Calculus 2 Quiz: Linear Partial Fractions
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Linear Partial FractionsQuestion 1 of 20

The equation of a curve is such that dydx=x−8x2−4\frac{dy}{dx} = \frac{x-8}{x^2-4}. What is the slope of the curve at x=3x=3?

−5-5
3/23/2
−5/2-5/2
−1-1
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Calculus 2 Quiz

Calculus 2 Quiz: Linear Partial Fractions

Practice Linear Partial Fractions 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 Linear Partial Fractions, 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

The equation of a curve is such that dydx=x−8x2−4\frac{dy}{dx} = \frac{x-8}{x^2-4}. What is the slope of the curve at x=3x=3?

  1. −5-5
  2. 3/23/2
  3. −5/2-5/2
  4. −1-1 (correct answer)
Explanation: This question tests your understanding that when you're given a derivative function, finding the slope at a specific point simply means evaluating that derivative at the given x-value. Since you're told that dydx=x−8x2−4\frac{dy}{dx} = \frac{x-8}{x^2-4}, the slope at any point is found by substituting the x-coordinate into this expression. At x=3x = 3, you substitute directly: dydx∣x=3=3−832−4=−59−4=−55=−1\frac{dy}{dx}\bigg|_{x=3} = \frac{3-8}{3^2-4} = \frac{-5}{9-4} = \frac{-5}{5} = -1 This confirms that answer D is correct. The wrong answers represent common algebraic mistakes. Answer A (-5) occurs if you forget to evaluate the denominator and only use the numerator 3−8=−53-8 = -5. Answer B (3/2) might result from sign errors—perhaps computing 8−33+1\frac{8-3}{3+1} by incorrectly flipping signs or miscomputing the denominator. Answer C (-5/2) could come from correctly finding the numerator as -5 but miscalculating the denominator as 2, possibly by computing 32−43^2 - 4 as 6−4=26 - 4 = 2 instead of 9−4=59 - 4 = 5. Key takeaway: When you see "find the slope at x = a" and you're given dydx\frac{dy}{dx}, simply substitute the x-value into the derivative. The derivative is the slope function. Double-check your arithmetic carefully, especially when working with negative numbers and computing squares.

Question 2

If 6x2−9=Ax−3+Bx+3\frac{6}{x^2-9} = \frac{A}{x-3} + \frac{B}{x+3}, which integral expression is correct?

  1. ∫(−1x−3+1x+3)dx\int (\frac{-1}{x-3} + \frac{1}{x+3}) dx
  2. ∫(3x−3−3x+3)dx\int (\frac{3}{x-3} - \frac{3}{x+3}) dx
  3. ∫(1x−3+1x+3)dx\int (\frac{1}{x-3} + \frac{1}{x+3}) dx
  4. ∫(1x−3−1x+3)dx\int (\frac{1}{x-3} - \frac{1}{x+3}) dx (correct answer)
Explanation: When you encounter a rational function with a quadratic denominator that factors, partial fraction decomposition is your key tool for integration. This technique breaks down complex fractions into simpler pieces that are easier to integrate. To find the values of A and B, start by factoring the denominator: x2−9=(x−3)(x+3)x^2 - 9 = (x-3)(x+3). So you have 6(x−3)(x+3)=Ax−3+Bx+3\frac{6}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3}. Multiply both sides by (x−3)(x+3)(x-3)(x+3) to get: 6=A(x+3)+B(x−3)6 = A(x+3) + B(x-3). To solve for A and B, substitute strategic values of x. When x=3x = 3: 6=A(6)+B(0)6 = A(6) + B(0), so A=1A = 1. When x=−3x = -3: 6=A(0)+B(−6)6 = A(0) + B(-6), so B=−1B = -1. Therefore: 6x2−9=1x−3+−1x+3=1x−3−1x+3\frac{6}{x^2-9} = \frac{1}{x-3} + \frac{-1}{x+3} = \frac{1}{x-3} - \frac{1}{x+3} The correct integral is D) ∫(1x−3−1x+3)dx\int \left(\frac{1}{x-3} - \frac{1}{x+3}\right) dx. Choice A has the wrong signs (both A and B are incorrect). Choice B uses the wrong coefficients (3 instead of 1, suggesting the student might have confused intermediate steps). Choice C has the wrong sign for the second term—it should be negative, not positive. Strategy tip: Always verify your partial fraction decomposition by adding the fractions back together before integrating. If they don't equal your original function, you've made an error in finding A and B.

Question 3

The integral ∫2x2+7x+3(x+1)(x+2)(x+3) dx\int \frac{2x^2 + 7x + 3}{(x + 1)(x + 2)(x + 3)} \, dx requires partial fraction decomposition. If the decomposition is Ax+1+Bx+2+Cx+3\frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}, what is the value of A+B+CA + B + C?

  1. 00
  2. 11
  3. 22 (correct answer)
  4. 33
Explanation: Since the degree of the numerator equals the degree of the denominator minus 1, we can find A+B+CA + B + C by examining the coefficient of x2x^2 on both sides after clearing denominators, or by substituting a convenient value. Method 1: The coefficient of x2x^2 in the numerator is 2. After clearing denominators: 2x2+7x+3=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)2x^2 + 7x + 3 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2). Expanding the right side, the coefficient of x2x^2 is A+B+CA + B + C. Therefore A+B+C=2A + B + C = 2. Method 2: We can find each coefficient individually. Setting x=−1x = -1: 2−7+3=A(1)(2)=2A2 - 7 + 3 = A(1)(2) = 2A, so A=−1A = -1. Setting x=−2x = -2: 8−14+3=B(−1)(1)=−B8 - 14 + 3 = B(-1)(1) = -B, so B=3B = 3. Setting x=−3x = -3: 18−21+3=C(−2)(−1)=2C18 - 21 + 3 = C(-2)(-1) = 2C, so C=0C = 0. Therefore A+B+C=−1+3+0=2A + B + C = -1 + 3 + 0 = 2.

Question 4

To evaluate ∫x−7(x−1)(x−4) dx\int \frac{x - 7}{(x - 1)(x - 4)} \, dx, a student sets up the partial fraction decomposition x−7(x−1)(x−4)=Ax−1+Bx−4\frac{x - 7}{(x - 1)(x - 4)} = \frac{A}{x - 1} + \frac{B}{x - 4} and correctly finds A=−2A = -2 and B=1B = 1. However, when writing the final answer, the student writes −2ln⁡∣x−1∣+ln⁡∣x−4∣+C-2 \ln|x - 1| + \ln|x - 4| + C. What error did the student make?

  1. The student should have A=2A = 2 and B=−1B = -1 instead of the given values (correct answer)
  2. The final answer should be −2ln⁡∣x−1∣−ln⁡∣x−4∣+C-2 \ln|x - 1| - \ln|x - 4| + C
  3. The final answer should be 2ln⁡∣x−1∣−ln⁡∣x−4∣+C2 \ln|x - 1| - \ln|x - 4| + C
  4. No error was made; the student's final answer is correct
Explanation: Let's check the student's coefficients. Setting up: x−7(x−1)(x−4)=Ax−1+Bx−4\frac{x - 7}{(x - 1)(x - 4)} = \frac{A}{x - 1} + \frac{B}{x - 4}. Clearing denominators: x−7=A(x−4)+B(x−1)x - 7 = A(x - 4) + B(x - 1). Setting x=1x = 1: 1−7=−6=A(1−4)=−3A1 - 7 = -6 = A(1 - 4) = -3A, so A=2A = 2. Setting x=4x = 4: 4−7=−3=B(4−1)=3B4 - 7 = -3 = B(4 - 1) = 3B, so B=−1B = -1. The student incorrectly found A=−2A = -2 and B=1B = 1. The correct values are A=2A = 2 and B=−1B = -1, giving the integral 2ln⁡∣x−1∣−ln⁡∣x−4∣+C2 \ln|x - 1| - \ln|x - 4| + C.

Question 5

To evaluate ∫1x2(x−1)dx\int \frac{1}{x^2(x-1)} dx, a student incorrectly sets up the partial fraction decomposition as Ax+Bx−1\frac{A}{x} + \frac{B}{x-1}. If the student proceeds with this incorrect setup, what value of A+BA+B would they find by clearing denominators and comparing the coefficients of the xx term?

  1. −1-1
  2. 11
  3. 00 (correct answer)
  4. No unique value can be determined.
Explanation: When you encounter a rational function like 1x2(x−1)\frac{1}{x^2(x-1)}, partial fraction decomposition requires accounting for all factors in the denominator. The correct setup should be Ax+Bx2+Cx−1\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} because the repeated factor x2x^2 needs both Ax\frac{A}{x} and Bx2\frac{B}{x^2} terms. However, let's follow the student's incorrect setup: 1x2(x−1)=Ax+Bx−1\frac{1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x-1} Clearing denominators by multiplying both sides by x2(x−1)x^2(x-1): 1=Ax(x−1)+Bx21 = Ax(x-1) + Bx^2 1=Ax2−Ax+Bx21 = Ax^2 - Ax + Bx^2 1=(A+B)x2−Ax+01 = (A+B)x^2 - Ax + 0 Comparing coefficients:
  • Constant term: 1=01 = 0 (impossible!)
  • Coefficient of xx: 0=−A0 = -A, so A=0A = 0
  • Coefficient of x2x^2: 0=A+B0 = A + B, so 0+B=00 + B = 0, giving B=0B = 0
Therefore, A+B=0+0=0A + B = 0 + 0 = 0, making (C) correct. (A) and (B) represent specific numerical values that might seem plausible but don't result from the coefficient comparison. (D) might seem tempting since the setup is fundamentally flawed (notice the impossible equation 1=01 = 0), but the algebra still produces definite coefficient values. Study tip: Always check that your partial fraction setup accounts for repeated factors properly. When you see xnx^n in the denominator, you need terms for x1,x2,...,xnx^1, x^2, ..., x^n. If your coefficient comparison leads to contradictions like 1=01 = 0, it's a red flag that your initial setup was incorrect.

Question 6

Evaluate ∫cos⁡(x)sin⁡2(x)−5sin⁡(x)+6dx\int \frac{\cos(x)}{\sin^2(x) - 5\sin(x) + 6} dx.

  1. arctan⁡(sin⁡(x)−5/2)+C\arctan(\sin(x)-5/2) + C
  2. ln⁡∣sin⁡(x)−2sin⁡(x)−3∣+C\ln|\frac{\sin(x)-2}{\sin(x)-3}| + C
  3. ln⁡∣(sin⁡(x)−3)(sin⁡(x)−2)∣+C\ln|(\sin(x)-3)(\sin(x)-2)| + C
  4. ln⁡∣sin⁡(x)−3sin⁡(x)−2∣+C\ln|\frac{\sin(x)-3}{\sin(x)-2}| + C (correct answer)
Explanation: When you encounter an integral with trigonometric functions where the denominator involves powers of sin(x) or cos(x), look for opportunities to use substitution. Here, the key insight is that the derivative of sin(x) is cos(x), which appears in the numerator. Let u=sin⁡(x)u = \sin(x), so du=cos⁡(x)dxdu = \cos(x)dx. The integral becomes: ∫duu2−5u+6\int \frac{du}{u^2 - 5u + 6} Factor the denominator: u2−5u+6=(u−2)(u−3)u^2 - 5u + 6 = (u-2)(u-3). This gives us: ∫du(u−2)(u−3)\int \frac{du}{(u-2)(u-3)} Use partial fraction decomposition: 1(u−2)(u−3)=Au−2+Bu−3\frac{1}{(u-2)(u-3)} = \frac{A}{u-2} + \frac{B}{u-3} Solving for A and B: 1=A(u−3)+B(u−2)1 = A(u-3) + B(u-2). Setting u=2u=2 gives A=−1A=-1, and setting u=3u=3 gives B=1B=1. So: ∫du(u−2)(u−3)=∫(−1u−2+1u−3)du=−ln⁡∣u−2∣+ln⁡∣u−3∣+C\int \frac{du}{(u-2)(u-3)} = \int \left(\frac{-1}{u-2} + \frac{1}{u-3}\right)du = -\ln|u-2| + \ln|u-3| + C Using logarithm properties: ln⁡∣u−3∣−ln⁡∣u−2∣=ln⁡∣u−3u−2∣\ln|u-3| - \ln|u-2| = \ln\left|\frac{u-3}{u-2}\right| Substituting back u=sin⁡(x)u = \sin(x): ln⁡∣sin⁡(x)−3sin⁡(x)−2∣+C\ln\left|\frac{\sin(x)-3}{\sin(x)-2}\right| + C Option A uses arctangent, which would arise from a quadratic that doesn't factor nicely. Option B has the fraction flipped. Option C lacks the fraction form entirely, missing the logarithm property application. Remember: When you see cos(x) in the numerator with powers of sin(x) in the denominator, try u=sin⁡(x)u = \sin(x) substitution first.

Question 7

The integral ∫dxx2+3x\int \frac{dx}{x^2+3x} is computed using partial fractions. What is the resulting antiderivative?

  1. 13ln⁡∣x2+3x∣+C\frac{1}{3}\ln|x^2+3x| + C
  2. 13ln⁡∣xx+3∣+C\frac{1}{3}\ln|\frac{x}{x+3}| + C (correct answer)
  3. 3ln⁡∣xx+3∣+C3\ln|\frac{x}{x+3}| + C
  4. 13ln⁡∣x(x+3)∣+C\frac{1}{3}\ln|x(x+3)| + C
Explanation: First, factor the denominator: x2+3x=x(x+3)x^2+3x = x(x+3). The partial fraction decomposition is 1x(x+3)=Ax+Bx+3\frac{1}{x(x+3)} = \frac{A}{x} + \frac{B}{x+3}. Solving 1=A(x+3)+Bx1 = A(x+3) + Bx gives A=1/3A=1/3 and B=−1/3B=-1/3. The integral is ∫(1/3x−1/3x+3)dx=13(ln⁡∣x∣−ln⁡∣x+3∣)+C\int (\frac{1/3}{x} - \frac{1/3}{x+3}) dx = \frac{1}{3}(\ln|x| - \ln|x+3|) + C. Using logarithm properties, this simplifies to 13ln⁡∣xx+3∣+C\frac{1}{3}\ln|\frac{x}{x+3}| + C. (A) results from incorrectly treating the integral as a simple ∫du/u\int du/u problem. (C) is a result of an error in calculating the coefficients A and B (missing the 1/3). (D) results from a sign error on the coefficient B.

Question 8

A student is asked to evaluate ∫46x+1x2−x−6dx\int_4^6 \frac{x+1}{x^2-x-6} dx. Their work is shown below.

Step 1: Decompose the integrand: x+1(x−3)(x+2)=Ax−3+Bx+2\frac{x+1}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}.

Step 2: Solve for coefficients: x+1=A(x+2)+B(x−3)x+1 = A(x+2) + B(x-3). This yields A=4/5A=4/5 and B=1/5B=1/5.

Step 3: Integrate: ∫46(4/5x−3+1/5x+2)dx=[45ln⁡(x−3)+15ln⁡(x+2)]46\int_4^6 (\frac{4/5}{x-3} + \frac{1/5}{x+2}) dx = [\frac{4}{5}\ln(x-3) + \frac{1}{5}\ln(x+2)]_4^6.

Step 4: Evaluate: (45ln⁡3+15ln⁡8)−(45ln⁡1+15ln⁡6)=45ln⁡3+35ln⁡2−15ln⁡6(\frac{4}{5}\ln 3 + \frac{1}{5}\ln 8) - (\frac{4}{5}\ln 1 + \frac{1}{5}\ln 6) = \frac{4}{5}\ln 3 + \frac{3}{5}\ln 2 - \frac{1}{5}\ln 6.

In which step did the student make their first error?

  1. Step 1
  2. Step 2
  3. Step 3 (correct answer)
  4. Step 4
Explanation: The first error occurs in Step 3. The general antiderivative of 1x−a\frac{1}{x-a} is ln⁡∣x−a∣\ln|x-a|, not ln⁡(x−a)\ln(x-a). While the arguments of the logarithms are positive on the interval of integration [4,6][4, 6], making the final numerical result correct, the written expression for the antiderivative is mathematically imprecise and incorrect. An antiderivative must be valid over the function's entire domain. For example, using this antiderivative on an interval where x−3x-3 is negative would lead to taking the logarithm of a negative number. Step 1 is correct; the factorization and decomposition form are correct. Step 2 is correct; the coefficients are calculated properly. Step 4 correctly evaluates the (incorrect) expression from Step 3.

Question 9

After performing partial fraction decomposition on 4x+1x2−1\frac{4x + 1}{x^2 - 1}, the resulting integral becomes ∫(Ax−1+Bx+1)dx\int \left(\frac{A}{x-1} + \frac{B}{x+1}\right) dx. Which of the following represents the antiderivative?

  1. 52ln⁡∣x−1∣−32ln⁡∣x+1∣+C\frac{5}{2} \ln|x - 1| - \frac{3}{2} \ln|x + 1| + C
  2. 32ln⁡∣x−1∣+52ln⁡∣x+1∣+C\frac{3}{2} \ln|x - 1| + \frac{5}{2} \ln|x + 1| + C
  3. 52ln⁡∣x−1∣+32ln⁡∣x+1∣+C\frac{5}{2} \ln|x - 1| + \frac{3}{2} \ln|x + 1| + C (correct answer)
  4. 32ln⁡∣x−1∣−52ln⁡∣x+1∣+C\frac{3}{2} \ln|x - 1| - \frac{5}{2} \ln|x + 1| + C
Explanation: First, decompose 4x+1x2−1=4x+1(x−1)(x+1)=Ax−1+Bx+1\frac{4x + 1}{x^2 - 1} = \frac{4x + 1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}. Multiplying by (x−1)(x+1)(x-1)(x+1): 4x+1=A(x+1)+B(x−1)4x + 1 = A(x+1) + B(x-1). Setting x=1x = 1: 5=2A5 = 2A, so A=52A = \frac{5}{2}. Setting x=−1x = -1: −3=−2B-3 = -2B, so B=32B = \frac{3}{2}. Therefore: ∫4x+1x2−1dx=∫(5/2x−1+3/2x+1)dx=52ln⁡∣x−1∣+32ln⁡∣x+1∣+C\int \frac{4x + 1}{x^2 - 1} dx = \int \left(\frac{5/2}{x-1} + \frac{3/2}{x+1}\right) dx = \frac{5}{2} \ln|x-1| + \frac{3}{2} \ln|x+1| + C.

Question 10

Consider the integral ∫x+2x2−3x+2 dx\int \frac{x + 2}{x^2 - 3x + 2} \, dx. After factoring the denominator and setting up partial fractions, what is the coefficient of ln⁡∣x−1∣\ln|x - 1| in the final antiderivative?

  1. −1-1
  2. −3-3 (correct answer)
  3. 11
  4. 33
Explanation: When you encounter a rational function where the degree of the numerator is less than the degree of the denominator, partial fraction decomposition is your go-to technique. This method breaks down complex fractions into simpler ones that are easier to integrate. First, factor the denominator: x2−3x+2=(x−1)(x−2)x^2 - 3x + 2 = (x-1)(x-2). Now set up partial fractions: x+2(x−1)(x−2)=Ax−1+Bx−2\frac{x+2}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} To find the coefficients, multiply both sides by (x−1)(x−2)(x-1)(x-2): x+2=A(x−2)+B(x−1)x+2 = A(x-2) + B(x-1) Using substitution: when x=1x = 1, we get 3=A(−1)3 = A(-1), so A=−3A = -3. When x=2x = 2, we get 4=B(1)4 = B(1), so B=4B = 4. Therefore: x+2(x−1)(x−2)=−3x−1+4x−2\frac{x+2}{(x-1)(x-2)} = \frac{-3}{x-1} + \frac{4}{x-2} Integrating: ∫x+2x2−3x+2dx=∫−3x−1dx+∫4x−2dx=−3ln⁡∣x−1∣+4ln⁡∣x−2∣+C\int \frac{x+2}{x^2-3x+2} dx = \int \frac{-3}{x-1} dx + \int \frac{4}{x-2} dx = -3\ln|x-1| + 4\ln|x-2| + C The coefficient of ln⁡∣x−1∣\ln|x-1| is −3-3, making (B) correct. Choice (A) gives −1-1, likely from incorrectly setting up the partial fractions. Choice (C) gives 11, probably from a sign error in the coefficient calculation. Choice (D) gives 33, which would result from forgetting the negative sign when A=−3A = -3. Study tip: Always verify your partial fraction setup by expanding back to check your coefficients before integrating. The substitution method for finding coefficients is usually the most reliable approach.

Question 11

A student attempts to decompose 4x−1x(x−3)\frac{4x - 1}{x(x - 3)} and writes 4x−1x(x−3)=Ax+Bx−3\frac{4x - 1}{x(x - 3)} = \frac{A}{x} + \frac{B}{x - 3} where A=13A = \frac{1}{3} and B=113B = \frac{11}{3}. To verify these values, which of the following checks should yield the identity 4x−1≡4x−14x - 1 \equiv 4x - 1?

  1. 13(x−3)+113(x)=13x−1+113x=4x−1\frac{1}{3}(x - 3) + \frac{11}{3}(x) = \frac{1}{3}x - 1 + \frac{11}{3}x = 4x - 1
  2. 13(x)+113(x−3)=13x+113x−11=4x−11\frac{1}{3}(x) + \frac{11}{3}(x - 3) = \frac{1}{3}x + \frac{11}{3}x - 11 = 4x - 11
  3. 13(x)+113(x−3)=13x+113x−333=4x−11\frac{1}{3}(x) + \frac{11}{3}(x - 3) = \frac{1}{3}x + \frac{11}{3}x - \frac{33}{3} = 4x - 11
  4. 13(x−3)+113(x)=13x−1+113x=123x−1=4x−1\frac{1}{3}(x - 3) + \frac{11}{3}(x) = \frac{1}{3}x - 1 + \frac{11}{3}x = \frac{12}{3}x - 1 = 4x - 1 (correct answer)
Explanation: When you encounter partial fraction decomposition verification, you're checking whether your constants produce the original numerator when you clear denominators. The key insight is understanding what each term represents after multiplying through. To verify the decomposition 4x−1x(x−3)=Ax+Bx−3\frac{4x - 1}{x(x - 3)} = \frac{A}{x} + \frac{B}{x - 3}, you multiply both sides by the common denominator x(x−3)x(x - 3). This gives you: 4x−1=A(x−3)+B(x)4x - 1 = A(x - 3) + B(x). Substituting A=13A = \frac{1}{3} and B=113B = \frac{11}{3}, you need: 4x−1=13(x−3)+113(x)4x - 1 = \frac{1}{3}(x - 3) + \frac{11}{3}(x). Answer choice D correctly sets up this verification. Working through: 13(x−3)+113(x)=13x−1+113x=13x+113x−1=123x−1=4x−1\frac{1}{3}(x - 3) + \frac{11}{3}(x) = \frac{1}{3}x - 1 + \frac{11}{3}x = \frac{1}{3}x + \frac{11}{3}x - 1 = \frac{12}{3}x - 1 = 4x - 1. This matches the original numerator perfectly. Answer choice A makes an arithmetic error: 13×(−3)=−1\frac{1}{3} \times (-3) = -1, but the calculation should continue to show 123x=4x\frac{12}{3}x = 4x, not stop at the intermediate step. Choices B and C reverse the terms, writing A(x)+B(x−3)A(x) + B(x-3) instead of A(x−3)+B(x)A(x-3) + B(x). This fundamental error leads to 4x−114x - 11 instead of 4x−14x - 1. Study tip: Always write out the full algebraic identity numerator=A×(denominator missing first factor)+B×(denominator missing second factor)\text{numerator} = A \times \text{(denominator missing first factor)} + B \times \text{(denominator missing second factor)} before substituting values. This prevents term reversal errors.

Question 12

The velocity of a particle is given by v(t)=1025−t2v(t) = \frac{10}{25-t^2} for t∈[0,4]t \in [0, 4]. If the particle's position at t=0t=0 is s(0)=ln⁡(5)s(0) = \ln(5), what is its position at t=4t=4?

  1. 3ln⁡33\ln 3
  2. ln⁡(45)\ln(45) (correct answer)
  3. 2ln⁡32\ln 3
  4. ln⁡(9/5)\ln(9/5)
Explanation: Position s(t)s(t) is the integral of velocity v(t)v(t). We need to calculate ∫1025−t2dt\int \frac{10}{25-t^2} dt. Using partial fractions: 10(5−t)(5+t)=A5−t+B5+t\frac{10}{(5-t)(5+t)} = \frac{A}{5-t} + \frac{B}{5+t}. Solving for coefficients: A=105+5=1A = \frac{10}{5+5} = 1 and B=105−(−5)=1B = \frac{10}{5-(-5)} = 1. So, s(t)=∫(15−t+15+t)dt=−ln⁡∣5−t∣+ln⁡∣5+t∣+C=ln⁡∣5+t5−t∣+Cs(t) = \int (\frac{1}{5-t} + \frac{1}{5+t}) dt = -\ln|5-t| + \ln|5+t| + C = \ln|\frac{5+t}{5-t}| + C. Using the initial condition s(0)=ln⁡5s(0)=\ln 5: ln⁡5=ln⁡∣55∣+C=ln⁡(1)+C=C\ln 5 = \ln|\frac{5}{5}| + C = \ln(1) + C = C. So C=ln⁡5C=\ln 5. The position function is s(t)=ln⁡∣5+t5−t∣+ln⁡5s(t) = \ln|\frac{5+t}{5-t}| + \ln 5. At t=4t=4, the position is s(4)=ln⁡∣5+45−4∣+ln⁡5=ln⁡(9)+ln⁡(5)=ln⁡(45)s(4) = \ln|\frac{5+4}{5-4}| + \ln 5 = \ln(9) + \ln(5) = \ln(45). (A) is the value of s(4)s(4) if C=0C=0. (C) results from a factor error in the integration. (D) results from subtracting ln⁡5\ln 5 instead of adding.

Question 13

To evaluate ∫3x+7(x−1)(x+2) dx\int \frac{3x + 7}{(x - 1)(x + 2)} \, dx using partial fractions, which of the following represents the correct decomposition and subsequent integration?

  1. 103ln⁡∣x−1∣−13ln⁡∣x+2∣+C\frac{10}{3} \ln|x - 1| - \frac{1}{3} \ln|x + 2| + C (correct answer)
  2. 103ln⁡∣x+2∣−13ln⁡∣x−1∣+C\frac{10}{3} \ln|x + 2| - \frac{1}{3} \ln|x - 1| + C
  3. 13ln⁡∣x−1∣+103ln⁡∣x+2∣+C\frac{1}{3} \ln|x - 1| + \frac{10}{3} \ln|x + 2| + C
  4. 3ln⁡∣x−1∣+7ln⁡∣x+2∣+C3 \ln|x - 1| + 7 \ln|x + 2| + C
Explanation: Setting up partial fractions: 3x+7(x−1)(x+2)=Ax−1+Bx+2\frac{3x + 7}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}. Multiplying both sides by (x−1)(x+2)(x - 1)(x + 2): 3x+7=A(x+2)+B(x−1)3x + 7 = A(x + 2) + B(x - 1). Setting x=1x = 1: 10=3A10 = 3A, so A=103A = \frac{10}{3}. Setting x=−2x = -2: 1=−3B1 = -3B, so B=−13B = -\frac{1}{3}. Therefore: ∫3x+7(x−1)(x+2) dx=∫10/3x−1 dx+∫−1/3x+2 dx=103ln⁡∣x−1∣−13ln⁡∣x+2∣+C\int \frac{3x + 7}{(x - 1)(x + 2)} \, dx = \int \frac{10/3}{x - 1} \, dx + \int \frac{-1/3}{x + 2} \, dx = \frac{10}{3} \ln|x - 1| - \frac{1}{3} \ln|x + 2| + C. Choice B swaps the coefficients, choice C has incorrect signs, and choice D uses the original numerator coefficients without proper decomposition.

Question 14

Evaluate the definite integral ∫23x+5x2+x−2dx\int_2^3 \frac{x+5}{x^2+x-2} dx

  1. ln⁡(16/5)\ln(16/5) (correct answer)
  2. ln⁡(5/16)\ln(5/16)
  3. ln⁡(25/32)\ln(25/32)
  4. ln⁡(32/25)\ln(32/25)
Explanation: First, factor the denominator: x2+x−2=(x+2)(x−1)x^2+x-2 = (x+2)(x-1). The integrand can be decomposed using partial fractions: x+5(x+2)(x−1)=Ax+2+Bx−1\frac{x+5}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}. To find the coefficients, we use the equation x+5=A(x−1)+B(x+2)x+5 = A(x-1) + B(x+2). Setting x=1x=1 gives 6=3B6 = 3B, so B=2B=2. Setting x=−2x=-2 gives 3=−3A3 = -3A, so A=−1A=-1. The integral becomes ∫23(−1x+2+2x−1)dx\int_2^3 (\frac{-1}{x+2} + \frac{2}{x-1}) dx. Evaluating this gives [−ln⁡∣x+2∣+2ln⁡∣x−1∣]23[-\ln|x+2| + 2\ln|x-1|]_2^3. Plugging in the limits: (−ln⁡(5)+2ln⁡(2))−(−ln⁡(4)+2ln⁡(1))=−ln⁡(5)+2ln⁡(2)+ln⁡(4)=−ln⁡(5)+2ln⁡(2)+2ln⁡(2)=4ln⁡(2)−ln⁡(5)=ln⁡(24)−ln⁡(5)=ln⁡(16/5)(-\ln(5) + 2\ln(2)) - (-\ln(4) + 2\ln(1)) = -\ln(5) + 2\ln(2) + \ln(4) = -\ln(5) + 2\ln(2) + 2\ln(2) = 4\ln(2) - \ln(5) = \ln(2^4) - \ln(5) = \ln(16/5). (B) results from a sign error during the evaluation of the definite integral. (C) results from swapping the coefficients A and B. (D) is the reciprocal of (C).

Question 15

The partial fraction decomposition of 3x2−7x+1x3−3x2+2x\frac{3x^2 - 7x + 1}{x^3 - 3x^2 + 2x} has the form Ax+Bx−1+Cx−2\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2}. What is the value of A+B+CA+B+C?

  1. 33 (correct answer)
  2. −7-7
  3. −3-3
  4. 11
Explanation: To find A+B+CA+B+C, one can either solve for each coefficient individually or use a shortcut. The shortcut involves recombining the partial fraction form: A(x−1)(x−2)+Bx(x−2)+Cx(x−1)x(x−1)(x−2)\frac{A(x-1)(x-2) + Bx(x-2) + Cx(x-1)}{x(x-1)(x-2)}. The numerator is A(x2−3x+2)+B(x2−2x)+C(x2−x)A(x^2-3x+2) + B(x^2-2x) + C(x^2-x). The coefficient of the x2x^2 term is A+B+CA+B+C. By equating this to the coefficient of the x2x^2 term in the original numerator, 3x2−7x+13x^2 - 7x + 1, we find that A+B+C=3A+B+C = 3. Alternatively, solving for each coefficient: A=1(−1)(−2)=1/2A = \frac{1}{(-1)(-2)} = 1/2. B=3−7+11(−1)=−3−1=3B = \frac{3-7+1}{1(-1)} = \frac{-3}{-1} = 3. C=3(4)−7(2)+12(1)=12−14+12=−1/2C = \frac{3(4)-7(2)+1}{2(1)} = \frac{12-14+1}{2} = -1/2. Then A+B+C=1/2+3−1/2=3A+B+C = 1/2 + 3 - 1/2 = 3. (B) is the coefficient of the x-term. (C) would be the result if there was a sign error in calculating B. (D) is the constant term.

Question 16

Evaluate ∫14x2−9dx\int \frac{1}{4x^2 - 9} dx.

  1. 112ln⁡∣2x−32x+3∣+C\frac{1}{12} \ln|\frac{2x-3}{2x+3}| + C (correct answer)
  2. 16ln⁡∣2x−32x+3∣+C\frac{1}{6} \ln|\frac{2x-3}{2x+3}| + C
  3. 112ln⁡∣4x2−9∣+C\frac{1}{12} \ln|4x^2-9| + C
  4. 16arctan⁡(2x3)+C\frac{1}{6} \arctan(\frac{2x}{3}) + C
Explanation: The denominator is a difference of squares, (2x)2−32=(2x−3)(2x+3)(2x)^2 - 3^2 = (2x-3)(2x+3). The partial fraction decomposition is 1(2x−3)(2x+3)=A2x−3+B2x+3\frac{1}{(2x-3)(2x+3)} = \frac{A}{2x-3} + \frac{B}{2x+3}. Solving 1=A(2x+3)+B(2x−3)1 = A(2x+3) + B(2x-3) gives A=1/6A=1/6 and B=−1/6B=-1/6. The integral is ∫(1/62x−3−1/62x+3)dx\int (\frac{1/6}{2x-3} - \frac{1/6}{2x+3}) dx. When integrating, remember the chain rule: ∫1ax+bdx=1aln⁡∣ax+b∣\int \frac{1}{ax+b} dx = \frac{1}{a}\ln|ax+b|. So, the result is 16(12ln⁡∣2x−3∣)−16(12ln⁡∣2x+3∣)+C=112(ln⁡∣2x−3∣−ln⁡∣2x+3∣)+C=112ln⁡∣2x−32x+3∣+C\frac{1}{6}(\frac{1}{2}\ln|2x-3|) - \frac{1}{6}(\frac{1}{2}\ln|2x+3|) + C = \frac{1}{12}(\ln|2x-3| - \ln|2x+3|) + C = \frac{1}{12} \ln|\frac{2x-3}{2x+3}| + C. (B) is a common error from forgetting the factor of 1/2 from the chain rule when integrating. (C) results from a sign error on the coefficient B. (D) is the result of incorrectly applying the arctan integration formula, which is for denominators of the form x2+a2x^2+a^2.

Question 17

The partial fraction decomposition of 4x2−3x−4x3−x\frac{4x^2-3x-4}{x^3-x} is given by Ax+Bx−1+Cx+1\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}. What is the value of the coefficient B?

  1. 44
  2. 3/23/2
  3. −3/2-3/2 (correct answer)
  4. 00
Explanation: The denominator factors as x(x−1)(x+1)x(x-1)(x+1). To find the coefficient B, we can use the Heaviside cover-up method. Multiply the entire rational function by (x−1)(x-1) and evaluate at x=1x=1: B=4x2−3x−4x(x+1)∣x=1=4(1)2−3(1)−41(1+1)=4−3−42=−32B = \frac{4x^2-3x-4}{x(x+1)} |_{x=1} = \frac{4(1)^2-3(1)-4}{1(1+1)} = \frac{4-3-4}{2} = \frac{-3}{2}. (A) is the value of A. (B) is the value of C. (D) would be the sum of B and C, which could be mistakenly calculated if a sign error was made.

Question 18

Let f(x)=1x2−a2f(x) = \frac{1}{x^2-a^2} where aa is a positive constant. Which of the following represents ∫f(x)dx\int f(x) dx?

  1. 12aln⁡∣x+ax−a∣+C\frac{1}{2a}\ln|\frac{x+a}{x-a}| + C
  2. 1aln⁡∣x−ax+a∣+C\frac{1}{a}\ln|\frac{x-a}{x+a}| + C
  3. 12aln⁡∣x−ax+a∣+C\frac{1}{2a}\ln|\frac{x-a}{x+a}| + C (correct answer)
  4. 1aarctan⁡(xa)+C\frac{1}{a}\arctan(\frac{x}{a}) + C
Explanation: When you encounter a rational function with a difference of squares in the denominator, partial fraction decomposition is your go-to technique. The expression 1x2−a2\frac{1}{x^2-a^2} factors as 1(x−a)(x+a)\frac{1}{(x-a)(x+a)}, which can be decomposed into partial fractions. Setting up the decomposition: 1(x−a)(x+a)=Ax−a+Bx+a\frac{1}{(x-a)(x+a)} = \frac{A}{x-a} + \frac{B}{x+a} Multiplying both sides by (x−a)(x+a)(x-a)(x+a): 1=A(x+a)+B(x−a)1 = A(x+a) + B(x-a) To find the constants, substitute strategic values. When x=ax = a: 1=A(2a)1 = A(2a), so A=12aA = \frac{1}{2a}. When x=−ax = -a: 1=B(−2a)1 = B(-2a), so B=−12aB = -\frac{1}{2a}. Therefore: 1x2−a2=12a(1x−a−1x+a)\frac{1}{x^2-a^2} = \frac{1}{2a}\left(\frac{1}{x-a} - \frac{1}{x+a}\right) Integrating: ∫1x2−a2dx=12a[ln⁡∣x−a∣−ln⁡∣x+a∣]+C=12aln⁡∣x−ax+a∣+C\int \frac{1}{x^2-a^2}dx = \frac{1}{2a}[\ln|x-a| - \ln|x+a|] + C = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C This confirms answer C is correct. Answer A has the correct coefficient 12a\frac{1}{2a} but incorrectly flips the fraction inside the logarithm. Answer B has the right fraction but wrong coefficient (missing the factor of 2). Answer D incorrectly applies the arctangent formula, which is used for 1x2+a2\frac{1}{x^2+a^2}, not 1x2−a2\frac{1}{x^2-a^2}. Study tip: Remember that x2−a2x^2-a^2 (difference) leads to partial fractions and logarithms, while x2+a2x^2+a^2 (sum) leads to arctangent. The sign makes all the difference in your integration approach.

Question 19

A rational function has a partial fraction decomposition of 3x−1x+2\frac{3}{x} - \frac{1}{x+2}. What is the value of this function at x=1x=1?

  1. 22
  2. 8/38/3 (correct answer)
  3. 33
  4. 2/32/3
Explanation: This question requires two steps: first, combine the fractions into a single rational function, and second, evaluate it at x=1x=1. The function is f(x)=3x−1x+2f(x) = \frac{3}{x} - \frac{1}{x+2}. Evaluating directly at x=1x=1 is simpler: f(1)=31−11+2=3−13=93−13=83f(1) = \frac{3}{1} - \frac{1}{1+2} = 3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}. Alternatively, one could combine the fractions first: 3(x+2)−1(x)x(x+2)=3x+6−xx(x+2)=2x+6x(x+2)\frac{3(x+2) - 1(x)}{x(x+2)} = \frac{3x+6-x}{x(x+2)} = \frac{2x+6}{x(x+2)}. Evaluating this at x=1x=1 gives 2(1)+61(1+2)=83\frac{2(1)+6}{1(1+2)} = \frac{8}{3}. (A) might result from a calculation error 3−1=23 - 1 = 2. (C) is the value of only the first term. (D) might result from an algebraic error when combining the fractions, such as 3−11(3)=2/3\frac{3-1}{1(3)} = 2/3.

Question 20

Which of the following integrals is most suitable for evaluation using the method of linear partial fractions after an appropriate substitution?

  1. ∫2x+1x2+x+1dx\int \frac{2x+1}{x^2+x+1} dx
  2. ∫1x(x−4)dx\int \frac{1}{\sqrt{x}(x-4)} dx (correct answer)
  3. ∫x3x2+1dx\int \frac{x^3}{x^2+1} dx
  4. ∫ln⁡(x2−4)dx\int \ln(x^2-4) dx
Explanation: (B) After the substitution u=xu = \sqrt{x}, we have x=u2x=u^2 and dx=2ududx = 2u du. The integral transforms into ∫2uduu(u2−4)=∫2(u−2)(u+2)du\int \frac{2u du}{u(u^2-4)} = \int \frac{2}{(u-2)(u+2)} du, which can be solved using partial fractions with distinct linear factors. (A) The numerator is the derivative of the denominator, so a simple substitution u=x2+x+1u=x^2+x+1 works. Also, the denominator is an irreducible quadratic, so partial fractions with linear factors would not apply. (C) This is an improper rational function. After long division, it becomes ∫(x−xx2+1)dx\int (x - \frac{x}{x^2+1}) dx, which does not require partial fractions. (D) This integral requires integration by parts.