Answer: $du$ is obtained by differentiating $u = g(x)$, giving $du = g'(x)\,dx$

The differential relationship: If $u = g(x)$, then by differentiation $du = g'(x)\,dx$. This connects the two differentials through the derivative of the substitution function. Why the correct answer works: This relationship is precisely what allows us to replace $g'(x)\,dx$ in the integrand with $du$. Why distractors fail: Option A is only true in the trivial case $u = x$. Option C confuses antidifferentiation with differentiation. Option D is wrong because they are linked by the substitution equation.

Answer: $\int x\sqrt{x+1}\,dx$ with $u = x + 1$

Analyze each substitution: For $\int x\sqrt{x+1}\,dx$ with $u = x+1$: $du = dx$ but we still have the factor $x$. We must write $x = u - 1$ to eliminate $x$ entirely. Why the correct answer works: Unlike the other options where $du$ absorbs all $x$-dependent factors, this integral requires rewriting $x = u - 1$ before integrating. Why distractors fail: Option A: $du = 3x^2\,dx$ absorbs the entire $x$-factor. Option C: $du = 5\,dx$, no leftover $x$. Option D: $du = \cos(x)\,dx$ absorbs the remaining factor.

Answer: The limits are converted to the corresponding $u$-values using the substitution equation

Changing limits in definite integrals: When substituting $u = g(x)$, the original limits $x = a$ and $x = b$ must be converted to $u = g(a)$ and $u = g(b)$. Why the correct answer works: Since the variable of integration changes from $x$ to $u$, the limits must also be expressed in terms of $u$ to maintain a consistent integral. Why distractors fail: Option A is wrong because keeping $x$-limits while integrating in $u$ is inconsistent (unless you back-substitute). Option C is wrong because the new limits depend on $g(x)$, not arbitrary values. Option D is wrong because limits are not automatically swapped—they follow from evaluating $g(a)$ and $g(b)$.

Answer: The integral equals $0$

Compute the new limits: With $u = \sin(x)$: when $x = 0$, $u = 0$; when $x = \pi$, $u = 0$. Evaluate the integral: $\int_0^0 u^3\,du = 0$. Any definite integral with equal upper and lower limits equals $0$. Why distractors fail: Option A is wrong because the integral converges. Option C is wrong because the substitution is valid—it correctly reveals the integral is $0$. Option D would result from incorrect limit handling.

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

Set up the substitution: Let $u = x^2+1$, so $du = 2x\,dx$ and $x\,dx = \frac{1}{2}du$. Limits: $x=0 \Rightarrow u=1$, $x=1 \Rightarrow u=2$. Evaluate: $\frac{1}{2}\int_1^2 u^{-2}\,du = \frac{1}{2}\left[-u^{-1}\right]_1^2 = \frac{1}{2}\left(-\frac{1}{2}+1\right) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$. Why distractors fail: Option A omits the $\frac{1}{2}$ constant from $du$. Option C results from an arithmetic error. Option D is too large for the given integrand on $[0,1]$.

Answer: $\frac{1}{12}$

Set up the substitution: Let $u = \cos(3x)$, so $du = -3\sin(3x)\,dx$, giving $\sin(3x)\,dx = -\frac{1}{3}du$. Change limits: When $x = 0$: $u = \cos(0) = 1$. When $x = \pi/6$: $u = \cos(\pi/2) = 0$. Evaluate: $\int_1^0 u^3 \cdot \left(-\frac{1}{3}\right)du = \frac{1}{3}\int_0^1 u^3\,du = \frac{1}{3}\cdot\frac{1}{4} = \frac{1}{12}$. Why distractors fail: Option B ($\frac{5}{48}$) results from an arithmetic error in evaluation. Option C ($\frac{1}{4}$) omits the $\frac{1}{3}$ factor from the chain rule. Option D ($\frac{1}{3}$) applies the chain rule factor but uses $\frac{u^2}{2}$ instead of $\frac{u^4}{4}$ for the antiderivative.

Answer: $\ln 2$

Set up the substitution: Let $u = e^x + 1$, so $du = e^x\,dx$. Limits: when $x = 0$, $u = 2$; when $x = \ln 3$, $u = 4$. Evaluate: $\int_2^4 \frac{du}{u} = \ln|u|\Big|_2^4 = \ln 4 - \ln 2 = \ln\frac{4}{2} = \ln 2$. Why distractors fail: Option B ($\ln 3$) confuses the upper $x$-limit with the answer. Option C ($\ln 4$) uses only the upper $u$-limit $\ln 4$ without subtracting $\ln 2$. Option D ($1$) has no basis from the computation.

Answer: $\frac{(x^2+x)^6}{6} + C$

Perform the substitution: With $u = x^2 + x$ and $du = (2x+1)\,dx$, the integral is $\int u^5\,du = \frac{u^6}{6} + C$. Back-substitute: Replace $u$ with $x^2 + x$ to get $\frac{(x^2+x)^6}{6} + C$. Why distractors fail: Option A uses exponent $5$ instead of $6$. Option B back-substitutes $2x+1$ instead of $x^2+x$. Option D divides by $12$ instead of $6$.

Answer: The inner function of a composite integrand

Core idea of u-substitution: The u-substitution technique reverses the chain rule. In a composite function $f(g(x))$, the substitution $u = g(x)$ targets the inner function. Why the correct answer works: Setting $u$ equal to the inner function allows us to rewrite the integral in terms of $u$, simplifying the composite structure. Why distractors fail: Option A is wrong because we choose the inner, not outermost, function. Option C is wrong because $du$ represents a derivative, not $u$ itself. Option D is wrong because $u$ is chosen before integration, not as a result of it.

Answer: $u = \ln(x)$

Identify the structure: Rewrite the integrand as $\frac{1}{\ln(x)}\cdot\frac{1}{x}$. Here, $\frac{1}{x}$ is the derivative of $\ln(x)$. Apply the substitution: Let $u = \ln(x)$, $du = \frac{1}{x}dx$. The integral becomes $\int \frac{du}{u} = \ln|u| + C = \ln|\ln(x)| + C$. Why distractors fail: Option A gives $du = dx$ and does not simplify the integral. Option B gives $du = -\frac{1}{x^2}dx$, which doesn't match. Option D gives $du = (\ln x + 1)dx$, which is more complex.

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

Perform the substitution: 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 the wrong sign (missing the negative from $du = -\sin(x)\,dx$). Option C and D involve $\sin(x)$ instead of $\cos(x)$.

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

Check each integral: For $\int \frac{2x}{x^2+9}\,dx$, let $u = x^2 + 9$, then $du = 2x\,dx$. The integral becomes $\int \frac{du}{u} = \ln|u| + C$. Why the correct answer works: The derivative of $x^2 + 9$ is $2x$, which appears exactly in the numerator. This is a perfect u-substitution candidate. Why distractors fail: Option A requires integration by parts (product of polynomial and exponential). Option C requires an inverse tangent formula, not u-substitution. Option D requires integration by parts (product of polynomial and logarithm).

Answer: Both substitutions are valid and yield the same numerical answer

Student A's approach: With $u = \tan(x)$, $du = \sec^2(x)\,dx$. The integral becomes $\int_0^1 u\,du = \frac{1}{2}$. Student B's approach: With $u = \sec(x)$, $du = \sec(x)\tan(x)\,dx$. The integral becomes $\int_1^{\sqrt{2}} u\,du = \frac{2-1}{2} = \frac{1}{2}$. Why distractors fail: Options A and B are wrong because both substitutions are valid. Option D is wrong because a definite integral has a unique value regardless of the method used.

Answer: The substitution still works because we can write $x\,dx = \frac{1}{2}\,du$ and multiply by the constant $\frac{1}{2}$

Adjusting for constant factors: Since $du = 2x\,dx$, we solve for $x\,dx = \frac{1}{2}du$. Constants can be freely moved in and out of integrals. Why the correct answer works: The integral becomes $\frac{1}{2}\int \sin(u)\,du$, which is straightforward. Missing constant factors are easily adjusted. Why distractors fail: Option A is wrong because a missing constant does not invalidate the substitution. Option C would not simplify the integral. Option D is wrong because constant multiples can always be adjusted.

Answer: $e^{\sqrt{x}} + C$

Set up the substitution: Let $u = \sqrt{x} = x^{1/2}$. Then $du = \frac{1}{2\sqrt{x}}\,dx$. Rewrite and integrate: The integrand $\frac{e^{\sqrt{x}}}{2\sqrt{x}}\,dx = e^u\,du$. So $\int e^u\,du = e^u + C = e^{\sqrt{x}} + C$. Why distractors fail: Option A incorrectly includes a factor of $\frac{1}{2}$. Option C incorrectly doubles the result. Option D adds an erroneous factor of $\sqrt{x}$.

Answer: Changing limits eliminates the need to re-express the antiderivative in terms of the original variable

Advantage of changing limits: When we convert the limits from $x$-values to $u$-values, the entire evaluation stays in terms of $u$. There is no need to undo the substitution. Why the correct answer works: This saves a step and reduces the chance of algebraic error, since we evaluate the $u$-antiderivative directly at the $u$-limits. Why distractors fail: Option A is false—back-substitution is perfectly valid. Option C is false—the FTC applies regardless of variable names. Option D is false—the antiderivative form is the same either way.

Answer: The chain rule

Connection to differentiation: The chain rule states $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. U-substitution reverses this by identifying $u = g(x)$ and $du = g'(x)\,dx$. Why the correct answer works: U-substitution directly undoes the chain rule, allowing us to integrate compositions of functions. Why distractors fail: Option A is wrong because the product rule is reversed by integration by parts. Option B is wrong because the quotient rule does not have a direct integration analog. Option D is wrong because the power rule is reversed by the basic power rule for antiderivatives.

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

Set up the substitution: Let $u = \ln x$, so $du = \frac{1}{x}dx$. Limits: $x = 1 \Rightarrow u = 0$; $x = e \Rightarrow u = 1$. Evaluate: $\int_0^1 u^3\,du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}$. Why distractors fail: Option A uses $\frac{u^3}{3}$ instead of $\frac{u^4}{4}$. Option B forgets to divide by $4$. Option D uses $\frac{u^2}{2}$.

Answer: Express $x^2$ as $u - 4$ using the substitution equation

Resolve the leftover variable: Since $u = x^2 + 4$, we have $x^2 = u - 4$. After replacing $x\,dx = \frac{1}{2}du$ and $x^2 = u - 4$, the integral becomes $\frac{1}{2}\int \frac{u - 4}{\sqrt{u}}\,du$. Why the correct answer works: Solving the substitution equation for leftover $x$-terms is a standard technique that keeps the integral within the u-substitution framework. Why distractors fail: Option A prematurely abandons a working method. Option C introduces unnecessary complexity—no second variable is needed. Option D yields $du = 3x^2\,dx$, which does not match the denominator structure.

Answer: The substitution $u = x^2$ gives $du = 2x\,dx$, but the integrand has no $x$-factor to match, so the substitution cannot eliminate $x$ from the integral

Why the substitution fails: Setting $u = x^2$ gives $du = 2x\,dx$, requiring a factor of $x$ in the integrand to replace with $du$. Since $e^{x^2}\,dx$ has no such factor, we cannot write $dx$ purely in terms of $du$ without introducing $\frac{1}{2x} = \frac{1}{2\sqrt{u}}$, leaving a non-trivial expression in $u$. The deeper issue: The integral $\int e^{x^2}\,dx$ has no closed-form antiderivative in terms of elementary functions. The student's answer $\frac{e^{x^2}}{2x}$ is incorrect, as differentiating it does not return $e^{x^2}$. Why distractors fail: Option B is incorrect because the derivative of $e^{x^2}$ does involve $x^2$ (via the chain rule, it's $2xe^{x^2}$). Option C is wrong because the issue is not a constant factor—it's the variable $x$. Option D would not simplify the integral either.

Answer: $\frac{62}{5}$

Set up the substitution: Let $u = x^3 + 1$, so $du = 3x^2\,dx$. Then $6x^2\,dx = 2\,du$. Change the limits: When $x = 0$: $u = 1$. When $x = 1$: $u = 2$. Evaluate the integral: The integral becomes $2\int_1^2 u^4\,du = 2\left[\frac{u^5}{5}\right]_1^2 = 2\cdot\frac{32 - 1}{5} = \frac{62}{5}$. Why distractors fail: Option A gives $\frac{31}{5}$, forgetting the factor of $2$ from $6x^2\,dx = 2\,du$. Option C gives $\frac{32}{5}$, which results from evaluating $\frac{u^5}{5}$ at the upper limit only without subtracting the lower limit, then multiplying by $2$. Option D gives $\frac{62}{3}$, which results from using $\frac{u^3}{3}$ instead of $\frac{u^5}{5}$ for the antiderivative.

Answer: $\frac{1}{2}$

Apply the substitution: With $u = \sin(x)$, $du = \cos(x)\,dx$. When $x = 0$: $u = 0$; when $x = \pi/2$: $u = 1$. Evaluate: $\int_0^1 u\,du = \left[\frac{u^2}{2}\right]_0^1 = \frac{1}{2}$. Why distractors fail: Option A equals $1$, which omits the $\frac{1}{2}$ from integration. Option B equals $\frac{1}{4}$, a common error from misapplying a double-angle formula. Option D equals $0$, which would result from incorrect limit handling.

Answer: Yes, the approach is valid and the answer $\frac{15}{4}$ is correct

Verify the antiderivative: With $u = x^2+1$ and $du = 2x\,dx$, the integral becomes $\int u^3\,du = \frac{u^4}{4}$. In terms of $x$: $\frac{(x^2+1)^4}{4}$. Check the evaluation: Using $x$-limits: $\frac{(1+1)^4}{4} - \frac{(0+1)^4}{4} = \frac{16-1}{4} = \frac{15}{4}$. Equivalently, with $u$-limits: $\frac{2^4 - 1^4}{4} = \frac{15}{4}$. Both approaches agree. Why distractors fail: Option A is wrong because evaluating an antiderivative at transformed limits is equivalent to changing limits first. Option C incorrectly divides by $8$. Option D is wrong because you can use either $x$- or $u$-limits as long as the antiderivative matches.

Answer: $u = e^{2x}+1$, giving $\int \frac{1}{2u}\,du$

Apply the substitution: Let $u = e^{2x}+1$. Then $du = 2e^{2x}\,dx$, so $e^{2x}\,dx = \frac{1}{2}du$. Rewrite the integral: $\int \frac{e^{2x}}{e^{2x}+1}\,dx = \int \frac{1}{u}\cdot\frac{1}{2}\,du = \int \frac{1}{2u}\,du = \frac{1}{2}\ln|u| + C$. Why distractors fail: Option B is technically valid but does not simplify the integral. Option C has $du = 2e^{2x}\,dx$, so $dx = \frac{du}{2u}$ and the integral becomes $\frac{1}{2}\int \frac{du}{u+1}$, not $\int \frac{du}{u+1}$. Option D omits the factor of $\frac{1}{2}$ from the chain rule.

Answer: $\frac{\ln(x)^3}{3} + C$

Set up the substitution: Let $u = \ln(x)$, so $du = \frac{1}{x}dx$. The integral becomes $\int u^2\,du$. Integrate and back-substitute: $\int u^2\,du = \frac{u^3}{3} + C = \frac{\ln(x)^3}{3} + C$. Why distractors fail: Option B uses the wrong exponent and power rule divisor. Option C incorrectly differentiates instead of integrating. Option D forgets the $\frac{1}{3}$ from the power rule.

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

Choose the substitution: Let $u = 1 - x^2$, so $du = -2x\,dx$, meaning $x\,dx = -\frac{1}{2}\,du$. Integrate: $\int \frac{x}{\sqrt{1 - x^2}}\,dx = -\frac{1}{2}\int u^{-1/2}\,du = -\frac{1}{2}\cdot 2u^{1/2} + C = -\sqrt{u} + C = -\sqrt{1-x^2} + C$. Why distractors fail: Option A is the antiderivative of $\frac{1}{\sqrt{1-x^2}}$ (without the $x$ in the numerator). Option C has the wrong sign. Option D is not a standard antiderivative form.

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

Choose the substitution: Let $u = \ln x$, so $du = \frac{1}{x}dx$. The integral becomes $\int \sec^2(u)\,du$. Integrate and back-substitute: $\int \sec^2(u)\,du = \tan(u) + C = \tan(\ln x) + C$. Why distractors fail: Option A is the derivative of $\sec(\ln x)$, not its integral-related form. Option C is the antiderivative of $\tan(u)$, not $\sec^2(u)$. Option D applies the power rule incorrectly to $\sec^2$.

Answer: $du = 3\,dx$

Computing du: Differentiating $u = 3x + 5$ with respect to $x$ gives $\frac{du}{dx} = 3$, so $du = 3\,dx$. Why the correct answer works: The derivative of $3x + 5$ is simply $3$, making $du = 3\,dx$. Why distractors fail: Option B confuses $u$ itself with $du$. Option C incorrectly uses only the constant term. Option D ignores the coefficient $3$ entirely.

Answer: The student forgot to account for the negative sign in $du = -2x\,dx$ when swapping the reversed limits

Trace the substitution: With $u = 4 - x^2$, $du = -2x\,dx$ and $x\,dx = -\frac{1}{2}du$. When $x = 0$: $u = 4$; when $x = 2$: $u = 0$. The sign issue: The integral becomes $-\frac{1}{2}\int_4^0 \sqrt{u}\,du = \frac{1}{2}\int_0^4 \sqrt{u}\,du$, which is positive. If the student kept the limits as $4$ to $0$ without flipping them and also kept the negative, they may have double-negated or mishandled the sign. Why distractors fail: Option A is wrong because $u = 4 - x^2$ is the correct choice. Option C describes a valid alternative approach, not an error. Option D is unrelated to the sign error.

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

Choose the substitution: Let $u = \ln x$, so $du = \frac{1}{x}\,dx$. The integrand $\frac{\cos(\ln x)}{x}\,dx$ becomes $\cos(u)\,du$. Integrate and back-substitute: $\int \cos(u)\,du = \sin(u) + C = \sin(\ln x) + C$. Why distractors fail: Option A has the wrong sign. Option C uses cosine instead of sine. Option D incorrectly includes a factor of $\frac{1}{x}$.

Answer: Yes, the result is correct

Verify the student's work: With $u = x^3+1$ and $du = 3x^2\,dx$, the integral becomes $\frac{1}{3}\int e^u\,du = \frac{1}{3}e^u + C$. Check back-substitution: Replacing $u$ with $x^3 + 1$ gives $\frac{1}{3}e^{x^3+1} + C$, which is what the student wrote. The answer is correct. Why distractors fail: Option A is wrong because the substitution is perfectly valid. Option B is wrong because the student did perform back-substitution. Option D is wrong because $\frac{1}{3}$ correctly comes from $du = 3x^2\,dx$.

Answer: $u = x^2$

Identifying the inner function: In $\cos(x^2)$, the inner function is $x^2$. Its derivative $2x$ appears as a factor in the integrand. Why the correct answer works: Setting $u = x^2$ gives $du = 2x\,dx$, which perfectly matches the remaining factor, simplifying the integral to $\int \cos(u)\,du$. Why distractors fail: Option A uses the outer function as $u$, which does not simplify the integral. Option B sets $u = 2x$, which does not address the composition. Option D is too complex and its derivative does not simplify the integrand.

Answer: $\int (u+3)u^{10}\,du$

Substitute all x-terms: With $u = x - 3$, we have $x = u + 3$ and $(x-3)^{10} = u^{10}$. So the integral becomes $\int (u+3)u^{10}\,du$. Why the correct answer works: Every instance of $x$ must be replaced. The factor $x$ becomes $u + 3$, and $(x-3)^{10}$ becomes $u^{10}$. Why distractors fail: Option A forgets to replace the standalone $x$ factor. Option C replaces $x$ with just the constant $3$. Option D uses $u - 3$ instead of $u + 3$.

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

Compute the new limits: When $x = 1$: $u = 1 + \sqrt{1} = 2$. When $x = 4$: $u = 1 + \sqrt{4} = 3$. Why the correct answer works: Substituting the original $x$-limits into $u = 1 + \sqrt{x}$ yields $u$-limits from $2$ to $3$. Why distractors fail: Option A uses the original $x$-values directly. Option C incorrectly computes $1 + 4 = 5$ instead of $1 + \sqrt{4} = 3$. Option D sets the lower limit to $0$ without justification.

Answer: $\int e^u\,du$

Substituting u and du: With $u = x^5$ and $du = 5x^4\,dx$, the factor $5x^4\,dx$ in the integrand is entirely replaced by $du$. Why the correct answer works: Replacing $x^5$ with $u$ and $5x^4\,dx$ with $du$, the integral $\int 5x^4 e^{x^5}\,dx$ becomes $\int e^u\,du$. Why distractors fail: Option A incorrectly retains a factor of $5u$. Option C incorrectly includes an extra $u$ factor. Option D incorrectly keeps the coefficient $5$ that was already absorbed into $du$.

Answer: To convert the answer from $u$ back to the original variable $x$

Role of back-substitution: After finding the antiderivative in terms of $u$, we replace $u$ with the original expression $g(x)$ so the final answer is in terms of the original variable. Why the correct answer works: An indefinite integral's answer must be expressed in the same variable as the original integrand. Replacing $u$ with $g(x)$ achieves this. Why distractors fail: Option B describes verification, which is a separate step. Option C is wrong because the constant $C$ is unaffected by back-substitution. Option D refers to definite integrals, not indefinite ones.

Answer: Both are correct; changing limits avoids back-substitution and is often more efficient

Compare both methods: With $u = 1+\sqrt{x}$, $du = \frac{1}{2\sqrt{x}}dx$. The integral becomes $\int \frac{du}{u}$. Method 1: change limits to $u=1,3$ and get $\ln 3 - \ln 1 = \ln 3$. Method 2: integrate, back-substitute, evaluate at $x$-limits. Why both are correct: Both approaches are mathematically equivalent. Changing limits is often preferred because it avoids the back-substitution step. Why distractors fail: Options A and B are wrong because both approaches are valid. Option D is wrong because $\frac{1}{2\sqrt{x}(1+\sqrt{x})}$ is integrable on $(0,4]$ and the integral converges.

Answer: It rewrites a composite integrand into a simpler form by changing the variable of integration

Purpose of the technique: U-substitution works by recognizing a composite structure $f(g(x))g'(x)$ and replacing it with a simpler integral in $u$. Why the correct answer works: By substituting $u = g(x)$, the composition is unwrapped and the integral often takes a standard, recognizable form. Why distractors fail: Option A is wrong because we still need to find an antiderivative in $u$. Option C is wrong because the result can be exponential, trigonometric, etc. Option D is wrong because degree reduction is not the general mechanism.