What is Improper rational functions partial fractions?
Definition: If the degree of numerator P(x) is greater than or equal to denominator Q(x), perform polynomial long division first to separate the proper rational part before applying partial fraction decomposition.
Example: For x+1x2+1, divide to get x−1+x+12. Integrate to 2x2−x+2ln∣x+1∣+C.
Reason: Partial fractions only apply to proper rational functions; division ensures the remainder is proper, allowing standard decomposition techniques to be applied correctly.
21
Easy
8
Medium
11
Hard
📝 All Improper rational functions partial fractions MCQs
Q1. For the improper rational function x2−1x3+2x2−1, what is the correct first step to evaluate its integral?
A.Perform long division to rewrite as x+2+x2−1x+1 ✅
B.Directly apply partial fraction decomposition without any modification
C.Use the substitution u=x2−1
D.Rewrite as x2−1(x2−1)(x+2)+(x+1)
💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: The integrand is improper because the degree of numerator (3) is greater than that of denominator (2). The correct first step is to perform polynomial long division, which yields x+2+x2−1x+1. This separates the polynomial part x+2 (which integrates easily) from a proper rational function that can then be decomposed using partial fractions. Directly applying partial fractions is incorrect because the method requires a proper fraction. The given rewrite in option D represents the same long division result but in a factored form.
Q2. Given the improper integral ∫x2+1x4+1dx, which approach would be most efficient?
A.Perform long division to get x2−1+x2+12 ✅
B.Use partial fractions directly after factoring denominator
C.Substitute u=x2
D.Use integration by parts with u=x4+1
💡 Difficulty: easy | ✅ Correct: A
📖 Explanation: The integrand x2+1x4+1 is improper. Long division yields x2−1+x2+12. This is efficient because ∫(x2−1)dx is elementary and ∫x2+12dx=2tan−1x+C. Direct partial fractions without division is impossible since the function is not proper. The other options either make the problem more complex or are irrelevant.
Q3. What is the result of performing long division on x2+13x3−2x2+5x−1?
A.3x−2+x2+12x+1 ✅
B.3x−2+x2+15x−1
C.3x−2+x2+12x−1
D.3x−2+x2+12x+1
💡 Difficulty: easy | ✅ Correct: A
📖 Explanation: Correct long division: 3x3−2x2+5x−1 divided by x2+1 gives quotient 3x−2 and remainder 2x+1, because (x2+1)(3x−2)=3x3−2x2+3x−2, subtracting this from the numerator yields 2x+1. This proper fraction can then be decomposed if needed. Options B and C have incorrect remainders.
Q4. Evaluate ∫x2+x+1x3+2x2+3x+4dx.
A.2x2+23x+21ln∣x2+x+1∣+35tan−1(32x+1)+C ✅
B.2x2+23x+21ln∣x2+x+1∣+35tan−1(32x+1)+C
C.x2+3x+ln∣x2+x+1∣+tan−1(32x+1)+C
D.2x2+23x+21ln∣x2+x+1∣+35tan−1(32x+1)+C
💡 Difficulty: easy | ✅ Correct: A
Q5. A student decomposes x2−12x3−3x2+1 as x−1Ax+B+x+1Cx+D without long division. What is the error?
A.The function is improper; partial fractions require proper rational functions ✅
B.The denominators are not factored completely
C.The coefficients should be constants, not linear expressions
D.There is no error; this is a valid decomposition
💡 Difficulty: medium | ✅ Correct: A
📖 Explanation: The degree of numerator (3) is greater than the degree of denominator (2), so the function is improper. Partial fraction decomposition is only valid for proper rational functions where numerator degree < denominator degree. The student must first perform long division to get a polynomial part plus a proper fraction. The attempt to directly decompose an improper function will lead to an incorrect result because the decomposition form assumes the fraction is proper. The error is not in the coefficients but in the applicability of the method.
Q6. If ∫x3+2x2+xx2+3x+2dx is evaluated, what is the role of the polynomial part after long division?
A.There is no polynomial part because the fraction is proper ✅
B.The polynomial part is x and integrates to 2x2
C.The polynomial part is 1 and integrates to x
D.The polynomial part is x+1 and integrates to 2x2+x
💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: The denominator x3+2x2+x=x(x2+2x+1)=x(x+1)2. The degree of numerator (2) is less than the degree of denominator (3), so the rational function is proper. Therefore, no long division is needed. The method of partial fractions can be applied directly. The polynomial part is zero. This is a common misconception: students often perform long division when the fraction is already proper, wasting time and potentially introducing errors.
Q7. What is the decomposition of x(x2+1)x3+2x2+x+1 after making it proper?
A.x+2+xA+x2+1Bx+C ✅
B.xA+x2+1Bx+C
C.x+2+xA+x2+1B
D.x+2+xA+x2+1Bx
💡 Difficulty: easy | ✅ Correct: A
Q8. After decomposing an improper rational function, the integral of the polynomial part ∫(3x2−2x+1)dx is:
A.x3−x2+x+C ✅
B.3x3−2x2+x+C
C.3x3−2x2+x+C
D.x3−x2+x+C
💡 Difficulty: easy | ✅ Correct: A
📖 Explanation: The integral of a polynomial is obtained by integrating term by term: ∫3x2dx=x3, ∫−2xdx=−x2, ∫1dx=x. So the antiderivative is x3−x2+x+C. This is a straightforward Easy of the power rule for integration, which is essential when handling the polynomial part of an improper rational function.
Q9. A physicist models the potential energy as U(x)=∫x2−1x3+2xdx. What is U(x) if U(0)=0?
A.2x2+23ln∣x−1∣−21ln∣x+1∣−21 ✅
B.2x2+23ln∣x−1∣+21ln∣x+1∣−23
C.2x2+23ln∣x−1∣−21ln∣x+1∣+C
D.2x2+23ln∣x−1∣−21ln∣x+1∣−21
💡 Difficulty: medium | ✅ Correct: A
Q10. What is the partial fraction decomposition of the proper fraction from x3+x2x2+3x+1 after any necessary long division?
A.xA+x2+1Bx+C ✅
B.xA+x2+1B
C.xA+x2+1Bx
D.x2+1Ax+B+xC
💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: The fraction x3+x2x2+3x+1 is proper (degree 2 vs 3). Factor denominator: x(x2+1). Since x2+1 is irreducible quadratic, the decomposition includes a term for the linear factor x (which is A/x) and a term for the irreducible quadratic x2+1 (which is (Bx+C)/(x2+1)). The numerator for the quadratic factor must be linear because the degree of the denominator is 2. Options B and C incorrectly assume a constant numerator for the quadratic term.
Q11. Given ∫x2+13x4−2x3+5x2−4x+1dx, after long division, what is the proper fraction to decompose?
A.x2+1−2x3+4x2−4x+1 ✅
B.x2+13x2−2x+2
C.x2+13x2−2x+2
D.x2+13x2−2x+2
💡 Difficulty: easy | ✅ Correct: A
Q12. In the integral ∫x2−4x3+1dx, what is the correct form after long division?
A.x+x2−44x+1 ✅
B.x+x2−4x+1
C.x+x2−44x+1
D.x+x2−44x+1
💡 Difficulty: easy | ✅ Correct: A
📖 Explanation: Long division: x3+1 divided by x2−4. x times x2−4=x3−4x. Subtract: (x3+1)−(x3−4x)=4x+1. So quotient = x, remainder = 4x+1. Thus x2−4x3+1=x+x2−44x+1. The proper fraction can then be decomposed: (x−2)(x+2)4x+1=x−2A+x+2B. Finding A and B: 4x+1=A(x+2)+B(x−2). x=2: 9=4A => A=9/4. x=-2: -7=-4B => B=7/4. So integral = 2x2+49ln∣x−2∣+47ln∣x+2∣+C.
Q13. A student integrates ∫x2−1x4+1dx by first decomposing x2−1x4+1=x−1A+x+1B. What is the flaw?
A.The fraction is improper; the numerator degree is greater than denominator ✅
B.The decomposition should include a polynomial part
C.The decomposition should include a polynomial part
D.The denominators should be (x−1)2 and (x+1)2
💡 Difficulty: medium | ✅ Correct: A
📖 Explanation: The fraction x2−1x4+1 is improper because the numerator degree (4) is greater than the denominator degree (2). Partial fraction decomposition as given is only valid for proper fractions. The correct approach is to first perform long division: x2−1x4+1=x2+1+x2−12 (since (x2−1)(x2+1)=x4−1, remainder 2). Then decompose x2−12=x−11−x+11. So integral = 3x3+x+ln∣x−1∣−ln∣x+1∣+C. The student's attempt ignores the polynomial part.
Q14. What is the integral of ∫x2−2x+12x3−3x2+4x−5dx?
A.x2+x+4ln∣x−1∣−x−12+C ✅
B.x2+x+4ln∣x−1∣+x−12+C
C.x2+x+4ln∣x−1∣−x−12+C
D.x2+x+4ln∣x−1∣−x−12+C
💡 Difficulty: easy | ✅ Correct: A
Q15. For what values of A and B does (x−1)(x+1)x2+2x+3=1+x−1A+x+1B hold?
A.A=2, B=1 ✅
B.A=1, B=2
C.A=3, B=0
D.A=0, B=3
💡 Difficulty: hard | ✅ Correct: A
Q16. How many terms are in the partial fraction decomposition of (x2+1)2(x−1)x5+1 after making it proper?
A.The decomposition has a polynomial part of degree 2 plus 3 fractional terms ✅
B.It has a polynomial part of degree 1 plus 4 fractional terms
C.It has a polynomial part of degree 0 plus 3 fractional terms
D.It has a polynomial part of degree 3 plus 2 fractional terms
💡 Difficulty: hard | ✅ Correct: A
Q17. Evaluate ∫−11x2+1x3+2x2+x+1dx using improper rational function techniques.
A.34+π ✅
B.34+2π
C.34
D.32+π
💡 Difficulty: easy | ✅ Correct: A
Q18. A student computes ∫x2−4x+4x3+1dx as 2x2+2x+4ln∣x−2∣−x−29+C. Is this correct? If not, what is the error?
A.Incorrect; the remainder after long division is wrong ✅
B.Incorrect; the partial fraction coefficients are wrong
C.Correct
D.Incorrect; the polynomial part should be x+4
💡 Difficulty: medium | ✅ Correct: A
📖 Explanation: Denominator x2−4x+4=(x−2)2. Long division: x3+1 divided by x2−4x+4. First term: x gives x3−4x2+4x. Subtract: 4x2−4x+1. Second term: +4 gives 4x2−16x+16. Subtract: 12x−15. So quotient = x+4, remainder = 12x−15. So integrand = x+4+(x−2)212x−15. Decompose (x−2)212x−15=x−2A+(x−2)2B. Multiply: 12x−15=A(x−2)+B. x=2: 9 = B. So B=9. Equate: 12x = Ax => A=12. Then -15 = -2A+B = -24+9 = -15 works. So (x−2)212x−15=x−212+(x−2)29. Integrate: ∫(x+4)dx=2x2+4x. ∫x−212dx=12ln∣x−2∣. ∫(x−2)29dx=−x−29. So correct integral = 2x2+4x+12ln∣x−2∣−x−29+C. The student's answer has 2x and 4ln and −9/(x−2). The polynomial part is wrong: it should be 4x, not 2x. So error is in the polynomial part.
Q19. Which of the following is the correct decomposition of x2−3x+2x3+2x2−5x−6?
A.x+5+x−16−x−22 ✅
B.x+5+x−16+x−22
C.x+5+x−16−x−22
D.x+5+x−16−x−22
💡 Difficulty: easy | ✅ Correct: A
Q20. For the improper integral ∫01x2+xx3+2x2+1dx, why is the value finite even though the integrand is improper?
A.After long division, the proper fraction has no singularities in [0,1]
B.The improper nature is resolved by cancellation of the (x) factor ✅
C.The integrand is actually proper
D.The integral diverges
💡 Difficulty: hard | ✅ Correct: B
Q21. In the decomposition of x3−xx4−1, what is the polynomial part after long division?
A.x ✅
B.x2
C.1
D.x+1
💡 Difficulty: easy | ✅ Correct: A
Q22. Which of the following is the correct antiderivative of x2+2x+1x3+3x2+3x+1?
A.2x2+x+C ✅
B.2x2+x+ln∣x+1∣+C
C.2x2+x+C
D.2x2+x−x+11+C
💡 Difficulty: easy | ✅ Correct: A
Q23. What is the integral of ∫x2+22x3+3x2+4x+5dx?
A.x2+3x+2ln(x2+2)+21tan−1(2x)+C ✅
B.x2+3x+2ln(x2+2)+21tan−1(2x)+C
C.x2+3x+2ln(x2+2)+21tan−1(2x)+C
D.x2+3x+2ln(x2+2)+21tan−1(2x)+C
💡 Difficulty: easy | ✅ Correct: A
Q24. A student evaluates ∫x2−1x3+2xdx and gets 2x2+ln∣x−1∣+ln∣x+1∣+C. What mistake did they make?
A.They forgot the polynomial part from long division ✅
B.They used the wrong partial fraction coefficients
C.They forgot the constant of integration
D.They integrated the improper fraction directly
💡 Difficulty: medium | ✅ Correct: A
Q25. For ∫x3+1x4+1dx, the long division yields quotient x and remainder −x+1. What is the correct integral?
Q26. Which of the following integrals requires polynomial long division before partial fractions?
A.∫x3−13x2+2x+1dx
B.∫x2−42x3−3x2+1dx ✅
C.∫x3+2x5x2+3dx
D.∫x2−3x+2x+1dx
💡 Difficulty: hard | ✅ Correct: B
📖 Explanation: An integral requires long division if the rational function is improper, i.e., the degree of the numerator is greater than or equal to the degree of the denominator. In option A: degree 2 vs 3 (proper). Option B: degree 3 vs 2 (improper, requires long division). Option C: degree 2 vs 3 (proper). Option D: degree 1 vs 2 (proper). So only option B requires long division. This tests the student's ability to recognize the condition for applying long division.
Q27. After long division, x2+x+1x3+2x2+3x+4=x+1+x2+x+1x+3. What is the integral?
A.2x2+x+21ln∣x2+x+1∣+35tan−1(32x+1)+C ✅
B.2x2+x+21ln∣x2+x+1∣+35tan−1(32x+1)+C
C.2x2+x+21ln∣x2+x+1∣+25tan−1(32x+1)+C
D.2x2+x+21ln∣x2+x+1∣+25tan−1(32x+1)+C
💡 Difficulty: easy | ✅ Correct: A
Q28. The integral ∫x2−1x4+1dx is evaluated. After long division, the proper fraction is x2−12. What is the final result?
A.3x3+x+ln∣x−1∣−ln∣x+1∣+C ✅
B.3x3+x+ln∣x−1∣+ln∣x+1∣+C
C.3x3+x+ln∣x−1∣−ln∣x+1∣+C
D.3x3+x+ln∣x−1∣−ln∣x+1∣+C
💡 Difficulty: easy | ✅ Correct: A
Q29. A student claims that ∫x2+xx3+1dx=2x2−x+ln∣x∣+ln∣x+1∣+C. Is this correct?
A.No, the polynomial part should be x−1 not x ✅
B.No, the polynomial part should be x−1 not x
C.No, the coefficients of the logs are wrong
D.Yes, it is correct
💡 Difficulty: medium | ✅ Correct: A
Q30. What is the integral of ∫x2+xx3+2x2+x+2dx?
A.2x2+x+ln∣x∣+ln∣x+1∣+C ✅
B.2x2+x+ln∣x∣+ln∣x+1∣+C
C.2x2+x+ln∣x∣+ln∣x+1∣+C
D.2x2+x+ln∣x∣+ln∣x+1∣+C
💡 Difficulty: easy | ✅ Correct: A
Q31. If ∫x2−1x3+ax2+bx+cdx=2x2+2x+ln∣x−1∣−ln∣x+1∣+C, what are a,b,c?
A.a=2, b=1, c=2 ✅
B.a=2, b=1, c=2
C.a=2, b=1, c=2
D.a=2, b=1, c=2
💡 Difficulty: medium | ✅ Correct: A
Q32. What is the partial fraction decomposition of x3+x3x2+2x+1 after making it proper? (The fraction is proper)
A.x1+x2+12x+1 ✅
B.x1+x2+12x+1
C.x1+x2+12x+1
D.x1+x2+12x+1
💡 Difficulty: easy | ✅ Correct: A
Q33. A graph of the integrand y=x2−3x+2x3+2x2−5x−6 has a hole at x=2. After simplifying, what is the integral?
A.2x2+5x+8ln∣x−1∣+C ✅
B.2x2+5x+8ln∣x−1∣+C
C.2x2+5x+8ln∣x−1∣+C
D.2x2+5x+8ln∣x−1∣+C
💡 Difficulty: hard | ✅ Correct: A
Q34. Which of the following is the correct decomposition of x2+1x4+2x3+3x2+2x+1 after long division?
A.x2+2x+2+x2+12x−1 ✅
B.x2+2x+2+x2+12x−1
C.x2+2x+2+x2+12x−1
D.x2+2x+2+x2+12x−1
💡 Difficulty: easy | ✅ Correct: A
Q35. A student incorrectly integrates ∫x−1x2+1dx as 3(x−1)3+2(x−1)+C. What is the correct result and what is the student's error?
A.Correct: 2x2+x+2ln∣x−1∣+C; student used substitution incorrectly ✅
B.Correct: 2x2+x+2ln∣x−1∣+C; student used substitution incorrectly
C.Correct: 2x2+x+2ln∣x−1∣+C; student used substitution incorrectly
D.Correct: 2x2+x+2ln∣x−1∣+C; student used substitution incorrectly
💡 Difficulty: medium | ✅ Correct: A
Q36. If ∫x2+2x+22x3+3x2+4x+5dx=x2+x+ln∣x2+2x+2∣+tan−1(x+1)+C, what is the value of the integral from 0 to 1?
A.2+ln5−ln2+4π−4π=2+ln(5/2) ✅
B.2+ln5−ln2+4π−4π=2+ln(5/2)
C.2+ln5−ln2
D.2+ln5−ln2+4π
💡 Difficulty: easy | ✅ Correct: A
Q37. What is the area under the curve y=x2+1x3+2x from x=0 to x=2?
A.2x2+ln(x2+1) evaluated from 0 to 2 = 2 + \ln 5 ✅
B.2x2+ln(x2+1) evaluated from 0 to 2 = 2 + \ln 5
C.2x2+ln(x2+1) evaluated from 0 to 2 = 2 + \ln 5
D.2x2+ln(x2+1) evaluated from 0 to 2 = 2 + \ln 5
💡 Difficulty: hard | ✅ Correct: A
Q38. For the improper rational function x2−4x3+1, what is the integral after long division and partial fractions?
A.2x2+49ln∣x−2∣+47ln∣x+2∣+C ✅
B.2x2+49ln∣x−2∣−47ln∣x+2∣+C
C.2x2+49ln∣x−2∣+47ln∣x+2∣+C
D.2x2+49ln∣x−2∣+47ln∣x+2∣+C
💡 Difficulty: easy | ✅ Correct: A
Q39. If the graph of y=x2−3x+2x3+2x2−5x−6 has vertical asymptotes at x=1 and x=2, what is the integral ∫x2−3x+2x3+2x2−5x−6dx?
A.2x2+5x+8ln∣x−1∣+C ✅
B.2x2+5x+8ln∣x−1∣+C
C.2x2+5x+8ln∣x−1∣+C
D.2x2+5x+8ln∣x−1∣+C
💡 Difficulty: hard | ✅ Correct: A
Q40. A student integrates ∫x2−1x4+1dx and gets 3x3+x+ln∣x−1∣−ln∣x+1∣+C. The student's answer is correct. If they wanted to check by differentiating, what would be the result?