Answer: $\frac{1}{2}\ln\!\left(\frac{3}{2}\right)$

Decompose: $\frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1}$. Integrate: $\int_2^3 \left(\frac{1/2}{x-1} - \frac{1/2}{x+1}\right) dx = \frac{1}{2}\left[\ln|x-1| - \ln|x+1|\right]_2^3 = \frac{1}{2}\left[\ln\frac{|x-1|}{|x+1|}\right]_2^3$. Evaluate bounds: At $x=3$: $\ln\frac{2}{4} = \ln\frac{1}{2}$. At $x=2$: $\ln\frac{1}{3}$. Result: $\frac{1}{2}\left(\ln\frac{1}{2} - \ln\frac{1}{3}\right) = \frac{1}{2}\ln\frac{1/2}{1/3} = \frac{1}{2}\ln\frac{3}{2}$. Why distractors fail: Option A ($\frac{1}{2}\ln(4/3)$) arises from an arithmetic error in evaluating the bounds. Option C is the negative of the correct answer. Option D is missing the factor of $\frac{1}{2}$ from the partial fraction coefficients.

Answer: Only Setup I is correct because repeated linear factors need separate constant-numerator terms for each power

Apply decomposition rules: $(x-2)^2$ is a repeated linear factor, requiring $\frac{A}{x-2} + \frac{B}{(x-2)^2}$ (constant numerators, separate terms for each power). $(x^2+9)$ is an irreducible quadratic, requiring $\frac{Cx+D}{x^2+9}$. Why Setup II fails: Setup II places $Ax + B$ over $(x-2)^2$, treating it as if it were an irreducible quadratic. While $Ax + B$ over $(x-2)^2$ can be algebraically rewritten as the sum in Setup I, it does not follow the standard partial fraction procedure and conflates the two separate terms. Why distractors fail: Option A is misleading — while Setup II is algebraically equivalent in a sense, it does not represent a proper partial fraction decomposition and will cause errors in systematic solving. Option C incorrectly applies the quadratic rule to a squared linear factor. Option D is wrong if $\deg(P) < 4$.

Answer: 5

Count contributions from each factor: $(x-3)$ contributes 1 constant ($A$). $(x+1)^2$ contributes 2 constants ($B$ and $C$). $(x^2+5)$ is an irreducible quadratic contributing 2 constants ($D$ and $E$). Why the correct answer works: Total: $1 + 2 + 2 = 5$ unknown constants. Why distractors fail: Option A (3) may only count one constant per factor. Option B (4) likely misses one constant from the quadratic or repeated factor. Option D (6) overcounts, perhaps adding an extra term.

Answer: $2x - 1 + \frac{-2x + 5}{x^2 + 1}$

Perform the division step by step: Step 1: $2x^3 \div x^2 = 2x$. Multiply: $2x(x^2 + 1) = 2x^3 + 2x$. Subtract: $(2x^3 - x^2 + 0x + 4) - (2x^3 + 2x) = -x^2 - 2x + 4$. Continue division: Step 2: $-x^2 \div x^2 = -1$. Multiply: $-1(x^2 + 1) = -x^2 - 1$. Subtract: $(-x^2 - 2x + 4) - (-x^2 - 1) = -2x + 5$. Write the result: $\frac{2x^3 - x^2 + 4}{x^2+1} = 2x - 1 + \frac{-2x + 5}{x^2+1}$. Verify: $(2x - 1)(x^2 + 1) + (-2x + 5) = 2x^3 + 2x - x^2 - 1 - 2x + 5 = 2x^3 - x^2 + 4$. ✓ Why distractors fail: Option A has an incorrect remainder of $-3x + 6$ from a subtraction error. Option B has the wrong quotient ($2x+1$ instead of $2x-1$). Option D has a sign error in the remainder.

Answer: The values are correct because $A + B = 4$, matching the numerator coefficient

Verify using the identity: $4x = A(x-3) + B(x+1)$. Setting $x = -1$: $-4 = A(-4)$, so $A = 1$. Setting $x = 3$: $12 = B(4)$, so $B = 3$. Quick check: Expanding: $A(x-3)+B(x+1) = (A+B)x + (-3A+B) = 4x + 0$, so $A+B=4$ and $-3A+B=0$. Both equations are satisfied by $A=1, B=3$. Why distractors fail: Option B is false — there is no rule requiring equal constants. Option C is incorrect since the cover-up method yields a positive $A$. Option D is wrong because the decomposition is a purely algebraic step verifiable without integration.

Answer: $-\ln|x+1| + \ln|x-4| + C$

Decompose into partial fractions: $\frac{5}{(x+1)(x-4)} = \frac{A}{x+1} + \frac{B}{x-4}$. Setting $x = -1$: $5 = A(-5)$, so $A = -1$. Setting $x = 4$: $5 = B(5)$, so $B = 1$. Integrate each term: $\int \left(\frac{-1}{x+1} + \frac{1}{x-4}\right) dx = -\ln|x+1| + \ln|x-4| + C$. Why distractors fail: Option A reverses the signs on both logarithms. Option B incorrectly makes both terms positive. Option C treats the integrand as $\frac{5}{u}$ with $u = (x+1)(x-4)$, ignoring that the derivative of $(x+1)(x-4)$ is not a constant.

Answer: $A = 0,\ B = 1,\ C = 0,\ D = 1$

Multiply both sides by $(x^2+1)^2$: $x^2 + 2 = (Ax+B)(x^2+1) + (Cx+D)$. Expanding: $x^2 + 2 = Ax^3 + Bx^2 + Ax + B + Cx + D$. Compare coefficients: $x^3$: $A = 0$. $x^2$: $B = 1$. $x^1$: $A + C = 0$, so $C = 0$. $x^0$: $B + D = 2$, so $D = 1$. Why distractors fail: Option B has $A = 1$ and misidentifies $B$. Option C incorrectly gives $C = 1$. Option D collapses everything into the second fraction, missing the first-level term.

Answer: $x + \frac{6x}{x^2-4}$

Perform long division: Divide $x^3 + 2x$ by $x^2 - 4$. The first term: $x^3 \div x^2 = x$. Multiply: $x(x^2 - 4) = x^3 - 4x$. Subtract: $(x^3 + 2x) - (x^3 - 4x) = 6x$. Write the result: $\frac{x^3 + 2x}{x^2-4} = x + \frac{6x}{x^2-4}$. Why distractors fail: Option B adds an extra constant 2. Option C incorrectly gets $x^2$ as the quotient. Option D uses $2x$ instead of $6x$ in the remainder — a subtraction error.

Answer: $A \ln|x - r| + C$

Recall the basic integral: $\int \frac{A}{x - r}\, dx = A \ln|x - r| + C$. Why distractors fail: Option A is the result of integrating $\frac{A}{(x-r)^2 + 1}$, not $\frac{A}{x-r}$. Option C is the derivative of $\frac{A}{x-r}$, not its antiderivative. Option D is the integral of $A$, not $\frac{A}{x-r}$.

Answer: $3\ln|x+1| - 3\ln|x| - \frac{3}{x} + C$

Set up partial fractions: $\frac{3}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$. Multiply both sides by $x^2(x+1)$: $3 = Ax(x+1) + B(x+1) + Cx^2$. Solve for constants: Set $x = 0$: $3 = B(1)$, so $B = 3$. Set $x = -1$: $3 = C(1)$, so $C = 3$. Compare $x^2$ coefficients: $0 = A + C$, so $A = -3$. Integrate: $\int \left(\frac{-3}{x} + \frac{3}{x^2} + \frac{3}{x+1}\right) dx = -3\ln|x| - \frac{3}{x} + 3\ln|x+1| + C = 3\ln|x+1| - 3\ln|x| - \frac{3}{x} + C$. Why distractors fail: Options A and C appear similar, but only C arranges the terms correctly matching $A=-3, B=3, C=3$. Option B reverses the signs on the logarithmic terms. Option D has incorrect signs throughout.

Answer: The cover-up method cannot determine $A$ because $(x-1)$ appears at multiple powers; it only directly gives $B$ and $C$

Limitations of cover-up: The cover-up method works by substituting a root to zero out all other terms. For $(x-1)^2$, setting $x=1$ zeros out $(x-1)^2$ and isolates $B$. Setting $x=-2$ isolates $C$. But there's no substitution that isolates $A$ alone. Why the correct answer works: The constant $A$ is associated with $(x-1)$ rather than $(x-1)^2$. The cover-up method directly determines only the constants for the highest power of each distinct root and for non-repeated factors. $A$ must be found by comparing coefficients or substituting a convenient value. Why distractors fail: Option A is wrong — the method works fine for three or more distinct factors. Option C is wrong — the method requires proper fractions, not improper ones. Option D overstates the limitation — it works partially even with repeated factors.

Answer: They omitted the term $\frac{B}{x^2}$ required for the repeated linear factor $x^2$

Identify the repeated factor: The factor $x^2$ means $x$ is repeated with multiplicity 2. The correct setup requires $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}$. Why the correct answer works: The student missed the $\frac{B}{x^2}$ term. A repeated linear factor $(x - r)^n$ requires terms for each power from 1 to $n$. Why distractors fail: Option A is wrong — $x^2$ is not an irreducible quadratic; it is a repeated linear factor. Option C is wrong — the fraction is already proper ($\deg(1) < \deg(3)$). Option D is irrelevant to the structural error.

Answer: A natural logarithm and an inverse tangent function

Standard technique: Split $Bx + C$ so that one part matches the derivative of the denominator (yielding $\ln$) and the remaining constant part, after completing the square, yields $\arctan$. Why the correct answer works: Writing $\frac{Bx + C}{x^2 + bx + c} = \frac{B}{2} \cdot \frac{2x + b}{x^2+bx+c} + \frac{C - Bb/2}{x^2+bx+c}$. The first part integrates to $\frac{B}{2}\ln(x^2+bx+c)$ and the second to a multiple of $\arctan$. Why distractors fail: Option A: exponentials do not arise from rational integrands. Option C: hyperbolic functions appear from different integral forms. Option D: inverse sine arises from $\frac{1}{\sqrt{a^2 - x^2}}$, not from irreducible quadratic denominators.

Answer: Performed polynomial long division first, since the numerator's degree exceeds the denominator's degree

Compare degrees: The numerator has degree 4 and the denominator has degree 2. Since $4 > 2$, the rational function is improper. Why the correct answer works: Long division must be performed to reduce the integrand to a polynomial plus a proper rational fraction before partial fractions can be applied. Why distractors fail: Option A would not simplify the improper fraction issue. Option C introduces an unconventional factoring that does not help. Option D is not the standard approach for rational functions.

Answer: 3

Apply the repeated linear factor rule: A repeated linear factor $(x-1)^3$ requires one term for each power: $\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$. Why the correct answer works: Regardless of $P(x)$ (as long as the fraction is proper), there are always 3 terms corresponding to the 3 powers of $(x-1)$. Why distractors fail: Option A only uses the highest power. Option B omits one level. Option D is incorrect — the number of terms is determined by the denominator's factorization, not the numerator.

Answer: $\frac{3}{5}\ln|x+1| + \frac{1}{5}\ln(x^2+4) + \frac{13}{10}\arctan\!\left(\frac{x}{2}\right) + C$

Set up decomposition: $\frac{x^2+3x+5}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+4}$. Find constants: Setting $x = -1$: $\frac{1 - 3 + 5}{1 + 4} = \frac{3}{5} = A$. Expanding $A(x^2+4) + (Bx+C)(x+1) = x^2+3x+5$ and comparing coefficients: $x^2$: $A + B = 1$, so $B = 2/5$. $x^1$: $B + C = 3$, so $C = 3 - 2/5 = 13/5$. $x^0$: $4A + C = 12/5 + 13/5 = 25/5 = 5$. ✓ Integrate the quadratic term: $\int \frac{(2/5)x + 13/5}{x^2+4}\,dx = \frac{1}{5}\int \frac{2x}{x^2+4}\,dx + \frac{13}{5}\int \frac{1}{x^2+4}\,dx = \frac{1}{5}\ln(x^2+4) + \frac{13}{5} \cdot \frac{1}{2}\arctan\!\left(\frac{x}{2}\right) = \frac{1}{5}\ln(x^2+4) + \frac{13}{10}\arctan\!\left(\frac{x}{2}\right)$. Combine: $\frac{3}{5}\ln|x+1| + \frac{1}{5}\ln(x^2+4) + \frac{13}{10}\arctan\!\left(\frac{x}{2}\right) + C$. Why distractors fail: Option B uses incorrect, unsimplified coefficients. Option C omits the arctan term entirely. Option D has the wrong coefficient ($7/10$ instead of $13/10$) on the arctan.

Answer: $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{(x^2+2)^2}$

Identify the factor type: $(x^2+2)^2$ is a repeated irreducible quadratic factor with multiplicity 2. Why the correct answer works: Each power of the irreducible quadratic gets its own term with a linear numerator: $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{(x^2+2)^2}$. Why distractors fail: Option A uses constant numerators, which is only correct for linear factors. Option C mixes constant and linear numerators inconsistently. Option D does not actually decompose the fraction into simpler parts.

Answer: $\frac{2x}{x^2 + 9}$

Analyze each integrand: Option C: $\frac{2x}{x^2+9}$. Here the numerator $2x$ is the derivative of the denominator $x^2 + 9$. This is a simple u-substitution integral: $\int \frac{2x}{x^2+9}\,dx = \ln(x^2+9) + C$. Why the other options need partial fractions: Option A: $x^2 + x - 6 = (x+3)(x-2)$, requiring decomposition. Option B: already factored denominator with distinct linear factors. Option D: $x^2 - 4 = (x-2)(x+2)$, requiring decomposition. Why distractors fail: Options A, B, and D all have denominators that factor into distinct linear factors with numerators that are not scalar multiples of the denominator's derivative, necessitating partial fractions.

Answer: $\frac{A}{x+2} + \frac{B}{(x+2)^2}$

Identify factor type: The denominator has a repeated linear factor $(x+2)^2$. Why the correct answer works: For a repeated linear factor $(x+2)^2$, we include one fraction for each power: $\frac{A}{x+2} + \frac{B}{(x+2)^2}$. Why distractors fail: Option A only accounts for the highest power, missing $\frac{A}{x+2}$. Option B incorrectly uses a linear numerator as if the denominator were an irreducible quadratic. Option D incorrectly places $Bx$ rather than a constant $B$ in the numerator of the second fraction.

Answer: $A = 1$

Set up the identity: $7 = A(x+6) + B(x-1)$. This must hold for all $x$. Use the cover-up method: Set $x = 1$: $7 = A(1+6) = 7A$, so $A = 1$. Why distractors fail: Option A ($A=7$) ignores the factor $(x+6)$ evaluated at $x=1$. Option B ($A=-1$) is a sign error. Option D ($7/6$) may arise from incorrectly substituting $x=0$ or other algebraic errors.

Answer: $\frac{x^2}{2} + \ln|x-1| + C$

Long division first: $\frac{x^3+1}{x^2-1} = x + \frac{x+1}{x^2-1}$. Simplify the remainder: $\frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}$ (for $x \neq -1$). Integrate: $\int \left(x + \frac{1}{x-1}\right) dx = \frac{x^2}{2} + \ln|x-1| + C$. Why distractors fail: Option A adds an extra $\ln|x+1|$ term. Option B adds an extra $x$ and uses $\ln|x^2-1|$ instead of simplifying. Option C subtracts $\frac{1}{2}\ln|x+1|$ unnecessarily.

Answer: Because the degree of the numerator equals the degree of the denominator, making it an improper fraction

Compare degrees: The numerator $x^2 + 3$ has degree 2, and the denominator $x^2 + x - 6$ also has degree 2. Since $\deg(P) \geq \deg(Q)$, the fraction is improper. Why the correct answer works: Partial fraction decomposition requires a proper fraction. Long division must be performed first to extract the polynomial part. Why distractors fail: Option A is wrong — $x^2 + x - 6 = (x+3)(x-2)$ factors over the reals. Option C is wrong — the numerator and denominator share no common factor. Option D is wrong — the roots $x = -3$ and $x = 2$ are both real.

Answer: $2\ln|x+1| + \ln|x+2| + C$

Decompose: $\frac{3x+5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$. Setting $x = -1$: $3(-1)+5 = 2 = A(1)$, so $A = 2$. Setting $x = -2$: $3(-2)+5 = -1 = B(-1)$, so $B = 1$. Integrate: $\int \frac{2}{x+1} + \frac{1}{x+2}\, dx = 2\ln|x+1| + \ln|x+2| + C$. Why distractors fail: Option B incorrectly uses the numerator coefficients directly. Option C swaps the values of $A$ and $B$. Option D has the wrong sign on the second term.

Answer: Perform polynomial long division because the degree of the numerator is greater than the degree of the denominator

Check degrees first: Before decomposing, compare the degree of the numerator (3) with the degree of the denominator (2). Since $3 \geq 2$, we must perform long division first. Why the correct answer works: Polynomial long division reduces the integrand to a polynomial plus a proper rational function, which can then be decomposed via partial fractions. Why distractors fail: Option A skips the required long division step. Option C is unnecessary since $x^2 - 1$ factors over the reals. Option D would not simplify this integrand appropriately for integration.

Answer: $\frac{Ax + B}{x^2 + 4}$

Recognize irreducible quadratic: $x^2 + 4$ does not factor over the real numbers, so it is an irreducible quadratic factor. Why the correct answer works: An irreducible quadratic factor requires a linear numerator $Ax + B$, giving $\frac{Ax + B}{x^2 + 4}$. Why distractors fail: Option A uses only a constant numerator, which is insufficient. Option B incorrectly factors $x^2 + 4$ over the reals. Option D introduces a second power that doesn't exist in the original denominator.

Answer: The degree of $P(x)$ must be strictly less than the degree of $Q(x)$

Definition of a proper rational function: Partial fraction decomposition applies directly only to proper rational functions, where $\deg(P) < \deg(Q)$. Why the correct answer works: The standard partial fraction technique requires $\frac{P(x)}{Q(x)}$ to be proper. If it is not, long division must be performed first to extract the polynomial part. Why distractors fail: Option A is too restrictive — $Q(x)$ can be any polynomial. Option C adds an unnecessary restriction that $Q(x)$ be linear. Option D describes an improper fraction, which requires long division before decomposition.

Answer: Recognize that $x^3 - 1 = (x-1)(x^2+x+1)$ and rewrite the integrand using this identity before decomposing

Recognize the denominator: $(x-1)(x^2+x+1) = x^3 - 1$. The numerator $3x^2 + 2$ has degree 2 while the denominator has degree 3, so it's proper. But noting $x^3 - 1$ may allow simplification. Strategic decomposition: Write $3x^2 + 2 = 3(x^2 + x + 1) - 3x - 3 + 2 = 3(x^2+x+1) - 3x - 1$. Then $\frac{3x^2+2}{(x-1)(x^2+x+1)} = \frac{3}{x-1} + \frac{-3x - 1}{x^2+x+1}$. Recognizing the factored form of the denominator simplifies the algebra significantly. Why distractors fail: Option B is wrong — $x^2 + x + 1$ is irreducible (discriminant $= 1 - 4 = -3 < 0$). Option C is wrong — degree of numerator (2) is less than degree of denominator (3). Option D is unnecessarily complicated and not a standard approach.

Answer: The Heaviside cover-up method, substituting the roots of each factor

Compare methods: For distinct linear factors, the Heaviside cover-up method is fastest: substitute $x = 2$ to find $A$ and $x = -3$ to find $B$ directly. Why the correct answer works: Setting $x = 2$: $10 = A(5)$, so $A = 2$. Setting $x = -3$: $10 = B(-5)$, so $B = -2$. Two substitutions, no system of equations needed. Why distractors fail: Option A works but is slower for distinct linear factors. Option C is unnecessarily complicated. Option D is not a standard technique for finding partial fraction constants.

Answer: The numerator over the irreducible quadratic should be $Bx + C$, not just $B$

Check the quadratic factor: The discriminant of $x^2 + x + 1$ is $1 - 4 = -3 < 0$, so it is irreducible over the reals. Correct decomposition form: An irreducible quadratic requires a linear numerator: $\frac{Bx + C}{x^2 + x + 1}$, not $\frac{B}{x^2 + x + 1}$. Why distractors fail: Option A is wrong — the discriminant is negative, so no further real factoring is possible. Option C is wrong — $\deg(2) < \deg(3)$, so the fraction is proper. Option D is wrong because the quadratic is indeed irreducible.

Answer: $x^2 + 4$ is irreducible and gets a linear numerator $Ax + B$; $x^2 - 4$ factors into $(x-2)(x+2)$ and gets separate constant numerators

Factor or determine irreducibility: $x^2 + 4$ has discriminant $0 - 16 < 0$, so it is irreducible over the reals. $x^2 - 4 = (x-2)(x+2)$ factors into distinct linear factors. Why the correct answer works: Irreducible quadratics require $\frac{Ax + B}{x^2+4}$. Factorable quadratics like $(x-2)(x+2)$ are decomposed as $\frac{A}{x-2} + \frac{B}{x+2}$. Why distractors fail: Option A confuses factorability with division requirements. Option C incorrectly treats both as irreducible. Option D reverses which is irreducible (over the reals, $x^2 - 4$ factors; its complex factoring is irrelevant to standard partial fractions).

Answer: Factor as $(x-1)(x+1)(x^2+1)$ and decompose into $\frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$

Complete factorization: $x^4 - 1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)$. Note $x^2 - 1$ factors further but $x^2 + 1$ does not. Why the correct answer works: The decomposition has two distinct linear factors giving constant numerators and one irreducible quadratic giving a linear numerator. This matches $\frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$. Why distractors fail: Option A stops factoring too early — $x^2 - 1$ must be factored into linear factors. Option C incorrectly squares the linear factors. Option D bypasses partial fractions entirely, which is not the standard or simplest approach.

Answer: $\ln(x^2 + 2x + 5) + \frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C$

Complete the square in the denominator: $x^2 + 2x + 5 = (x+1)^2 + 4$. The denominator is an irreducible quadratic. Split the numerator: Write $2x + 3 = (2x + 2) + 1$. Then $\int \frac{2x+2}{(x+1)^2+4}\,dx + \int \frac{1}{(x+1)^2+4}\,dx$. Integrate each part: The first integral: let $u = (x+1)^2 + 4$, $du = 2(x+1)\,dx$, giving $\ln|(x+1)^2+4| + C_1 = \ln(x^2+2x+5)+C_1$. The second: $\int \frac{1}{(x+1)^2+4}\,dx = \frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C_2$. Why distractors fail: Option B has the wrong coefficient on the arctan term ($1$ instead of $\frac{1}{2}$). Option C omits the arctan entirely. Option D uses $\frac{1}{4}$ instead of $\frac{1}{2}$.

Answer: 6 constants

Check properness: Numerator degree: 5. Denominator degree: $2 + 4 = 6$. Since $5 < 6$, no long division is needed. Set up the decomposition: $(x-1)^2$ contributes: $\frac{A}{x-1} + \frac{B}{(x-1)^2}$ → 2 constants. $(x^2+4)^2$ contributes: $\frac{Cx+D}{x^2+4} + \frac{Ex+F}{(x^2+4)^2}$ → 4 constants. Total count: $2 + 4 = 6$ unknown constants. Why distractors fail: Option A (4) likely misses the repeated quadratic terms. Option B (5) miscounts one term. Option D (8) overcounts, perhaps adding unnecessary terms.

Answer: $\frac{A}{x - 1} + \frac{B}{x + 4}$

Identify factor types: The denominator consists of two distinct linear factors: $(x - 1)$ and $(x + 4)$. Why the correct answer works: For distinct linear factors, each gets a constant numerator: $\frac{A}{x-1} + \frac{B}{x+4}$. Why distractors fail: Option B does not decompose the fraction at all. Option C incorrectly uses a linear numerator $Bx + C$ over a linear factor. Option D introduces a redundant third term over the original denominator.

Answer: $A = 2,\ B = 1,\ C = 3$

Set up the identity: $6x^2 - 3x - 15 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)$. Find A: set x = 1: $6 - 3 - 15 = -12$ and $(1+2)(1-3) = 3 \cdot (-2) = -6$, so $A = -12 / -6 = 2$. Find B: set x = -2: $6(4) - 3(-2) - 15 = 24 + 6 - 15 = 15$ and $(-2-1)(-2-3) = (-3)(-5) = 15$, so $B = 15/15 = 1$. Find C: set x = 3: $6(9) - 3(3) - 15 = 54 - 9 - 15 = 30$ and $(3-1)(3+2) = 2 \cdot 5 = 10$, so $C = 30/10 = 3$. Verification: Expand: $2(x+2)(x-3) + 1(x-1)(x-3) + 3(x-1)(x+2) = 2(x^2 - x - 6) + (x^2 - 4x + 3) + 3(x^2 + x - 2) = 2x^2 - 2x - 12 + x^2 - 4x + 3 + 3x^2 + 3x - 6 = 6x^2 - 3x - 15$. ✓ Why distractors fail: Options B, C, and D permute the values of $A$, $B$, and $C$. Each constant is uniquely determined by substituting the root of the corresponding factor, so swapping values will not satisfy the identity.

Answer: $\frac{3/2}{x-1} + \frac{(-3/2)x + 1/2}{x^2+1}$

Set up the decomposition: $\frac{2x+1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+1}$. Find A by cover-up: Set $x = 1$: $\frac{2(1)+1}{1^2+1} = \frac{3}{2}$, so $A = \frac{3}{2}$. Find B and C: $2x + 1 = \frac{3}{2}(x^2+1) + (Bx+C)(x-1)$. Expanding and comparing the $x^2$ coefficient: $0 = \frac{3}{2} + B$, so $B = -\frac{3}{2}$. Comparing constants: $1 = \frac{3}{2} - C$, so $C = \frac{1}{2}$. Why distractors fail: Option A uses only a constant numerator over the irreducible quadratic, omitting the required $Bx$ term. Option C uses incorrect coefficient values that do not satisfy the identity. Option D has the correct $A$ but swaps or miscalculates $B$ and $C$.

Answer: Differentiate the result and simplify to see if it equals the original integrand

Fundamental verification: The most reliable way to verify an indefinite integral is to differentiate the answer and check if it equals the original integrand. Why the correct answer works: Differentiating $\frac{1}{2}\ln|x-1| - \ln|x-2| + \frac{1}{2}\ln|x-3|$ gives $\frac{1/2}{x-1} - \frac{1}{x-2} + \frac{1/2}{x-3}$. Recombining this over a common denominator should yield $\frac{1}{(x-1)(x-2)(x-3)}$ if the integration was done correctly. Why distractors fail: Option B only checks a single point and may not catch errors. Option C is impractical and approximate. Option D states a rule that does not generally hold.