IB Mathematics: Analysis and Approaches Quiz: Partial Fractions In Integration
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Partial Fractions In IntegrationQuestion 1 of 20

Evaluate the definite integral ∫342xx2−x−2dx\int_3^4 \frac{2x}{x^2-x-2} dx.

23ln⁡(5)\frac{2}{3}\ln(5)
23ln⁡(10)\frac{2}{3}\ln(10)
43ln⁡(5)−2ln⁡(2)\frac{4}{3}\ln(5) - 2\ln(2)
23ln⁡(80)\frac{2}{3}\ln(80)
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Partial Fractions In Integration

Practice Partial Fractions In Integration in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Partial Fractions In Integration, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

Evaluate the definite integral ∫342xx2−x−2dx\int_3^4 \frac{2x}{x^2-x-2} dx.

  1. 23ln⁡(5)\frac{2}{3}\ln(5) (correct answer)
  2. 23ln⁡(10)\frac{2}{3}\ln(10)
  3. 43ln⁡(5)−2ln⁡(2)\frac{4}{3}\ln(5) - 2\ln(2)
  4. 23ln⁡(80)\frac{2}{3}\ln(80)
Explanation: First, factor the denominator: x2−x−2=(x−2)(x+1)x^2-x-2 = (x-2)(x+1). Decompose the fraction: 2x(x−2)(x+1)=Ax−2+Bx+1\frac{2x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}. This gives 2x=A(x+1)+B(x−2)2x = A(x+1) + B(x-2). If x=2x=2, 4=3A⇒A=4/34 = 3A \Rightarrow A = 4/3. If x=−1x=-1, −2=−3B⇒B=2/3-2 = -3B \Rightarrow B = 2/3. So we need to evaluate ∫34(4/3x−2+2/3x+1)dx\int_3^4 (\frac{4/3}{x-2} + \frac{2/3}{x+1}) dx. The antiderivative is [43ln⁡∣x−2∣+23ln⁡∣x+1∣]34[\frac{4}{3}\ln|x-2| + \frac{2}{3}\ln|x+1|]_3^4. Evaluating at the limits: At x=4x=4: 43ln⁡(2)+23ln⁡(5)\frac{4}{3}\ln(2) + \frac{2}{3}\ln(5). At x=3x=3: 43ln⁡(1)+23ln⁡(4)=0+23ln⁡(22)=43ln⁡(2)\frac{4}{3}\ln(1) + \frac{2}{3}\ln(4) = 0 + \frac{2}{3}\ln(2^2) = \frac{4}{3}\ln(2). Subtracting the lower limit from the upper limit: (43ln⁡(2)+23ln⁡(5))−(43ln⁡(2))=23ln⁡(5)(\frac{4}{3}\ln(2) + \frac{2}{3}\ln(5)) - (\frac{4}{3}\ln(2)) = \frac{2}{3}\ln(5).

Question 2

The integration of P(x)Q(x)\frac{P(x)}{Q(x)} by partial fractions, where P(x)P(x) and Q(x)Q(x) are polynomials, results in an antiderivative containing an arctan⁡\arctan term. Which of the following must be true about the denominator Q(x)Q(x)?

  1. Q(x)Q(x) has a repeated linear factor.
  2. Q(x)Q(x) has an irreducible quadratic factor. (correct answer)
  3. The degree of P(x)P(x) is greater than or equal to the degree of Q(x)Q(x).
  4. Q(x)Q(x) has only distinct linear factors.
Explanation: The arctan⁡\arctan function arises from integrals of the form ∫1u2+a2du\int \frac{1}{u^2+a^2} du. In the context of partial fractions, this type of integral results from a term of the form Bx+Cx2+px+q\frac{Bx+C}{x^2+px+q}, where the denominator is an irreducible quadratic factor (i.e., it has no real roots, p2−4q<0p^2-4q < 0). After completing the square, the denominator can be written in the form (x+k)2+a2(x+k)^2+a^2, which leads to an arctan⁡\arctan term upon integration. A repeated linear factor (A) leads to terms of the form C(x−a)−nC(x-a)^{-n} and logarithmic terms. An improper fraction (C) requires long division first, but this does not guarantee an arctan⁡\arctan term. Distinct linear factors (D) only produce logarithmic terms.

Question 3

The integration of P(x)Q(x)\frac{P(x)}{Q(x)} by partial fractions, where P(x)P(x) and Q(x)Q(x) are polynomials, results in an antiderivative containing an arctan⁡\arctan term. Which of the following must be true about the denominator Q(x)Q(x)?

  1. Q(x)Q(x) has a repeated linear factor.
  2. Q(x)Q(x) has an irreducible quadratic factor. (correct answer)
  3. The degree of P(x)P(x) is greater than or equal to the degree of Q(x)Q(x).
  4. Q(x)Q(x) has only distinct linear factors.
Explanation: The arctan⁡\arctan function arises from integrals of the form ∫1u2+a2du\int \frac{1}{u^2+a^2} du. In the context of partial fractions, this type of integral results from a term of the form Bx+Cx2+px+q\frac{Bx+C}{x^2+px+q}, where the denominator is an irreducible quadratic factor (i.e., it has no real roots, p2−4q<0p^2-4q < 0). After completing the square, the denominator can be written in the form (x+k)2+a2(x+k)^2+a^2, which leads to an arctan⁡\arctan term upon integration. A repeated linear factor (A) leads to terms of the form C(x−a)−nC(x-a)^{-n} and logarithmic terms. An improper fraction (C) requires long division first, but this does not guarantee an arctan⁡\arctan term. Distinct linear factors (D) only produce logarithmic terms.

Question 4

Which of the following integrals requires polynomial long division before applying partial fraction decomposition?

  1. ∫x2+1x3−1dx\int \frac{x^2+1}{x^3-1} dx
  2. ∫x3+1x3−1dx\int \frac{x^3+1}{x^3-1} dx (correct answer)
  3. ∫x+1(x2−1)(x+2)dx\int \frac{x+1}{(x^2-1)(x+2)} dx
  4. ∫1x3−x2dx\int \frac{1}{x^3-x^2} dx
Explanation: Polynomial long division is required when the degree of the numerator is greater than or equal to the degree of the denominator (i.e., the rational function is improper). A: Degree of numerator (2) < degree of denominator (3). Not required. B: Degree of numerator (3) is equal to degree of denominator (3). Required. C: Degree of numerator (1) < degree of denominator (3). Not required. D: Degree of numerator (0) < degree of denominator (3). Not required. Therefore, only the integral in option B requires long division before proceeding with partial fractions.

Question 5

Which expression is equivalent to ∫x3+2x2x2−1dx\int \frac{x^3+2x^2}{x^2-1} dx?

  1. 12x2+2x+12ln⁡∣x2−1∣+C\frac{1}{2}x^2+2x + \frac{1}{2}\ln|x^2-1| + C
  2. 12x2+2x+32ln⁡∣x−1∣+12ln⁡∣x+1∣+C\frac{1}{2}x^2+2x + \frac{3}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + C
  3. 12x2+2x+32ln⁡∣x−1∣−12ln⁡∣x+1∣+C\frac{1}{2}x^2+2x + \frac{3}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C (correct answer)
  4. (x4/4)+(2x3/3)(x3/3)−x+C\frac{(x^4/4)+ (2x^3/3)}{(x^3/3)-x} + C
Explanation: The integrand is an improper rational function, so first perform polynomial long division. Dividing x3+2x2x^3+2x^2 by x2−1x^2-1 gives a quotient of x+2x+2 and a remainder of x+2x+2. So, x3+2x2x2−1=x+2+x+2x2−1\frac{x^3+2x^2}{x^2-1} = x+2 + \frac{x+2}{x^2-1}. Now, decompose the remainder term: x+2(x−1)(x+1)=Ax−1+Bx+1\frac{x+2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}. This gives x+2=A(x+1)+B(x−1)x+2 = A(x+1) + B(x-1). If x=1x=1, 3=2A⇒A=3/23 = 2A \Rightarrow A=3/2. If x=−1x=-1, 1=−2B⇒B=−1/21 = -2B \Rightarrow B=-1/2. The integral becomes ∫(x+2+3/2x−1−1/2x+1)dx\int (x+2 + \frac{3/2}{x-1} - \frac{1/2}{x+1}) dx. Integrating gives 12x2+2x+32ln⁡∣x−1∣−12ln⁡∣x+1∣+C\frac{1}{2}x^2+2x + \frac{3}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C.

Question 6

Find the indefinite integral ∫4x2−4dx\int \frac{4}{x^2-4} dx.

  1. ln⁡∣x2−4∣+C\ln|x^2-4| + C
  2. ln⁡∣x+2x−2∣+C\ln|\frac{x+2}{x-2}| + C
  3. ln⁡∣x−2x+2∣+C\ln|\frac{x-2}{x+2}| + C (correct answer)
  4. 4ln⁡∣x2−4∣+C4\ln|x^2-4| + C
Explanation: First, decompose the integrand into partial fractions. The denominator factors as x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2). Let 4(x−2)(x+2)=Ax−2+Bx+2\frac{4}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}. Multiplying by the denominator gives 4=A(x+2)+B(x−2)4 = A(x+2) + B(x-2). To find A, let x=2x=2: 4=A(2+2)⇒4=4A⇒A=14 = A(2+2) \Rightarrow 4 = 4A \Rightarrow A=1. To find B, let x=−2x=-2: 4=B(−2−2)⇒4=−4B⇒B=−14 = B(-2-2) \Rightarrow 4 = -4B \Rightarrow B=-1. So the integral becomes ∫(1x−2−1x+2)dx\int (\frac{1}{x-2} - \frac{1}{x+2}) dx. Integrating term by term gives ln⁡∣x−2∣−ln⁡∣x+2∣+C\ln|x-2| - \ln|x+2| + C. Using the logarithm property ln⁡a−ln⁡b=ln⁡(a/b)\ln a - \ln b = \ln(a/b), the result is ln⁡∣x−2x+2∣+C\ln|\frac{x-2}{x+2}| + C.

Question 7

Using the substitution u=exu = e^x, find ∫exe2x−3ex+2dx\int \frac{e^x}{e^{2x} - 3e^x + 2} dx.

  1. ln⁡∣ex−1ex−2∣+C\ln|\frac{e^x-1}{e^x-2}| + C
  2. ln⁡∣ex−2ex−1∣+C\ln|\frac{e^x-2}{e^x-1}| + C (correct answer)
  3. ln⁡∣e2x−3ex+2∣+C\ln|e^{2x} - 3e^x + 2| + C
  4. arctan⁡(ex−3/2)+C\arctan(e^x - 3/2) + C
Explanation: Let u=exu = e^x. Then du=exdxdu = e^x dx and e2x=u2e^{2x} = u^2. The integral transforms to ∫1u2−3u+2du=∫1(u−1)(u−2)du\int \frac{1}{u^2-3u+2} du = \int \frac{1}{(u-1)(u-2)} du. Decompose the new integrand: 1(u−1)(u−2)=Au−1+Bu−2\frac{1}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}. 1=A(u−2)+B(u−1)1 = A(u-2) + B(u-1). If u=1u=1, 1=−A⇒A=−11 = -A \Rightarrow A=-1. If u=2u=2, 1=B1 = B. The integral in uu is ∫(−1u−1+1u−2)du=−ln⁡∣u−1∣+ln⁡∣u−2∣+C=ln⁡∣u−2u−1∣+C\int (\frac{-1}{u-1} + \frac{1}{u-2}) du = -\ln|u-1| + \ln|u-2| + C = \ln|\frac{u-2}{u-1}| + C. Substituting back u=exu = e^x gives ln⁡∣ex−2ex−1∣+C\ln|\frac{e^x-2}{e^x-1}| + C.

Question 8

Given that ∫ax+bx2−3x−4dx=2ln⁡∣x−4∣+ln⁡∣x+1∣+C\int \frac{ax+b}{x^2-3x-4} dx = 2\ln|x-4| + \ln|x+1| + C, find the value of aa.

  1. -1
  2. 1
  3. 2
  4. 3 (correct answer)
Explanation: The result of the integration tells us about the partial fraction decomposition of the integrand. The integral is ∫(2x−4+1x+1)dx\int (\frac{2}{x-4} + \frac{1}{x+1}) dx. To find the original integrand ax+bx2−3x−4\frac{ax+b}{x^2-3x-4}, we combine the fractions: 2x−4+1x+1=2(x+1)+1(x−4)(x−4)(x+1)\frac{2}{x-4} + \frac{1}{x+1} = \frac{2(x+1) + 1(x-4)}{(x-4)(x+1)}. =2x+2+x−4x2−3x−4=3x−2x2−3x−4= \frac{2x+2+x-4}{x^2-3x-4} = \frac{3x-2}{x^2-3x-4}. By comparing 3x−2x2−3x−4\frac{3x-2}{x^2-3x-4} with ax+bx2−3x−4\frac{ax+b}{x^2-3x-4}, we can see that a=3a=3 and b=−2b=-2. The question asks for the value of aa, which is 3.

Question 9

Find ∫2(x−1)(x2+1)dx\int \frac{2}{(x-1)(x^2+1)} dx.

  1. ln⁡∣x−1∣−12ln⁡(x2+1)−arctan⁡(x)+C\ln|x-1| - \frac{1}{2}\ln(x^2+1) - \arctan(x) + C (correct answer)
  2. ln⁡∣x−1∣+12ln⁡(x2+1)+arctan⁡(x)+C\ln|x-1| + \frac{1}{2}\ln(x^2+1) + \arctan(x) + C
  3. ln⁡∣x−1∣−ln⁡(x2+1)−arctan⁡(x)+C\ln|x-1| - \ln(x^2+1) - \arctan(x) + C
  4. ln⁡∣x−1∣−12ln⁡(x2+1)+arctan⁡(x)+C\ln|x-1| - \frac{1}{2}\ln(x^2+1) + \arctan(x) + C
Explanation: Decompose the integrand: 2(x−1)(x2+1)=Ax−1+Bx+Cx2+1\frac{2}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}. 2=A(x2+1)+(Bx+C)(x−1)2 = A(x^2+1) + (Bx+C)(x-1). Let x=1x=1: 2=A(12+1)⇒2=2A⇒A=12 = A(1^2+1) \Rightarrow 2 = 2A \Rightarrow A=1. Substitute A=1A=1 and expand: 2=(x2+1)+(Bx+C)(x−1)=x2+1+Bx2−Bx+Cx−C2 = (x^2+1) + (Bx+C)(x-1) = x^2+1+Bx^2-Bx+Cx-C. 2=(1+B)x2+(−B+C)x+(1−C)2 = (1+B)x^2 + (-B+C)x + (1-C). Comparing coefficients of x2x^2: 0=1+B⇒B=−10 = 1+B \Rightarrow B=-1. Comparing constant terms: 2=1−C⇒C=−12 = 1-C \Rightarrow C=-1. The integral is ∫(1x−1+−x−1x2+1)dx=∫1x−1dx−∫xx2+1dx−∫1x2+1dx\int (\frac{1}{x-1} + \frac{-x-1}{x^2+1}) dx = \int \frac{1}{x-1}dx - \int \frac{x}{x^2+1}dx - \int \frac{1}{x^2+1}dx. Integrating each part gives ln⁡∣x−1∣−12ln⁡(x2+1)−arctan⁡(x)+C\ln|x-1| - \frac{1}{2}\ln(x^2+1) - \arctan(x) + C.

Question 10

The integral ∫2x2+4x(x2+4)dx\int \frac{2x^2+4}{x(x^2+4)} dx is equal to:

  1. ln⁡∣x∣+arctan⁡(x/2)+C\ln|x| + \arctan(x/2) + C
  2. ln⁡∣x∣−12ln⁡(x2+4)+C\ln|x| - \frac{1}{2}\ln(x^2+4) + C
  3. ln⁡∣x∣+12ln⁡(x2+4)+C\ln|x| + \frac{1}{2}\ln(x^2+4) + C (correct answer)
  4. ln⁡∣x(x2+4)∣+C\ln|x(x^2+4)| + C
Explanation: Decompose the fraction: 2x2+4x(x2+4)=Ax+Bx+Cx2+4\frac{2x^2+4}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4}. 2x2+4=A(x2+4)+(Bx+C)x=(A+B)x2+Cx+4A2x^2+4 = A(x^2+4) + (Bx+C)x = (A+B)x^2+Cx+4A. Comparing coefficients: Constant term: 4=4A⇒A=14=4A \Rightarrow A=1. Coefficient of xx: C=0C=0. Coefficient of x2x^2: 2=A+B⇒2=1+B⇒B=12=A+B \Rightarrow 2=1+B \Rightarrow B=1. The integral becomes ∫(1x+xx2+4)dx\int (\frac{1}{x} + \frac{x}{x^2+4}) dx. =∫1xdx+∫xx2+4dx= \int \frac{1}{x} dx + \int \frac{x}{x^2+4} dx. This evaluates to ln⁡∣x∣+12ln⁡(x2+4)+C\ln|x| + \frac{1}{2}\ln(x^2+4) + C. The second part is a u-substitution where u=x2+4u=x^2+4.

Question 11

Find ∫2(x−1)(x2+1)dx\int \frac{2}{(x-1)(x^2+1)} dx.

  1. ln⁡∣x−1∣−12ln⁡(x2+1)−arctan⁡(x)+C\ln|x-1| - \frac{1}{2}\ln(x^2+1) - \arctan(x) + C (correct answer)
  2. ln⁡∣x−1∣+12ln⁡(x2+1)+arctan⁡(x)+C\ln|x-1| + \frac{1}{2}\ln(x^2+1) + \arctan(x) + C
  3. ln⁡∣x−1∣−ln⁡(x2+1)−arctan⁡(x)+C\ln|x-1| - \ln(x^2+1) - \arctan(x) + C
  4. ln⁡∣x−1∣−12ln⁡(x2+1)+arctan⁡(x)+C\ln|x-1| - \frac{1}{2}\ln(x^2+1) + \arctan(x) + C
Explanation: Decompose the integrand: 2(x−1)(x2+1)=Ax−1+Bx+Cx2+1\frac{2}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}. 2=A(x2+1)+(Bx+C)(x−1)2 = A(x^2+1) + (Bx+C)(x-1). Let x=1x=1: 2=A(12+1)⇒2=2A⇒A=12 = A(1^2+1) \Rightarrow 2 = 2A \Rightarrow A=1. Substitute A=1A=1 and expand: 2=(x2+1)+(Bx+C)(x−1)=x2+1+Bx2−Bx+Cx−C2 = (x^2+1) + (Bx+C)(x-1) = x^2+1+Bx^2-Bx+Cx-C. 2=(1+B)x2+(−B+C)x+(1−C)2 = (1+B)x^2 + (-B+C)x + (1-C). Comparing coefficients of x2x^2: 0=1+B⇒B=−10 = 1+B \Rightarrow B=-1. Comparing constant terms: 2=1−C⇒C=−12 = 1-C \Rightarrow C=-1. The integral is ∫(1x−1+−x−1x2+1)dx=∫1x−1dx−∫xx2+1dx−∫1x2+1dx\int (\frac{1}{x-1} + \frac{-x-1}{x^2+1}) dx = \int \frac{1}{x-1}dx - \int \frac{x}{x^2+1}dx - \int \frac{1}{x^2+1}dx. Integrating each part gives ln⁡∣x−1∣−12ln⁡(x2+1)−arctan⁡(x)+C\ln|x-1| - \frac{1}{2}\ln(x^2+1) - \arctan(x) + C.

Question 12

Evaluate ∫121x3+xdx\int_1^2 \frac{1}{x^3+x} dx.

  1. 12ln⁡(25)\frac{1}{2}\ln(\frac{2}{5})
  2. ln⁡(45)\ln(\frac{4}{5})
  3. 12ln⁡(85)\frac{1}{2}\ln(\frac{8}{5}) (correct answer)
  4. 12ln⁡(10)\frac{1}{2}\ln(10)
Explanation: First, factor the denominator: x3+x=x(x2+1)x^3+x = x(x^2+1). Decompose the fraction: 1x(x2+1)=Ax+Bx+Cx2+1\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}. 1=A(x2+1)+(Bx+C)x=(A+B)x2+Cx+A1 = A(x^2+1) + (Bx+C)x = (A+B)x^2+Cx+A. Comparing coefficients: A=1A=1, C=0C=0, A+B=0⇒B=−1A+B=0 \Rightarrow B=-1. The integral is ∫12(1x−xx2+1)dx\int_1^2 (\frac{1}{x} - \frac{x}{x^2+1}) dx. The antiderivative is [ln⁡∣x∣−12ln⁡(x2+1)]12[\ln|x| - \frac{1}{2}\ln(x^2+1)]_1^2. Evaluate at the limits: At x=2x=2: ln⁡(2)−12ln⁡(5)\ln(2) - \frac{1}{2}\ln(5). At x=1x=1: ln⁡(1)−12ln⁡(2)=−12ln⁡(2)\ln(1) - \frac{1}{2}\ln(2) = -\frac{1}{2}\ln(2). Subtracting: (ln⁡(2)−12ln⁡(5))−(−12ln⁡(2))=32ln⁡(2)−12ln⁡(5)(\ln(2) - \frac{1}{2}\ln(5)) - (-\frac{1}{2}\ln(2)) = \frac{3}{2}\ln(2) - \frac{1}{2}\ln(5). Using logarithm properties: 12(3ln⁡(2)−ln⁡(5))=12(ln⁡(23)−ln⁡(5))=12ln⁡(85)\frac{1}{2}(3\ln(2) - \ln(5)) = \frac{1}{2}(\ln(2^3) - \ln(5)) = \frac{1}{2}\ln(\frac{8}{5}).

Question 13

A suitable substitution for evaluating ∫cos⁡(x)sin⁡2(x)−4dx\int \frac{\cos(x)}{\sin^2(x) - 4} dx leads to an integral in terms of uu. What is the result of this integration in terms of uu?

  1. 14ln⁡∣u−2u+2∣+C\frac{1}{4}\ln|\frac{u-2}{u+2}| + C (correct answer)
  2. 14ln⁡∣u2−4∣+C\frac{1}{4}\ln|u^2-4| + C
  3. ln⁡∣u−2u+2∣+C\ln|\frac{u-2}{u+2}| + C
  4. 12arctan⁡(u2)+C\frac{1}{2}\arctan(\frac{u}{2}) + C
Explanation: Let u=sin⁡(x)u = \sin(x), so du=cos⁡(x)dxdu = \cos(x) dx. The integral becomes ∫1u2−4du\int \frac{1}{u^2-4} du. This requires partial fractions: 1(u−2)(u+2)=Au−2+Bu+2\frac{1}{(u-2)(u+2)} = \frac{A}{u-2} + \frac{B}{u+2}. 1=A(u+2)+B(u−2)1 = A(u+2) + B(u-2). If u=2u=2, 1=4A⇒A=1/41 = 4A \Rightarrow A=1/4. If u=−2u=-2, 1=−4B⇒B=−1/41 = -4B \Rightarrow B=-1/4. The integral is ∫(1/4u−2−1/4u+2)du=14(ln⁡∣u−2∣−ln⁡∣u+2∣)+C\int (\frac{1/4}{u-2} - \frac{1/4}{u+2}) du = \frac{1}{4}(\ln|u-2| - \ln|u+2|) + C. Using logarithm properties, this simplifies to 14ln⁡∣u−2u+2∣+C\frac{1}{4}\ln|\frac{u-2}{u+2}| + C.

Question 14

Evaluate ∫e2e31x(ln⁡x)2−xln⁡xdx\int_{e^2}^{e^3} \frac{1}{x(\ln x)^2 - x \ln x} dx.

  1. ln⁡(2/3)\ln(2/3)
  2. ln⁡(3/4)\ln(3/4)
  3. ln⁡(4/3)\ln(4/3) (correct answer)
  4. ln⁡(3/2)\ln(3/2)
Explanation: First, simplify the integrand: 1x((ln⁡x)2−ln⁡x)=1xln⁡x(ln⁡x−1)\frac{1}{x((\ln x)^2 - \ln x)} = \frac{1}{x\ln x(\ln x - 1)}. Let u=ln⁡xu = \ln x. Then du=1xdxdu = \frac{1}{x} dx. The limits of integration also change: when x=e2x=e^2, u=ln⁡(e2)=2u=\ln(e^2)=2. When x=e3x=e^3, u=ln⁡(e3)=3u=\ln(e^3)=3. The integral becomes ∫231u(u−1)du\int_2^3 \frac{1}{u(u-1)} du. Decompose into partial fractions: 1u(u−1)=Au+Bu−1\frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1}. 1=A(u−1)+Bu1 = A(u-1) + Bu. If u=0u=0, 1=−A⇒A=−11 = -A \Rightarrow A=-1. If u=1u=1, 1=B1 = B. The integral is ∫23(−1u+1u−1)du=[−ln⁡∣u∣+ln⁡∣u−1∣]23=[ln⁡∣u−1u∣]23\int_2^3 (\frac{-1}{u} + \frac{1}{u-1}) du = [-\ln|u| + \ln|u-1|]_2^3 = [\ln|\frac{u-1}{u}|]_2^3. Evaluating at the limits: ln⁡(3−13)−ln⁡(2−12)=ln⁡(23)−ln⁡(12)=ln⁡(2/31/2)=ln⁡(43)\ln(\frac{3-1}{3}) - \ln(\frac{2-1}{2}) = \ln(\frac{2}{3}) - \ln(\frac{1}{2}) = \ln(\frac{2/3}{1/2}) = \ln(\frac{4}{3}).

Question 15

Evaluate the definite integral ∫342xx2−x−2dx\int_3^4 \frac{2x}{x^2-x-2} dx.

  1. 23ln⁡(5)\frac{2}{3}\ln(5) (correct answer)
  2. 23ln⁡(10)\frac{2}{3}\ln(10)
  3. 43ln⁡(5)−2ln⁡(2)\frac{4}{3}\ln(5) - 2\ln(2)
  4. 23ln⁡(80)\frac{2}{3}\ln(80)
Explanation: First, factor the denominator: x2−x−2=(x−2)(x+1)x^2-x-2 = (x-2)(x+1). Decompose the fraction: 2x(x−2)(x+1)=Ax−2+Bx+1\frac{2x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}. This gives 2x=A(x+1)+B(x−2)2x = A(x+1) + B(x-2). If x=2x=2, 4=3A⇒A=4/34 = 3A \Rightarrow A = 4/3. If x=−1x=-1, −2=−3B⇒B=2/3-2 = -3B \Rightarrow B = 2/3. So we need to evaluate ∫34(4/3x−2+2/3x+1)dx\int_3^4 (\frac{4/3}{x-2} + \frac{2/3}{x+1}) dx. The antiderivative is [43ln⁡∣x−2∣+23ln⁡∣x+1∣]34[\frac{4}{3}\ln|x-2| + \frac{2}{3}\ln|x+1|]_3^4. Evaluating at the limits: At x=4x=4: 43ln⁡(2)+23ln⁡(5)\frac{4}{3}\ln(2) + \frac{2}{3}\ln(5). At x=3x=3: 43ln⁡(1)+23ln⁡(4)=0+23ln⁡(22)=43ln⁡(2)\frac{4}{3}\ln(1) + \frac{2}{3}\ln(4) = 0 + \frac{2}{3}\ln(2^2) = \frac{4}{3}\ln(2). Subtracting the lower limit from the upper limit: (43ln⁡(2)+23ln⁡(5))−(43ln⁡(2))=23ln⁡(5)(\frac{4}{3}\ln(2) + \frac{2}{3}\ln(5)) - (\frac{4}{3}\ln(2)) = \frac{2}{3}\ln(5).

Question 16

Find ∫5x(x2+4)dx\int \frac{5}{x(x^2+4)} dx.

  1. 54ln⁡∣x∣−54ln⁡(x2+4)+C\frac{5}{4}\ln|x| - \frac{5}{4}\ln(x^2+4) + C
  2. 54ln⁡∣x∣+58ln⁡(x2+4)+C\frac{5}{4}\ln|x| + \frac{5}{8}\ln(x^2+4) + C
  3. 54ln⁡∣x∣−58ln⁡(x2+4)+C\frac{5}{4}\ln|x| - \frac{5}{8}\ln(x^2+4) + C (correct answer)
  4. 54ln⁡∣x∣−58arctan⁡(x2)+C\frac{5}{4}\ln|x| - \frac{5}{8}\arctan(\frac{x}{2}) + C
Explanation: The decomposition for the irreducible quadratic factor is 5x(x2+4)=Ax+Bx+Cx2+4\frac{5}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4}. Multiplying by the denominator gives 5=A(x2+4)+(Bx+C)x5 = A(x^2+4) + (Bx+C)x. Expanding gives 5=(A+B)x2+Cx+4A5 = (A+B)x^2 + Cx + 4A. Comparing coefficients: Constant term: 5=4A⇒A=5/45 = 4A \Rightarrow A=5/4. Coefficient of xx: 0=C0 = C. Coefficient of x2x^2: 0=A+B⇒B=−A=−5/40 = A+B \Rightarrow B = -A = -5/4. The integral is ∫(5/4x−5/4xx2+4)dx\int (\frac{5/4}{x} - \frac{5/4 x}{x^2+4}) dx. This equals 54ln⁡∣x∣−54∫xx2+4dx\frac{5}{4}\ln|x| - \frac{5}{4}\int \frac{x}{x^2+4} dx. For the second integral, use substitution u=x2+4u=x^2+4, du=2xdxdu=2x dx. The integral becomes ∫12udu=12ln⁡∣u∣=12ln⁡(x2+4)\int \frac{1}{2u} du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+4). The final result is 54ln⁡∣x∣−54(12ln⁡(x2+4))+C=54ln⁡∣x∣−58ln⁡(x2+4)+C\frac{5}{4}\ln|x| - \frac{5}{4}(\frac{1}{2}\ln(x^2+4)) + C = \frac{5}{4}\ln|x| - \frac{5}{8}\ln(x^2+4) + C.

Question 17

Using the substitution u=exu = e^x, find ∫exe2x−3ex+2dx\int \frac{e^x}{e^{2x} - 3e^x + 2} dx.

  1. ln⁡∣ex−1ex−2∣+C\ln|\frac{e^x-1}{e^x-2}| + C
  2. ln⁡∣ex−2ex−1∣+C\ln|\frac{e^x-2}{e^x-1}| + C (correct answer)
  3. ln⁡∣e2x−3ex+2∣+C\ln|e^{2x} - 3e^x + 2| + C
  4. arctan⁡(ex−3/2)+C\arctan(e^x - 3/2) + C
Explanation: Let u=exu = e^x. Then du=exdxdu = e^x dx and e2x=u2e^{2x} = u^2. The integral transforms to ∫1u2−3u+2du=∫1(u−1)(u−2)du\int \frac{1}{u^2-3u+2} du = \int \frac{1}{(u-1)(u-2)} du. Decompose the new integrand: 1(u−1)(u−2)=Au−1+Bu−2\frac{1}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}. 1=A(u−2)+B(u−1)1 = A(u-2) + B(u-1). If u=1u=1, 1=−A⇒A=−11 = -A \Rightarrow A=-1. If u=2u=2, 1=B1 = B. The integral in uu is ∫(−1u−1+1u−2)du=−ln⁡∣u−1∣+ln⁡∣u−2∣+C=ln⁡∣u−2u−1∣+C\int (\frac{-1}{u-1} + \frac{1}{u-2}) du = -\ln|u-1| + \ln|u-2| + C = \ln|\frac{u-2}{u-1}| + C. Substituting back u=exu = e^x gives ln⁡∣ex−2ex−1∣+C\ln|\frac{e^x-2}{e^x-1}| + C.

Question 18

The integral ∫2x2−3x+2x(x−1)2dx\int \frac{2x^2-3x+2}{x(x-1)^2} dx is equal to Aln⁡∣x∣+Bln⁡∣x−1∣+Cx−1+KA\ln|x| + B\ln|x-1| + \frac{C}{x-1} + K. What is the value of A+BA+B?

  1. -1
  2. 0
  3. 1
  4. 2 (correct answer)
Explanation: The partial fraction decomposition is 2x2−3x+2x(x−1)2=Ax+Bx−1+C(x−1)2\frac{2x^2-3x+2}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}. Multiply by the denominator: 2x2−3x+2=A(x−1)2+Bx(x−1)+Cx2x^2-3x+2 = A(x-1)^2 + Bx(x-1) + Cx. To find A, let x=0x=0: 2=A(−1)2⇒A=22 = A(-1)^2 \Rightarrow A=2. To find C, let x=1x=1: 2(1)2−3(1)+2=C(1)⇒1=C2(1)^2 - 3(1) + 2 = C(1) \Rightarrow 1 = C. To find B, we can compare the coefficients of the x2x^2 term on both sides. On the left, the coefficient is 2. On the right, it is A+BA+B. So, A+B=2A+B = 2. Since we are asked for A+BA+B, the value is 2. We don't even need to find B explicitly, but if we did: 2+B=2⇒B=02+B=2 \Rightarrow B=0.

Question 19

Given that ∫ax+bx2−3x−4dx=2ln⁡∣x−4∣+ln⁡∣x+1∣+C\int \frac{ax+b}{x^2-3x-4} dx = 2\ln|x-4| + \ln|x+1| + C, find the value of aa.

  1. -1
  2. 1
  3. 2
  4. 3 (correct answer)
Explanation: The result of the integration tells us about the partial fraction decomposition of the integrand. The integral is ∫(2x−4+1x+1)dx\int (\frac{2}{x-4} + \frac{1}{x+1}) dx. To find the original integrand ax+bx2−3x−4\frac{ax+b}{x^2-3x-4}, we combine the fractions: 2x−4+1x+1=2(x+1)+1(x−4)(x−4)(x+1)\frac{2}{x-4} + \frac{1}{x+1} = \frac{2(x+1) + 1(x-4)}{(x-4)(x+1)}. =2x+2+x−4x2−3x−4=3x−2x2−3x−4= \frac{2x+2+x-4}{x^2-3x-4} = \frac{3x-2}{x^2-3x-4}. By comparing 3x−2x2−3x−4\frac{3x-2}{x^2-3x-4} with ax+bx2−3x−4\frac{ax+b}{x^2-3x-4}, we can see that a=3a=3 and b=−2b=-2. The question asks for the value of aa, which is 3.

Question 20

Evaluate ∫e2e31x(ln⁡x)2−xln⁡xdx\int_{e^2}^{e^3} \frac{1}{x(\ln x)^2 - x \ln x} dx.

  1. ln⁡(2/3)\ln(2/3)
  2. ln⁡(3/4)\ln(3/4)
  3. ln⁡(4/3)\ln(4/3) (correct answer)
  4. ln⁡(3/2)\ln(3/2)
Explanation: First, simplify the integrand: 1x((ln⁡x)2−ln⁡x)=1xln⁡x(ln⁡x−1)\frac{1}{x((\ln x)^2 - \ln x)} = \frac{1}{x\ln x(\ln x - 1)}. Let u=ln⁡xu = \ln x. Then du=1xdxdu = \frac{1}{x} dx. The limits of integration also change: when x=e2x=e^2, u=ln⁡(e2)=2u=\ln(e^2)=2. When x=e3x=e^3, u=ln⁡(e3)=3u=\ln(e^3)=3. The integral becomes ∫231u(u−1)du\int_2^3 \frac{1}{u(u-1)} du. Decompose into partial fractions: 1u(u−1)=Au+Bu−1\frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1}. 1=A(u−1)+Bu1 = A(u-1) + Bu. If u=0u=0, 1=−A⇒A=−11 = -A \Rightarrow A=-1. If u=1u=1, 1=B1 = B. The integral is ∫23(−1u+1u−1)du=[−ln⁡∣u∣+ln⁡∣u−1∣]23=[ln⁡∣u−1u∣]23\int_2^3 (\frac{-1}{u} + \frac{1}{u-1}) du = [-\ln|u| + \ln|u-1|]_2^3 = [\ln|\frac{u-1}{u}|]_2^3. Evaluating at the limits: ln⁡(3−13)−ln⁡(2−12)=ln⁡(23)−ln⁡(12)=ln⁡(2/31/2)=ln⁡(43)\ln(\frac{3-1}{3}) - \ln(\frac{2-1}{2}) = \ln(\frac{2}{3}) - \ln(\frac{1}{2}) = \ln(\frac{2/3}{1/2}) = \ln(\frac{4}{3}).