Answer: $\frac{(3x+1)^6}{18} + C$

Choose u: Let $u = 3x + 1$. Then $du = 3\,dx$, so $dx = \frac{1}{3}du$. Integrate: $\int u^5 \cdot \frac{1}{3}\,du = \frac{1}{3} \cdot \frac{u^6}{6} = \frac{u^6}{18}$. Back-substitute: $\frac{(3x+1)^6}{18} + C$. Why distractors fail: Option A divides only by 6 (from the power rule) but misses the $\frac{1}{3}$ from the chain rule adjustment. Option C divides only by 3 but misses the 6. Option D is the derivative of $(3x+1)^5$, not its antiderivative.

Answer: $u = \sqrt{x}$, resulting in $\int 2e^u\,du$

Set up the substitution: Let $u = \sqrt{x} = x^{1/2}$. Then $du = \frac{1}{2\sqrt{x}}\,dx$, so $\frac{dx}{\sqrt{x}} = 2\,du$. Transform the integral: $\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = \int e^u \cdot 2\,du = 2\int e^u\,du = 2e^u + C = 2e^{\sqrt{x}} + C$. Why distractors fail: Option B: if $u = e^{\sqrt{x}}$, then $du = \frac{e^{\sqrt{x}}}{2\sqrt{x}}dx$, giving $\int 2\,du$, not $\int \frac{1}{u}du$. Option C misses the factor of 2 from the constant adjustment. Option D substitutes $u = x$ which changes nothing.

Answer: $g(x)$

Identify the inner function: In a composite function $f(g(x))$, the inner function is $g(x)$. U-substitution sets $u = g(x)$ so that $du = g'(x)\,dx$. Why the correct answer works: Setting $u = g(x)$ transforms the integral into $\int f(u)\,du$, which is typically simpler to evaluate. Why distractors fail: Option A ($f(x)$) is the outer function applied to $x$, not the inner function. Option B ($g'(x)$) is the derivative of the inner function, which becomes part of $du$. Option D ($f(g(x))$) is the entire composite, not the substitution variable.

Answer: $u = x^3 + 5$, $du = 3x^2\,dx$

Identify the inner function: The integrand contains $\sqrt{x^3 + 5}$, a composite function. The inner function is $x^3 + 5$. Verify du matches: With $u = x^3 + 5$, we get $du = 3x^2\,dx$. The factor $x^2\,dx$ in the integrand is proportional to $du$ (off by a constant $\frac{1}{3}$), so the substitution works cleanly. Why distractors fail: Option B sets $u$ to the entire square root, making the substitution unnecessarily complex. Option C sets $u = x^2$, which does not simplify the square root. Option D uses $u = x^3$, which misses the $+5$ and would leave a $\sqrt{u+5}$ instead of $\sqrt{u}$.

Answer: Set $u = x^3 + 2$ so that $du = 3x^2\,dx$ and the integral becomes $\frac{1}{3}\int u^{-5}\,du$.

Identify the structure: The integrand is $\frac{x^2}{(x^3+2)^5}$. The derivative of $x^3+2$ is $3x^2$, and $x^2$ is present in the numerator up to a constant. Why the correct answer works: With $u = x^3+2$, $du = 3x^2\,dx$, so $x^2\,dx = \frac{1}{3}du$. The integral becomes $\frac{1}{3}\int u^{-5}\,du = \frac{1}{3} \cdot \frac{u^{-4}}{-4} + C = -\frac{1}{12(x^3+2)^4} + C$. Why distractors fail: Option A is impractical—expanding a fifth power creates enormous algebra. Option C: partial fractions apply to rational functions with factorable denominators, not powers of irreducible cubics. Option D: setting $u = x^2$ does not simplify the $(x^3+2)^5$ in the denominator.

Answer: $\frac{(\ln(x))^2}{2} + C$

Choose u: Let $u = \ln(x)$. Then $du = \frac{1}{x}\,dx$. Rewrite and integrate: $\int \frac{\ln(x)}{x}\,dx = \int u\,du = \frac{u^2}{2} + C$. Back-substitute: $\frac{(\ln(x))^2}{2} + C$. Why distractors fail: Option B ($\ln(\ln(x))$) would be the result of $\int \frac{1}{x\ln(x)}\,dx$, not $\int \frac{\ln(x)}{x}\,dx$. Option C: $\frac{\ln(x^2)}{2} = \ln(x)$, which is not the same as $\frac{(\ln(x))^2}{2}$. Option D does not follow from any correct integration.

Answer: The integral $\int e^{x^2}\,dx$ cannot be evaluated using u-substitution because $du$ requires a factor not present in the integrand.

Why u-sub fails here: With $u = x^2$, $du = 2x\,dx$. The integrand $e^{x^2}\,dx$ contains no $x$ factor, and you cannot simply introduce one. U-substitution requires that $du$ (up to a constant multiple) be present in the integrand. Why the correct answer works: This integral has no elementary closed-form antiderivative. It is related to the error function and cannot be solved by any standard Calculus 1 technique. Why distractors fail: Option A is invalid—you cannot multiply and divide by a variable ($2x$) because that changes the integrand in a non-trivial way (unlike constant adjustment). Option B: setting $u = e^{x^2}$ gives $du = 2x e^{x^2}dx$, which is even more complex. Option D: setting $u = x$ changes nothing about the difficulty.

Answer: To reverse the chain rule for evaluating integrals of composite functions

Core principle of u-substitution: U-substitution is a technique for evaluating integrals where the integrand involves a composite function. It works by reversing the chain rule of differentiation. Why the correct answer works: The chain rule states $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. U-substitution reverses this: when we see an integrand of the form $f'(g(x)) \cdot g'(x)$, we set $u = g(x)$ to simplify the integral. Why distractors fail: Option A is wrong because reversing the product rule corresponds to integration by parts, not u-substitution. Option C describes differentiation, not integration. Option D describes L'Hôpital's Rule.

Answer: $\sin(x^2) + C$

Choose u: Let $u = x^2$. Then $du = 2x\,dx$. Rewrite and integrate: The integral becomes $\int \cos(u)\,du = \sin(u) + C$. Back-substitute: Replacing $u$ gives $\sin(x^2) + C$. Why distractors fail: Option B has the wrong sign. Option C introduces an extra factor of 2 (double-counting). Option D incorrectly multiplies by $x^2$, which does not arise from the integration.

Answer: Yes — both approaches are valid and yield the same result.

Two valid approaches for definite integrals: When applying u-substitution to a definite integral $\int_a^b f(g(x))g'(x)\,dx$, you may either convert the limits so that $x = a \to u = g(a)$ and $x = b \to u = g(b)$, or integrate in $u$, back-substitute to $x$, and evaluate at the original limits $a$ and $b$. Why the correct answer works: Both methods are mathematically equivalent. Changing limits avoids back-substitution; back-substituting avoids recalculating limits. The student's statement is correct. Why distractors fail: Options A and B each insist on only one method, but both are valid. Option D incorrectly adds a one-to-one requirement—u-substitution for definite integrals works even when $g(x)$ is not one-to-one, because the integral accounts for signed area.

Answer: $-\ln|\cos(x)| + C$

Rewrite the integrand: $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Choose u: Let $u = \cos(x)$, so $du = -\sin(x)\,dx$, meaning $\sin(x)\,dx = -du$. Integrate: $\int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos(x)| + C$. Why distractors fail: Option A has a positive sign instead of negative. Option C uses $\sin(x)$ in the logarithm, which would result from integrating $\cot(x)$, not $\tan(x)$. Option D is the derivative of $\tan(x)$, not its antiderivative.

Answer: It uses the fact that integration and differentiation are inverse operations together with the chain rule to simplify composite integrands.

Conceptual basis: U-substitution is grounded in the chain rule. If $F'(u) = f(u)$, then $\frac{d}{dx}[F(g(x))] = f(g(x)) \cdot g'(x)$. Integrating both sides gives $\int f(g(x))\cdot g'(x)\,dx = F(g(x)) + C$. Why the correct answer works: The method directly leverages the inverse relationship between differentiation and integration, specifically reversing the chain rule to simplify the integrand. Why distractors fail: Option A is incorrect because u-substitution does not always produce polynomials. Option C conflates the FTC with the substitution technique—the FTC connects definite integrals to antiderivatives but does not explain why the substitution is valid. Option D describes trigonometric substitution, a different technique.

Answer: $\frac{1}{132}$

Choose u and express x in terms of u: Let $u = 1 - x$, so $x = 1 - u$ and $dx = -du$. When $x = 0$, $u = 1$; when $x = 1$, $u = 0$. Rewrite the integral: $\int_1^0 (1-u)u^{10}(-du) = \int_0^1 (1-u)u^{10}\,du = \int_0^1 (u^{10} - u^{11})\,du$. Evaluate: $\left[\frac{u^{11}}{11} - \frac{u^{12}}{12}\right]_0^1 = \frac{1}{11} - \frac{1}{12} = \frac{12 - 11}{132} = \frac{1}{132}$. Why distractors fail: Option B ($\frac{1}{11}$) results from integrating only $u^{10}$ and forgetting the $x = 1-u$ factor. Option C ($\frac{1}{12}$) comes from integrating only $u^{11}$. Option D ($\frac{1}{110}$) is a common arithmetic error.

Answer: Replace $u$ with the original expression in terms of $x$

Back-substitution requirement: For indefinite integrals, the final answer must be expressed in terms of the original variable $x$, not $u$. Why the correct answer works: Since $u = g(x)$, after integrating in terms of $u$, we substitute back to get the answer in terms of $x$, then add the constant $C$. Why distractors fail: Option A applies to definite integrals, not indefinite ones. Option C (differentiating to check) is a useful verification step but not the required final step of the method. Option D is nonsensical—the constant of integration $C$ is added, not multiplied.

Answer: $u = x^4$ works because $du = 4x^3\,dx$ matches the remaining factors; $u = \sin(x^4)$ leads to a more complicated integral.

Analyze u = x⁴: If $u = x^4$, then $du = 4x^3\,dx$. The integral becomes $\int \sin(u)\,du = -\cos(u) + C = -\cos(x^4) + C$. This is clean and direct. Analyze u = sin(x⁴): If $u = \sin(x^4)$, then $du = \cos(x^4) \cdot 4x^3\,dx$. The integrand has $\sin(x^4) \cdot 4x^3\,dx$ but no $\cos(x^4)$ factor, so this substitution does not simplify the integral. Why distractors fail: Option A is wrong because the two substitutions are not equivalent in effectiveness. Option C is wrong because setting $u = \sin(x^4)$ requires $\cos(x^4)$ in the integrand, which is absent. Option D is false; $u = x^4$ works perfectly.

Answer: $\sin(\ln(x)) + C$

Choose u: Let $u = \ln(x)$. Then $du = \frac{1}{x}\,dx$. Rewrite and integrate: $\int \cos(u)\,du = \sin(u) + C = \sin(\ln(x)) + C$. Why distractors fail: Option B has the wrong sign. Option C incorrectly multiplies by $\ln(x)$. Option D incorrectly introduces a $\frac{1}{x}$ factor in the result.

Answer: $e - 1$

Choose u and change limits: Let $u = x^3$, so $du = 3x^2\,dx$. When $x = 0$, $u = 0$; when $x = 1$, $u = 1$. Evaluate the transformed integral: $\int_0^1 e^u\,du = e^u\Big|_0^1 = e^1 - e^0 = e - 1$. Why distractors fail: Option B forgets to subtract $e^0 = 1$. Option C incorrectly keeps the factor of 3 that was already absorbed into $du$. Option D uses the wrong upper limit ($u = 3$ instead of $u = 1$).

Answer: $\int \sin(x)\cos(x)\,dx$

Identify the function-derivative pair: In $\int \sin(x)\cos(x)\,dx$, we can let $u = \sin(x)$ and $du = \cos(x)\,dx$ (or vice versa). The integrand is a product of a function and its derivative. Why the correct answer works: The substitution transforms it to $\int u\,du = \frac{u^2}{2} + C = \frac{\sin^2(x)}{2} + C$, a clean application of u-substitution. Why distractors fail: Option A ($\int xe^x\,dx$) requires integration by parts since $x$ is not the derivative of the exponent. Option C does not have a clean function-derivative pair in the numerator and denominator. Option D ($\int \ln(x)\,dx$) also typically requires integration by parts.

Answer: $\frac{1}{4}$

Choose u and change limits: Let $u = \ln(x)$, $du = \frac{1}{x}dx$. When $x = 1$, $u = 0$; when $x = e$, $u = 1$. Integrate: $\int_0^1 u^3\,du = \frac{u^4}{4}\Big|_0^1 = \frac{1}{4}$. Why distractors fail: Option A ($\frac{1}{3}$) uses the wrong power rule exponent (dividing by 3 instead of 4). Option B ($1$) omits the division by 4 entirely. Option D ($\frac{3}{4}$) may result from an error like $1 - \frac{1}{4}$.

Answer: The student computes $\frac{(\pi/2)^2}{2} \approx 1.23$ instead of the correct value $\frac{1}{2}$.

Identify the error: After substituting $u = \sin(x)$, the limits should change: $x=0 \to u=0$ and $x=\pi/2 \to u=1$. The correct integral is $\int_0^1 u\,du = \frac{1}{2}$. What the student computes incorrectly: By keeping $x$-limits and treating them as $u$-limits, the student computes $\frac{u^2}{2}\Big|_0^{\pi/2} = \frac{(\pi/2)^2}{2} \approx 1.23$, which is wrong. Why distractors fail: Option A is false—the limits $[0, \pi/2]$ for $u$ and $[0,1]$ for $u$ are different. Option C: the limits are not reversed; the answer is positive but wrong. Option D: $\sin(\pi/2) = 1 \neq 0$.

Answer: The chain rule gives $\frac{d}{dx}[F(g(x))] = F'(g(x))g'(x)$; integrating both sides yields the u-substitution formula.

Derive the connection: The chain rule states $\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x)$. Integrating both sides: $F(g(x)) + C = \int F'(g(x)) \cdot g'(x)\,dx$. Setting $f = F'$ gives $\int f(g(x))g'(x)\,dx = F(g(x)) + C$, which is exactly u-substitution with $u = g(x)$. Why the correct answer works: Option B correctly states this derivation: the chain rule for differentiation, when integrated, produces the u-substitution formula. Why distractors fail: Option A confuses the direction—u-sub reverses the chain rule, it doesn't apply it to derivatives. Option C is factually wrong; they are directly related. Option D: the chain rule is not used during back-substitution (that is just algebraic replacement).

Answer: $\frac{1}{2}\sin(x^2) + C$

Choose u and find du: Let $u = x^2$, so $du = 2x\,dx$, which means $x\,dx = \frac{1}{2}du$. Rewrite and integrate: $\int x\cos(x^2)\,dx = \frac{1}{2}\int \cos(u)\,du = \frac{1}{2}\sin(u) + C$. Back-substitute: Replacing $u$ gives $\frac{1}{2}\sin(x^2) + C$. Why distractors fail: Option A omits the $\frac{1}{2}$ constant adjustment. Option C multiplies by 2 instead of $\frac{1}{2}$. Option D uses cosine instead of sine for the antiderivative of cosine.

Answer: Both are correct because their answers differ only by a constant.

Verify Student A: $u = \sin(x)$, $du = \cos(x)\,dx$ → $\int u\,du = \frac{\sin^2(x)}{2} + C_1$. ✓ Verify Student B: $u = \cos(x)$, $du = -\sin(x)\,dx$ → $\int -u\,du = -\frac{\cos^2(x)}{2} + C_2$. ✓ Show the answers are equivalent: Using $\sin^2(x) + \cos^2(x) = 1$, we get $\frac{\sin^2(x)}{2} = \frac{1 - \cos^2(x)}{2} = -\frac{\cos^2(x)}{2} + \frac{1}{2}$. The answers differ by the constant $\frac{1}{2}$, absorbed into the arbitrary constant $C$. Why distractors fail: Options A and B each claim only one is correct, but both are valid. Option D is incorrect—u-substitution works perfectly here.

Answer: $\frac{1}{4}$

Choose u and change limits: Let $u = 1 + e^x$, $du = e^x\,dx$. When $x = 0$, $u = 2$; when $x = \ln(3)$, $u = 1 + 3 = 4$. Integrate: $\int_2^4 u^{-2}\,du = \left[-u^{-1}\right]_2^4 = -\frac{1}{4} - \left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. Why distractors fail: Option B ($\frac{1}{8}$) could arise from squaring the limits incorrectly or misapplying the power rule for $u^{-2}$. Option C ($\frac{3}{8}$) is a common arithmetic mistake when combining fractions. Option D ($\frac{1}{2}$) results from forgetting to subtract $\frac{1}{4}$ and only evaluating the antiderivative at the lower limit $u = 2$.

Answer: $\ln|1 + e^x| + C$

Choose u: Let $u = 1 + e^x$. Then $du = e^x\,dx$. Rewrite and integrate: $\int \frac{e^x}{1+e^x}\,dx = \int \frac{1}{u}\,du = \ln|u| + C$. Back-substitute: $\ln|1 + e^x| + C$. Why distractors fail: Option A would be the result of differentiating, not integrating. Option C incorrectly multiplies by $e^x$. Option D has the wrong sign.

Answer: $u = 2$ to $u = 3$

Convert limits using the substitution: With $u = 1 + \sqrt{x}$: when $x = 1$, $u = 1 + \sqrt{1} = 2$; when $x = 4$, $u = 1 + \sqrt{4} = 3$. Why the correct answer works: The new limits are $u = 2$ (lower) and $u = 3$ (upper). Why distractors fail: Option A uses the original $x$-limits without converting. Option C incorrectly computes $u(4) = 1 + 4 = 5$ (using $x$ instead of $\sqrt{x}$). Option D gets the upper limit wrong for the same reason as Option C.

Answer: $-\sqrt{1 - x^2} + C$

Choose u: Let $u = 1 - x^2$. Then $du = -2x\,dx$, so $x\,dx = -\frac{1}{2}du$. Rewrite and integrate: $\int \frac{-\frac{1}{2}du}{\sqrt{u}} = -\frac{1}{2}\int u^{-1/2}\,du = -\frac{1}{2} \cdot 2u^{1/2} + C = -\sqrt{u} + C$. Back-substitute: $-\sqrt{1 - x^2} + C$. Why distractors fail: Option A ($\arcsin(x)$) is the antiderivative of $\frac{1}{\sqrt{1-x^2}}$ (no $x$ in the numerator). The presence of $x$ in the numerator changes the integral fundamentally. Option C has the wrong sign. Option D combines the wrong function with the wrong sign.

Answer: No — the correct result is $\frac{5(x^3+1)^5}{15} + C = \frac{(x^3+1)^5}{3} + C$, so the answer is actually correct.

Perform the substitution: Let $u = x^3 + 1$, $du = 3x^2\,dx$, so $x^2\,dx = \frac{1}{3}du$. Compute the integral: $\int 5x^2(x^3+1)^4\,dx = 5 \cdot \frac{1}{3}\int u^4\,du = \frac{5}{3} \cdot \frac{u^5}{5} = \frac{u^5}{3} = \frac{(x^3+1)^5}{3} + C$. Verify by differentiation: $\frac{d}{dx}\left[\frac{(x^3+1)^5}{3}\right] = \frac{5(x^3+1)^4 \cdot 3x^2}{3} = 5x^2(x^3+1)^4$. ✓ Why distractors fail: Option A says yes without justification—though the answer happens to be correct, the more informative choice is B which shows why. Option C ($\frac{5(x^3+1)^5}{3}$) fails to cancel the 5 from the coefficient with the 5 from the power rule. Option D ($\frac{(x^3+1)^5}{15}$) applies the constant adjustment from $du$ but also incorrectly divides by 5 again.

Answer: No — the correct answer is $\frac{1}{3}\tan(3x) + C$ because the chain rule requires dividing by the inner derivative.

Apply u-substitution: Let $u = 3x$, $du = 3\,dx$, so $dx = \frac{1}{3}du$. Then $\int \sec^2(u) \cdot \frac{1}{3}\,du = \frac{1}{3}\tan(u) + C = \frac{1}{3}\tan(3x) + C$. Verify by differentiation: $\frac{d}{dx}\left[\frac{1}{3}\tan(3x)\right] = \frac{1}{3} \cdot \sec^2(3x) \cdot 3 = \sec^2(3x)$. ✓ Why distractors fail: Option A is wrong because differentiating $\tan(3x)$ gives $3\sec^2(3x)$, not $\sec^2(3x)$. Option C multiplies by 3 instead of dividing, making the error worse. Option D is false—u-substitution works perfectly here.