Answer: $u = \ln x$, $dv = x^{1/2}dx$; remaining integral is $\int \frac{2}{3}x^{1/2}\,dx$

Set up IBP using LIATE: $\ln x$ is logarithmic (L), highest LIATE priority. Let $u = \ln x$, $dv = x^{1/2}\,dx$. Then $du = \frac{1}{x}\,dx$, $v = \frac{2}{3}x^{3/2}$. Compute the remaining integral: $\int v\,du = \int \frac{2}{3}x^{3/2} \cdot \frac{1}{x}\,dx = \int \frac{2}{3}x^{1/2}\,dx$. This matches Option A. Why distractors fail: Option B reverses the LIATE priority and produces a messy integral. Option C incorrectly simplifies $x^{3/2}/x$ as $x^{3/2}$ instead of $x^{1/2}$. Option D uses a non-standard grouping.

Answer: $e^x(x^2 - 2x + 2) + C$

Apply the tabular method: Derivatives of $x^2$: $x^2 \to 2x \to 2 \to 0$. Antiderivatives of $e^x$: $e^x \to e^x \to e^x$. Signs: $+, -, +$. Compute the result: $\int x^2 e^x\,dx = (+)x^2 e^x + (-)2x e^x + (+)2e^x + C = e^x(x^2 - 2x + 2) + C$. Why distractors fail: Option B has the wrong signs on the $2x$ and/or constant terms. Option C has $-2$ instead of $+2$ for the constant. Option D has incorrect coefficients throughout.

Answer: Yes — the answer $-x^2\cos x + 2x\sin x + 2\cos x + C$ is correct.

Verify the tabular computation: Diagonal products with signs: $(+)(x^2)(-\cos x) = -x^2\cos x$; $(-)(2x)(-\sin x) = +2x\sin x$; $(+)(2)(\cos x) = +2\cos x$. Check by differentiation: $\frac{d}{dx}[-x^2\cos x + 2x\sin x + 2\cos x] = -2x\cos x + x^2\sin x + 2\sin x + 2x\cos x - 2\sin x = x^2\sin x$. ✓ Why distractors fail: Option A is wrong; the signs are correctly alternated. Option B is wrong; the derivative column reaches $2$ (constant), and one more derivative gives $0$, but the table already captures all needed terms via three diagonal products. Option D references an unnecessary additional step.

Answer: $xe^x - e^x + C$

Set up integration by parts: Let $u = x$, $dv = e^x\,dx$. Then $du = dx$, $v = e^x$. Apply the formula: $\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C$. Why distractors fail: Option A has a plus sign instead of minus. Option C treats it as if no IBP were needed. Option D has an incorrect polynomial factor.

Answer: $\int x^4 e^{-x}\,dx$

Identify when the tabular method is ideal: The tabular method works best when one factor is a polynomial (whose repeated derivatives eventually reach zero) and the other is easily integrated repeatedly (like $e^{-x}$). Why the correct answer works: $\int x^4 e^{-x}\,dx$ has $x^4$ (differentiates to 0 in 5 steps) and $e^{-x}$ (integrates repeatedly with ease), making Option C ideal for the tabular method. Why distractors fail: Option A ($e^x \sin x$) cycles and never terminates — tabular method doesn't directly apply. Option B ($\ln x$) needs only one IBP step. Option D is solved by simple $u$-substitution, not IBP.

Answer: The resulting integral $\int v\,du$ is more complex than the original.

Analyze the student's choice: With $u = \cos(3x)$, $du = -3\sin(3x)\,dx$, and $v = \frac{x^2}{2}$. The new integral is $\int \frac{x^2}{2}\cdot(-3\sin(3x))\,dx$, which involves a higher power of $x$ — it's harder, not simpler. Why the correct answer works: Option A correctly identifies that this poor choice increases the complexity of the remaining integral. Why distractors fail: Option B is false; $v = x^2/2$ is easy to find. Option C is false; IBP works fine with trig functions. Option D is incorrect; the issue is increasing complexity, not cycling.

Answer: IBP integrates the product rule $d(uv) = u\,dv + v\,du$ and rearranges to isolate $\int u\,dv$.

Derive IBP from the product rule: The product rule gives $d(uv) = u\,dv + v\,du$. Integrating both sides: $uv = \int u\,dv + \int v\,du$. Rearranging: $\int u\,dv = uv - \int v\,du$. Why the correct answer works: Option B correctly describes this derivation: integrate the product rule and rearrange. Why distractors fail: Option A confuses IBP with $u$-substitution (reverse chain rule). Option C invokes the quotient rule, which is unrelated. Option D states a false property — integrals of products don't factor like that.

Answer: $u = (\ln x)^2$, $dv = dx$; this requires two applications of IBP in total.

Identify the setup: For $\int (\ln x)^2\,dx$, set $u = (\ln x)^2$ and $dv = dx$. Then $du = \frac{2\ln x}{x}\,dx$ and $v = x$. This gives $x(\ln x)^2 - 2\int \ln x\,dx$, which itself requires one more IBP. Why the correct answer works: Option A correctly identifies both the assignment and the fact that two total IBP steps are needed. Why distractors fail: Option B splits $\ln x$ as both $u$ and $dv$, which doesn't follow a standard approach. Option C reverses the assignment making it harder. Option D's $dv = x\,dx$ doesn't match the integrand $\int (\ln x)^2\,dx$.

Answer: $u = \arctan x$; result is $x\arctan x - \frac{1}{2}\ln(1+x^2) + C$

Apply LIATE and set up IBP: $\arctan x$ is inverse trigonometric (I in LIATE), which has high priority. Let $u = \arctan x$, $dv = dx$, so $du = \frac{1}{1+x^2}\,dx$ and $v = x$. Evaluate the remaining integral: $\int \arctan x\,dx = x\arctan x - \int \frac{x}{1+x^2}\,dx = x\arctan x - \frac{1}{2}\ln(1+x^2) + C$. Why distractors fail: Option B has the wrong sign on the logarithm. Option C incorrectly treats the integral like a power rule for $\arctan$. Option D uses the wrong $u$ choice and an incorrect result.

Answer: $\int u\,dv = uv - \int v\,du$

Recall the IBP formula: Integration by parts states $\int u\,dv = uv - \int v\,du$. This follows from the product rule for differentiation reversed. Why the correct answer works: Option B matches the standard formula exactly: the product $uv$ minus the integral $\int v\,du$. Why distractors fail: Option A has a plus sign instead of minus. Option C has $\int u\,dv$ on both sides, which is circular. Option D introduces a spurious factor of $\frac{1}{2}$.

Answer: 4 rows

Track successive derivatives: Starting from $x^3$: first derivative is $3x^2$, second is $6x$, third is $6$, fourth is $0$. That is 4 differentiations to reach zero. Why the correct answer works: Option C (4 rows of differentiation) is correct because a degree-3 polynomial requires exactly 4 derivatives to reach 0. Why distractors fail: Option A (2) and Option B (3) stop too early. Option D (5) overshoots — the derivative is already 0 after 4 steps.

Answer: The student's choice of $u$ and $dv$ increased the polynomial degree, signaling a poor assignment that should be reversed.

Analyze the result of the student's choice: Setting $u = e^x$ gives $du = e^x\,dx$, and $v = \frac{x^3}{3}$. The new integral $\int \frac{x^3}{3} e^x\,dx$ has a higher polynomial degree than the original — the problem is getting harder. Why the correct answer works: Option B correctly diagnoses the issue: the wrong assignment caused the polynomial degree to increase. Reversing (setting $u = x^2$) would reduce it each step. Why distractors fail: Option A is wrong; continuing this way makes things progressively worse. Option C is false; $\int e^x x^2\,dx$ is elementary. Option D is incorrect; this integral is not a cycling type.

Answer: An integral where repeated IBP reproduces the original integral, allowing it to be solved algebraically

Define cycling integrals: A cycling integral occurs when applying IBP twice (or more) yields the original integral on the right side. You can then move it to the left side and solve algebraically. Why the correct answer works: Option C accurately describes the cycling phenomenon: the original integral reappears, and you solve for it. Why distractors fail: Option A confuses cycling with divergence. Option B incorrectly combines two unrelated techniques. Option D references complex analysis, which is unrelated.

Answer: $+, -, +, -$

Recall the sign pattern in the tabular method: The tabular method always begins with a positive sign and alternates: $+, -, +, -, \ldots$ Why the correct answer works: Option B gives the standard alternating sign pattern starting with $+$. Why distractors fail: Option A doesn't alternate properly. Option C starts with a negative sign. Option D has consecutive negative signs.

Answer: $\frac{e^{2x}(2\cos x + \sin x)}{5} + C$

First IBP: Let $u = \cos x$, $dv = e^{2x}dx$. Then $I = \frac{e^{2x}\cos x}{2} + \frac{1}{2}\int e^{2x}\sin x\,dx$. Second IBP: For $\int e^{2x}\sin x\,dx$, let $u = \sin x$, $dv = e^{2x}dx$. Get $\frac{e^{2x}\sin x}{2} - \frac{1}{2}\int e^{2x}\cos x\,dx = \frac{e^{2x}\sin x}{2} - \frac{1}{2}I$. Solve for I: $I = \frac{e^{2x}\cos x}{2} + \frac{1}{2}\left(\frac{e^{2x}\sin x}{2} - \frac{1}{2}I\right) = \frac{e^{2x}\cos x}{2} + \frac{e^{2x}\sin x}{4} - \frac{I}{4}$. So $\frac{5I}{4} = \frac{e^{2x}(2\cos x + \sin x)}{4}$, giving $I = \frac{e^{2x}(2\cos x + \sin x)}{5} + C$. Why distractors fail: Options B, C, and D have incorrect sign combinations or wrong denominators resulting from algebraic errors in the cycling process.

Answer: The product rule

Connect IBP to differentiation: The product rule states $\frac{d}{dx}[u\cdot v] = u\,v' + v\,u'$. Integrating both sides and rearranging gives the IBP formula. Why the correct answer works: Option C is correct because integrating the product rule yields $\int u\,dv = uv - \int v\,du$. Why distractors fail: The chain rule gives rise to $u$-substitution, not IBP. The quotient rule and power rule do not produce the IBP formula.

Answer: Although IBP appears to cycle, the original integral reappears after two applications and can be solved algebraically.

Identify the cycling integral technique: When IBP is applied twice to $\int e^x \sin x\,dx$, the original integral reappears on the right side. This allows us to solve for it algebraically. Why the correct answer works: Option C correctly identifies that cycling is not a problem — it's a feature that allows algebraic resolution. Why distractors fail: Option A is false; the integral does have a closed form. Option B is incorrect; $u$-substitution doesn't apply here. Option D is wrong; the tabular method doesn't naturally handle cycling integrals since neither factor differentiates to zero.

Answer: Because differentiating $\ln x$ produces the simpler expression $\frac{1}{x}$, reducing the complexity of the remaining integral

Explain the rationale for choosing $u$: The goal in IBP is to choose $u$ so that $du$ is simpler and $v$ (from $dv$) is easy to find. Differentiating $\ln x$ gives $\frac{1}{x}$, which simplifies the resulting integral. Why the correct answer works: Option B correctly identifies that the derivative of $\ln x$ simplifies the problem, which is exactly why LIATE ranks logarithmic functions highest. Why distractors fail: Option A is false — $\int \ln x\,dx$ can be computed (it equals $x\ln x - x + C$). Option C is not a valid mathematical reason. Option D is false — setting $dv = \ln x\,dx$ is allowed, just less efficient.

Answer: $\pi$

Set up IBP: Let $u = x$, $dv = \sin x\,dx$, so $du = dx$, $v = -\cos x$. Apply and evaluate: $\int_0^{\pi} x\sin x\,dx = [-x\cos x]_0^{\pi} + \int_0^{\pi}\cos x\,dx = [-\pi(-1) - 0] + [\sin x]_0^{\pi} = \pi + 0 = \pi$. Why distractors fail: Option A (0) would require the integrand to be odd on a symmetric interval, which isn't the case here. Option C ($2\pi$) and Option D ($-\pi$) result from sign or evaluation errors.

Answer: Both students, provided each is consistent in their choice across both IBP steps.

Understand consistency in cycling integrals: For cycling integrals like $\int e^x \sin x\,dx$, either the exponential or the trigonometric function can be $u$. The critical requirement is consistency: the same type must be $u$ in both steps. Why the correct answer works: Option C correctly states that both approaches work as long as the student is consistent, and both yield the same final answer. Why distractors fail: Options A and B each claim only one approach is valid, which is false. Option D is incorrect — the standard IBP method with algebraic solving works fine.

Answer: $k \neq 1$

Analyze the algebraic step: If $I = A + kI$, then $(1-k)I = A$, so $I = \frac{A}{1-k}$. This requires $k \neq 1$. Why the correct answer works: Option C ($k \neq 1$) is the necessary and sufficient condition for the equation to be solvable for $I$. Why distractors fail: Option A ($k=1$) would make the equation $0 = A$, which is generally contradictory. Option B ($k=-1$) is one specific valid value, not the general requirement. Option D is unnecessarily restrictive.

Answer: $x\ln x - x + C$

Set up IBP: Let $u = \ln x$, $dv = dx$. Then $du = \frac{1}{x}\,dx$, $v = x$. Apply the formula: $\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C$. Why distractors fail: Option A gives the derivative of $\ln x$ divided by $x$. Option C has the wrong sign on the second term. Option D applies a power-rule-like approach to $\ln x$ that is incorrect.

Answer: Functions higher on the LIATE list tend to simplify when differentiated, while those lower tend to remain manageable when integrated.

Explain the heuristic rationale: The purpose of LIATE is to select $u$ as the function that becomes simpler when differentiated (e.g., $\ln x \to 1/x$, polynomials lose a degree) and $dv$ as the function that doesn't become more complex when integrated (e.g., $e^x$ stays $e^x$, trig stays trig). Why the correct answer works: Option B captures this dual property: $u$ simplifies upon differentiation, and $dv$ remains tractable upon integration. Why distractors fail: Option A is false — all LIATE functions can be integrated. Option C has no basis in the formula. Option D confuses convergence properties with selection heuristics.

Answer: Because the integrand can be written as $\ln x \cdot \frac{1}{x}$ where $\frac{1}{x}$ is the derivative of $\ln x$, enabling a direct substitution $u = \ln x$.

Recognize the substitution structure: In $\int \frac{\ln x}{x}\,dx$, the factor $\frac{1}{x}$ is exactly $\frac{d}{dx}[\ln x]$. Setting $u = \ln x$ gives $du = \frac{1}{x}\,dx$, so the integral becomes $\int u\,du = \frac{u^2}{2} + C$. Why the correct answer works: Option B correctly identifies the substitution structure that makes IBP unnecessary. Why distractors fail: Option A is irrelevant — $\ln x$ is differentiable on its domain. Option C states a false restriction. Option D incorrectly credits a 'power rule' rather than substitution.

Answer: The tabular method is merely a compact organizational tool that produces the same result as applying IBP five times.

Understand the tabular method's role: The tabular method is a shorthand for carrying out repeated IBP. It doesn't change the mathematics; it simply organizes the alternating signs and diagonal products in a table. Why the correct answer works: Option B correctly describes the tabular method as an organizational tool that yields the same result as repeated standard IBP. Why distractors fail: Option A is false — both methods always give the same result. Option C is misleading; the tabular method actually reduces sign errors through systematic organization. Option D is false; the method works for any polynomial degree.

Answer: $\int 2x\cos x\,dx$

Compute $v$ and $du$: With $u = x^2$, $du = 2x\,dx$. With $dv = \sin x\,dx$, $v = -\cos x$. Apply the full IBP formula: $\int x^2 \sin x\,dx = uv - \int v\,du = -x^2\cos x - \int (-\cos x)(2x)\,dx = -x^2\cos x + \int 2x\cos x\,dx$. The remaining integral to evaluate is $\int 2x\cos x\,dx$. Why distractors fail: Option A uses $\sin x$ instead of $-\cos x$ for $v$, forgetting to integrate $dv$. Option B gives $\int v\,du = \int -2x\cos x\,dx$ before accounting for the minus sign in the IBP formula $uv - \int v\,du$; the double negative produces the positive form in Option C. Option D incorrectly retains $x^2$ instead of differentiating it to $2x$.

Answer: $\int (\ln x)^n\,dx = x(\ln x)^n - n\int (\ln x)^{n-1}\,dx$

Apply IBP: Let $u = (\ln x)^n$, $dv = dx$. Then $du = \frac{n(\ln x)^{n-1}}{x}\,dx$ and $v = x$. Compute: $\int (\ln x)^n\,dx = x(\ln x)^n - \int x \cdot \frac{n(\ln x)^{n-1}}{x}\,dx = x(\ln x)^n - n\int (\ln x)^{n-1}\,dx$. Why distractors fail: Option A has a plus instead of minus. Option C attempts a power-rule-style formula that doesn't apply to $\ln x$. Option D doesn't simplify $x \cdot \frac{1}{x}$ and keeps the wrong exponent.

Answer: $\frac{x^3}{3}\ln x - \frac{x^3}{9} + C$

Set up IBP: Let $u = \ln x$, $dv = x^2\,dx$. Then $du = \frac{1}{x}\,dx$, $v = \frac{x^3}{3}$. Apply the formula: $\int x^2 \ln x\,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3}\cdot\frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C$. Why distractors fail: Option A has $\frac{x^3}{3}$ instead of $\frac{x^3}{9}$ for the second term. Option C has the wrong sign. Option D has $\frac{1}{9}$ in the first term instead of $\frac{1}{3}$.

Answer: $u = x$, $dv = e^x\,dx$

Apply the LIATE guideline: In $\int x e^x\,dx$, $x$ is algebraic (A) and $e^x$ is exponential (E). LIATE says algebraic comes before exponential, so set $u = x$. Why the correct answer works: With $u = x$ and $dv = e^x\,dx$, we get $du = dx$ and $v = e^x$, giving $xe^x - \int e^x\,dx = xe^x - e^x + C$, a simpler integral. Why distractors fail: Option A reverses the roles, making the resulting integral harder. Option C sets $u = xe^x$ which makes $du$ more complex. Option D sets $dv = xe^x\,dx$ which is not easily integrable.

Answer: Differentiate the claimed answer and check if it equals $x^2 e^{3x}$.

Identify the verification strategy: For any antiderivative, the most direct verification is to differentiate it and confirm it matches the original integrand. Why the correct answer works: Option B uses the Fundamental Theorem of Calculus: if $F'(x) = f(x)$, then $\int f(x)\,dx = F(x) + C$. Differentiation is quick and definitive. Why distractors fail: Option A works but is far from the quickest check. Option C checks only one point, which is insufficient. Option D only checks the table format, not the mathematical correctness.

Answer: $\int x \ln x\,dx$

Examine each integral: Options A, B, and D can all be solved with a simple $u$-substitution. Option C has a product of an algebraic function and a logarithm, which is the hallmark of an IBP integral. Why the correct answer works: For $\int x \ln x\,dx$, set $u = \ln x$ and $dv = x\,dx$. There is no substitution that simplifies this product directly. Why distractors fail: Option A: $u = x^2+1$. Option B: $u = \sin x$. Option D: $u = x^2+4$. All are standard $u$-substitution problems.

Answer: Yes, it is always valid but produces $x F(x) - \int F(x)\,dx$ where $F' = f$, which is not generally simpler.

Test the assignment: With $u = x$, $du = dx$, and $dv = f(x)\,dx$, we need $v = F(x)$ where $F' = f$. The IBP formula gives $\int f(x)\,dx = xF(x) - \int F(x)\,dx$. Assess usefulness: This is mathematically valid, but it requires knowing $F(x)$ (an antiderivative of $f$) to set up $v$, and the resulting integral $\int F(x)\,dx$ is typically harder than the original. So while legal, it is rarely useful in this generic form. Why distractors fail: Options A and D make false claims — there is no rule that $u$ must already appear in the integrand. Option B incorrectly states the result as $xf(x) - \int xf'(x)\,dx$, which would arise from setting $u = x$ and $dv = f'(x)\,dx$ (differentiating $f$ rather than integrating it), confusing the roles of $u$ and $v$.

Answer: $-\frac{x^3}{2}e^{-2x} - \frac{3x^2}{4}e^{-2x} + \cdots$

Set up the tabular method: Derivatives of $x^3$: $x^3, 3x^2, 6x, 6, 0$. Antiderivatives of $e^{-2x}$: $-\frac{1}{2}e^{-2x}, \frac{1}{4}e^{-2x}, -\frac{1}{8}e^{-2x}, \frac{1}{16}e^{-2x}$. Signs: $+, -, +, -$. Compute the first two terms: First term: $(+)(x^3)(-\frac{1}{2}e^{-2x}) = -\frac{x^3}{2}e^{-2x}$. Second term: $(-)(3x^2)(\frac{1}{4}e^{-2x}) = -\frac{3x^2}{4}e^{-2x}$. Why distractors fail: Option A has a plus for the second term (wrong sign from the alternating pattern). Option C has a plus for the first term (wrong). Option D has $\frac{3x^2}{2}$ instead of $\frac{3x^2}{4}$.

Answer: The strategy is flawed — some products (like $x\cos(x^2)$) are better handled by substitution, and IBP should be reserved for products where substitution fails.

Examine the claim critically: While IBP is designed for products, not all products need it. For instance, $\int x\cos(x^2)\,dx$ is efficiently solved with $u = x^2$, making it a simple substitution problem. Why the correct answer works: Option C correctly identifies that the strategy is overly broad: substitution should be tried first when one factor is the derivative of the other's inner function. Why distractors fail: Options A and B overstate IBP's applicability. Option D is false — IBP is specifically for integrands that are products.

Answer: It causes the second IBP to undo the first, yielding the trivial identity $I = I$.

Trace through the inconsistent choices: If the first step uses $u = e^x$, $dv = \sin x\,dx$, we get boundary terms and $\int e^x \cos x\,dx$. If the second step reverses to $u = \cos x$ (the trig part), $dv = e^x\,dx$, the IBP undoes the first step, returning us to $\int e^x \sin x\,dx$ with nothing gained. Why the correct answer works: Option B correctly identifies that switching the role assignments causes the second step to reverse the first, yielding $I = I$ — a tautology. Why distractors fail: Option A: the issue isn't just a sign error, it's a complete reversal. Option C: the integral doesn't diverge. Option D: it doesn't change the technique needed.

Answer: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential

Recall the LIATE order: The standard LIATE order is: Logarithmic → Inverse trigonometric → Algebraic → Trigonometric → Exponential. Why the correct answer works: Option B lists the categories in the correct LIATE order. Why distractors fail: Option A swaps Algebraic and Inverse trig. Option C replaces Logarithmic with Linear. Option D swaps Exponential and Trigonometric at the end.

Answer: Logarithmic functions

Identify the LIATE hierarchy: LIATE stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. These are listed in priority order for choosing $u$. Why the correct answer works: The 'L' in LIATE represents Logarithmic functions, which are typically chosen as $u$ first because they simplify upon differentiation. Why distractors fail: Option A (Linear) is a subset of algebraic functions, not what L stands for. Options C and D (Laplace, Limit) are not part of the LIATE mnemonic.

Answer: $1 - 2e^{-1}$

Set up IBP: Let $u = x$, $dv = e^{-x}\,dx$, so $du = dx$, $v = -e^{-x}$. Apply the formula and evaluate: $\int_0^1 xe^{-x}\,dx = \left[-xe^{-x}\right]_0^1 + \int_0^1 e^{-x}\,dx = -e^{-1} + \left[-e^{-x}\right]_0^1 = -e^{-1} + (-e^{-1}+1) = 1 - 2e^{-1}$. Why distractors fail: Option B omits one of the $e^{-1}$ terms. Option C reverses the sign. Option D only evaluates the boundary term.

Answer: Alternating positive and negative signs

Recall the tabular method structure: In the tabular method, one column contains successive derivatives of $u$ and the other contains successive antiderivatives of $dv$. A column of alternating signs ($+, -, +, -, \ldots$) is applied to each diagonal product. Why the correct answer works: Option B correctly identifies the alternating $+/-$ sign pattern that is the hallmark of the tabular method. Why distractors fail: Option A describes an operation not used. Option C mischaracterizes the columns — each column does only one operation. Option D describes something unrelated to the table structure.

Answer: $I_n = x^n e^x - n I_{n-1}$

Derive the reduction formula: Let $u = x^n$, $dv = e^x\,dx$. Then $du = nx^{n-1}\,dx$, $v = e^x$. So $I_n = x^n e^x - \int nx^{n-1}e^x\,dx = x^n e^x - nI_{n-1}$. Why the correct answer works: Option B matches the derived recurrence: $I_n = x^n e^x - nI_{n-1}$. Why distractors fail: Option A has $+n$ instead of $-n$. Option C incorrectly differentiates $x^n$ into the boundary term. Option D uses $I_{n+1}$ instead of $I_{n-1}$, going in the wrong direction.

Answer: $I = e^x(\sin x + \cos x)/2 + C$ after solving $2I = e^x(\sin x + \cos x)$

First IBP application: Let $u = e^x$, $dv = \cos x\,dx$. Then $I = e^x \sin x - \int e^x \sin x\,dx$. Second IBP application: For $\int e^x \sin x\,dx$, let $u = e^x$, $dv = \sin x\,dx$. Then $\int e^x \sin x\,dx = -e^x \cos x + \int e^x \cos x\,dx = -e^x \cos x + I$. Solve for I: Substituting: $I = e^x \sin x - (-e^x \cos x + I) = e^x \sin x + e^x \cos x - I$. So $2I = e^x(\sin x + \cos x)$, giving $I = \frac{e^x(\sin x + \cos x)}{2} + C$. Why distractors fail: Options A, B, and C present intermediate or incorrect equations that don't properly solve for $I$. Only Option D gives the final resolved result.