1)+C=x−5(x−2−2)+C=x+52x−2+C\int 1 dx - 5\int x^{-3} dx = x - 5\left(\frac{x^{-3+1}}{-3+1}\right) + C = x - 5\left(\frac{x^{-2}}{-2}\right) + C = x + \frac{5}{2}x^{-2} + C

Question 7

Find the indefinite integral ∫dx2x\int \frac{dx}{2x}.

  1. 12ln⁡∣x∣+C\frac{1}{2}\ln|x| + C (correct answer)
  2. ln⁡∣2x∣+C\ln|2x| + C
  3. −12x2+C\frac{-1}{2x^2} + C
  4. 12ln⁡∣x∣+C1\frac{1}{2}\ln|x| + C_1
Explanation: The integral can be rewritten by factoring out the constant 12\frac{1}{2}: ∫dx2x=12∫1xdx\int \frac{dx}{2x} = \frac{1}{2} \int \frac{1}{x} dx The integral of 1x\frac{1}{x} is ln⁡∣x∣\ln|x| (we use absolute value bars since xx can be negative). Therefore, 12∫1xdx=12ln⁡∣x∣+C\frac{1}{2} \int \frac{1}{x} dx = \frac{1}{2}\ln|x| + C Distractor B is incorrect because ln⁡∣2x∣=ln⁡(2)+ln⁡∣x∣\ln|2x| = \ln(2) + \ln|x|, which would be the antiderivative of 1x\frac{1}{x}, not 12x\frac{1}{2x}. Distractor C incorrectly applies the power rule.

Question 8

Let F(x)F(x) be an antiderivative of a continuous function f(x)f(x). Which of the following is an antiderivative of the function g(x)=f(x)+2g(x) = f(x) + 2?

  1. F(x)+2xF(x) + 2x (correct answer)
  2. F(x)+2F(x) + 2
  3. F(x+2)F(x+2)
  4. 2F(x)2F(x)
Explanation: To find an antiderivative of g(x)=f(x)+2g(x) = f(x) + 2, we integrate g(x)g(x) with respect to xx: ∫g(x)dx=∫(f(x)+2)dx\int g(x) dx = \int (f(x) + 2) dx Using the sum rule for integrals: ∫f(x)dx+∫2dx\int f(x) dx + \int 2 dx We are given that F(x)F(x) is an antiderivative of f(x)f(x), so ∫f(x)dx=F(x)+C\int f(x) dx = F(x) + C. The integral of 2 is 2x2x. Combining these, the most general antiderivative is F(x)+2x+CF(x) + 2x + C. Therefore, F(x)+2xF(x) + 2x is an antiderivative of g(x)g(x).

Question 9

If F(x)F(x) and G(x)G(x) are two different functions such that F′(x)=G′(x)F'(x) = G'(x) for all xx in (−∞,∞)(-\infty, \infty), which of the following must be true?

  1. F(x)=G(x)F(x) = G(x) for all xx.
  2. F(x)−G(x)=CF(x) - G(x) = C for some non-zero constant CC.
  3. The graphs of F(x)F(x) and G(x)G(x) are vertical translations of each other. (correct answer)
  4. The graphs of F(x)F(x) and G(x)G(x) must intersect at exactly one point.
Explanation: If two functions F(x)F(x) and G(x)G(x) have the same derivative on an interval, then they must differ by a constant. Let H(x)=F(x)−G(x)H(x) = F(x) - G(x). Then H′(x)=F′(x)−G′(x)=0H'(x) = F'(x) - G'(x) = 0. A function whose derivative is zero everywhere must be a constant function, so H(x)=CH(x) = C for some constant CC. This means F(x)−G(x)=CF(x) - G(x) = C, or F(x)=G(x)+CF(x) = G(x) + C. Graphically, adding a constant CC to a function shifts its graph vertically. Therefore, the graphs of F(x)F(x) and G(x)G(x) are vertical translations of each other.

Question 10

Let f(x)f(x) be a function defined as f(x)={2x+1if x<13x2if x≥1f(x) = \begin{cases} 2x + 1 & \text{if } x < 1 \\ 3x^2 & \text{if } x \ge 1 \end{cases}. If F(x)F(x) is a continuous antiderivative of f(x)f(x) such that F(2)=10F(2) = 10, find F(0)F(0).

  1. 0
  2. 1 (correct answer)
  3. 2
  4. 3
Explanation: We find the antiderivative for each piece. For x<1x < 1, F(x)=∫(2x+1)dx=x2+x+C1F(x) = \int (2x+1) dx = x^2 + x + C_1. For x≥1x \ge 1, F(x)=∫(3x2)dx=x3+C2F(x) = \int (3x^2) dx = x^3 + C_2. We use the condition F(2)=10F(2)=10 with the second piece, since 2≥12 \ge 1: F(2)=23+C2=8+C2=10F(2) = 2^3 + C_2 = 8 + C_2 = 10, which implies C2=2C_2 = 2. So, for x≥1x \ge 1, F(x)=x3+2F(x) = x^3 + 2. Since F(x)F(x) must be continuous at x=1x=1, the values of the two pieces must be equal at x=1x=1. So, 12+1+C1=13+21^2 + 1 + C_1 = 1^3 + 2, which simplifies to 2+C1=32 + C_1 = 3, giving C1=1C_1 = 1. Thus, for x<1x < 1, F(x)=x2+x+1F(x) = x^2 + x + 1. We need to find F(0)F(0). Since 0<10 < 1, we use the first piece: F(0)=02+0+1=1F(0) = 0^2 + 0 + 1 = 1.

Question 11

Find the indefinite integral of g(x)=(x+x)2x2g(x) = \frac{(x + \sqrt{x})^2}{x^2} for x>0x > 0.

  1. x+2x+ln⁡(x)+Cx + 2\sqrt{x} + \ln(x) + C
  2. x+ln⁡(x)+Cx + \ln(x) + C
  3. x+4x+ln⁡(x)+Cx + 4\sqrt{x} + \ln(x) + C (correct answer)
  4. x−4x−1/2−x−1+Cx - 4x^{-1/2} - x^{-1} + C
Explanation: When you encounter a rational function like this, your first instinct should be to simplify before integrating. The expression (x+x)2x2\frac{(x + \sqrt{x})^2}{x^2} looks complex, but expanding the numerator will reveal a much simpler form. Let's expand (x+x)2=x2+2xx+x(x + \sqrt{x})^2 = x^2 + 2x\sqrt{x} + x. Now we can rewrite our function as: g(x)=x2+2xx+xx2=x2x2+2xxx2+xx2=1+2xx+1xg(x) = \frac{x^2 + 2x\sqrt{x} + x}{x^2} = \frac{x^2}{x^2} + \frac{2x\sqrt{x}}{x^2} + \frac{x}{x^2} = 1 + \frac{2\sqrt{x}}{x} + \frac{1}{x} Simplifying further: g(x)=1+2x−1/2+x−1g(x) = 1 + 2x^{-1/2} + x^{-1} Now integration is straightforward: ∫g(x)dx=∫(1+2x−1/2+x−1)dx=x+2⋅x1/21/2+ln⁡(x)+C=x+4x+ln⁡(x)+C\int g(x)dx = \int (1 + 2x^{-1/2} + x^{-1})dx = x + 2 \cdot \frac{x^{1/2}}{1/2} + \ln(x) + C = x + 4\sqrt{x} + \ln(x) + C This confirms answer C is correct. Looking at the wrong answers: A has 2x2\sqrt{x} instead of 4x4\sqrt{x}, suggesting the student forgot to apply the power rule correctly to 2x−1/22x^{-1/2}. B is missing the x\sqrt{x} term entirely, indicating the middle term was overlooked during expansion. D shows the antiderivative in unsimplified form with negative exponents rather than converting back to radicals. Study tip: Always simplify rational expressions before integrating. Expand polynomials in the numerator and split fractions into separate terms—this transforms intimidating expressions into basic power functions you can integrate term by term.

Question 12

Let F(x)F(x) and G(x)G(x) be two distinct antiderivatives of a continuous function f(x)f(x) on the interval (−∞,∞)(-\infty, \infty). Which of the following statements must be true about the graphs of y=F(x)y=F(x) and y=G(x)y=G(x)?

  1. The graphs must intersect at least at one point.
  2. The graphs have the same slope at every value of xx. (correct answer)
  3. The graphs are horizontal translations of each other.
  4. The graphs have the same concavity at every value of xx.
Explanation: If F(x)F(x) and G(x)G(x) are both antiderivatives of f(x)f(x), then F′(x)=f(x)F'(x) = f(x) and G′(x)=f(x)G'(x) = f(x). This means F′(x)=G′(x)F'(x) = G'(x) for all xx. Since the derivative of a function gives the slope of its tangent line, the graphs of y=F(x)y=F(x) and y=G(x)y=G(x) must have the same slope at every value of xx. This is the fundamental property that distinguishes antiderivatives - they differ only by a constant, so they are vertical translations of each other with identical slopes at corresponding points.

Question 13

Let F(x)F(x) be the most general antiderivative of f(x)=3⋅5xf(x) = 3 \cdot 5^x. Which of the following is F(x)F(x)?

  1. 3⋅5xln⁡(5)+C3 \cdot 5^x \ln(5) + C
  2. 3⋅5x+1x+1+C3 \cdot \frac{5^{x+1}}{x+1} + C
  3. 35xln⁡5+C3 \frac{5^x}{\ln 5} + C (correct answer)
  4. 3⋅5x+C3 \cdot 5^x + C
Explanation: To find the antiderivative of f(x)=3⋅5xf(x) = 3 \cdot 5^x, we use the constant multiple rule and the rule for integrating exponential functions of the form axa^x. The rule is ∫axdx=axln⁡a+C\int a^x dx = \frac{a^x}{\ln a} + C. ∫3⋅5xdx=3∫5xdx=3(5xln⁡5)+C=3⋅5xln⁡5+C\int 3 \cdot 5^x dx = 3 \int 5^x dx = 3 \left(\frac{5^x}{\ln 5}\right) + C = \frac{3 \cdot 5^x}{\ln 5} + C

Question 14

Which of the following is the most general antiderivative of f(x)=51−x2f(x) = \frac{5}{\sqrt{1-x^2}}?

  1. 5ln⁡(1−x2)+C5\ln(\sqrt{1-x^2}) + C
  2. 5arcsin⁡(x)+C5\arcsin(x) + C (correct answer)
  3. 5x1−x2+C\frac{5x}{\sqrt{1-x^2}} + C
  4. −10x(1−x2)−3/2+C-10x(1-x^2)^{-3/2} + C
Explanation: The integral is ∫51−x2dx\int \frac{5}{\sqrt{1-x^2}} dx. We can factor out the constant 5: 5∫11−x2dx5 \int \frac{1}{\sqrt{1-x^2}} dx The integral ∫11−x2dx\int \frac{1}{\sqrt{1-x^2}} dx is the standard form for the arcsine function. Therefore: 5∫11−x2dx=5arcsin⁡(x)+C5 \int \frac{1}{\sqrt{1-x^2}} dx = 5\arcsin(x) + C

Question 15

Evaluate ∫x+1xdx\int \frac{x+1}{x} dx.

  1. ln⁡∣x+1∣+C\ln|x+1| + C
  2. x2/2+xx2/2+C\frac{x^2/2 + x}{x^2/2} + C
  3. x+ln⁡∣x∣+Cx + \ln|x| + C (correct answer)
  4. 1+ln⁡∣x∣+C1 + \ln|x| + C
Explanation: To solve this integral, first simplify the integrand by splitting the fraction: ∫x+1xdx=∫(xx+1x)dx=∫(1+1x)dx\int \frac{x+1}{x} dx = \int \left(\frac{x}{x} + \frac{1}{x}\right) dx = \int \left(1 + \frac{1}{x}\right) dx Now, integrate term by term: ∫1dx+∫1xdx=x+ln⁡∣x∣+C\int 1 dx + \int \frac{1}{x} dx = x + \ln|x| + C This is a common problem where algebraic simplification is required before integration.

Question 16

An object's velocity is given by v(t)=cos⁡(2t)v(t) = \cos(2t). Find the most general function for the object's position, s(t)s(t), given that s′(t)=v(t)s'(t) = v(t).

  1. −2sin⁡(2t)+C-2\sin(2t) + C
  2. −12sin⁡(2t)+C-\frac{1}{2}\sin(2t) + C
  3. sin⁡(2t)+C\sin(2t) + C
  4. 12sin⁡(2t)+C\frac{1}{2}\sin(2t) + C (correct answer)
Explanation: To find the position function s(t)s(t), we must find the antiderivative of the velocity function v(t)v(t): s(t)=∫cos⁡(2t)dts(t) = \int \cos(2t) dt This requires a reversal of the chain rule. The integral of cos⁡(u)\cos(u) is sin⁡(u)\sin(u). If we guess sin⁡(2t)\sin(2t), its derivative is cos⁡(2t)⋅2\cos(2t) \cdot 2. Since our integrand is missing the factor of 2, we must compensate by multiplying by 12\frac{1}{2}. Therefore: s(t)=12sin⁡(2t)+Cs(t) = \frac{1}{2}\sin(2t) + C

Question 17

Find the indefinite integral: ∫x3−4x+6xdx\int \frac{x^3 - 4x + 6}{x} dx

  1. x33−4x+6ln⁡(x)+C\frac{x^3}{3} - 4x + 6\ln(x) + C
  2. 14x4−2x2+6x12x2+C\frac{\frac{1}{4}x^4 - 2x^2 + 6x}{\frac{1}{2}x^2} + C
  3. x33−4x+6ln⁡∣x∣+C\frac{x^3}{3} - 4x + 6\ln|x| + C (correct answer)
  4. x33−4+6ln⁡∣x∣+C\frac{x^3}{3} - 4 + 6\ln|x| + C
Explanation: To evaluate the integral, first simplify the integrand by dividing each term in the numerator by xx: ∫x3−4x+6xdx=∫(x3x−4xx+6x)dx=∫(x2−4+6x)dx\int \frac{x^3 - 4x + 6}{x} dx = \int \left(\frac{x^3}{x} - \frac{4x}{x} + \frac{6}{x}\right) dx = \int \left(x^2 - 4 + \frac{6}{x}\right) dx Now, integrate term-by-term: ∫x2dx−∫4dx+∫6xdx=x33−4x+6ln⁡∣x∣+C\int x^2 dx - \int 4 dx + \int \frac{6}{x} dx = \frac{x^3}{3} - 4x + 6\ln|x| + C The absolute value is necessary because the domain of ln⁡(u)\ln(u) is u>0u>0, while the original integrand is defined for all x≠0x \neq 0.

Question 18

Find the most general antiderivative of f(θ)=1+tan⁡2θf(\theta) = 1 + \tan^2\theta.

  1. θ+tan⁡3θ3+C\theta + \frac{\tan^3\theta}{3} + C
  2. tan⁡θ+C\tan\theta + C (correct answer)
  3. sec⁡θ+C\sec\theta + C
  4. θ+sec⁡2θ+C\theta + \sec^2\theta + C
Explanation: This problem requires knowledge of trigonometric identities. Recall the Pythagorean identity: sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta. By substituting this into the integral, we get: ∫(1+tan⁡2θ)dθ=∫sec⁡2θdθ\int (1 + \tan^2\theta) d\theta = \int \sec^2\theta d\theta The integral of sec⁡2θ\sec^2\theta is a standard result, as the derivative of tan⁡θ\tan\theta is sec⁡2θ\sec^2\theta. Therefore: ∫sec⁡2θdθ=tan⁡θ+C\int \sec^2\theta d\theta = \tan\theta + C

Question 19

Find the indefinite integral ∫(e2+π)dx\int (e^2 + \pi) dx.

  1. e2x+πx+Ce^2 x + \pi x + C (correct answer)
  2. e33+πx+C\frac{e^3}{3} + \pi x + C
  3. e2+π+Ce^2 + \pi + C
  4. 2ex+C2ex + C
Explanation: The integrand, e2+πe^2 + \pi, is a constant because ee and π\pi are constant numbers. Let k=e2+πk = e^2 + \pi. The integral is ∫kdx\int k dx. The integral of a constant kk with respect to xx is kx+Ckx + C. Therefore: ∫(e2+π)dx=(e2+π)x+C=e2x+πx+C\int (e^2 + \pi) dx = (e^2 + \pi)x + C = e^2 x + \pi x + C A common mistake is to treat ee as a variable and apply the power rule.

Question 20

Which of the following is an antiderivative of f(x)=2x2−3x2+1f(x) = \frac{2x^2 - 3}{x^2 + 1}?

  1. 2x−5ln⁡(x2+1)2x - 5\ln(x^2+1)
  2. x2−3arctan⁡(x)x^2 - 3\arctan(x)
  3. 2x+5arctan⁡(x)2x + 5\arctan(x)
  4. 2x−5arctan⁡(x)2x - 5\arctan(x) (correct answer)
Explanation: The integrand is an improper rational function. We can rewrite it using algebraic manipulation. f(x)=2x2−3x2+1=2(x2+1)−2−3x2+1=2(x2+1)−5x2+1=2(x2+1)x2+1−5x2+1=2−5x2+1f(x) = \frac{2x^2 - 3}{x^2 + 1} = \frac{2(x^2 + 1) - 2 - 3}{x^2 + 1} = \frac{2(x^2 + 1) - 5}{x^2 + 1} = \frac{2(x^2+1)}{x^2+1} - \frac{5}{x^2+1} = 2 - \frac{5}{x^2+1}. Now, we integrate this simplified form: ∫(2−5x2+1)dx=∫2dx−5∫1x2+1dx\int (2 - \frac{5}{x^2+1}) dx = \int 2 dx - 5\int \frac{1}{x^2+1} dx. This evaluates to 2x−5arctan⁡(x)+C2x - 5\arctan(x) + C. Any function of the form 2x−5arctan⁡(x)+C2x - 5\arctan(x) + C is an antiderivative.