Answer: The first yields $\arcsin(x) + C$ and the second yields $\arctan(x) + C$.

Match each form to its inverse trig antiderivative: $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x) + C$ and $\int \frac{1}{1+x^2}\,dx = \arctan(x) + C$. The square root in the denominator is the distinguishing feature. Why the correct answer works: Option B correctly pairs each integrand with its standard inverse trig antiderivative. Why distractors fail: Option A incorrectly assigns the same result to both. Option C reverses the pairings. Option D incorrectly claims both yield $\arcsin$.

Answer: The claim is incorrect; only $n = -1$ requires $\ln|x| + C$, while all other negative exponents use the power rule.

Clarify when logarithm is needed: The power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ works for all $n \neq -1$, including negative values like $n = -2, -3, -0.5$, etc. Why the correct answer works: Option B correctly identifies that only the single value $n = -1$ is excluded from the power rule and requires $\ln|x| + C$. Why distractors fail: Option A overgeneralizes; e.g., $\int x^{-2}\,dx = -x^{-1} + C$, not $\ln|x| + C$. Option C creates a false distinction between integer and fractional exponents. Option D is entirely incorrect.

Answer: It represents the entire family of antiderivatives that differ by a constant.

Understand why +C is needed: If $F(x)$ is an antiderivative of $f(x)$, then $F(x) + k$ is also an antiderivative for any constant $k$, because the derivative of a constant is zero. Why the correct answer works: Option B correctly states that $C$ represents all possible vertical shifts of a single antiderivative, forming a family of functions. Why distractors fail: Option A is incorrect because $C$ is not about rounding. Option C is wrong because the antiderivative need not pass through the origin. Option D conflates indefinite and definite integrals; $C$ is specific to indefinite integrals.

Answer: $x^3 + \frac{5x^2}{2} - 7x + C$

Integrate term by term: $\int 3x^2\,dx = 3 \cdot \frac{x^3}{3} = x^3$. $\int 5x\,dx = 5 \cdot \frac{x^2}{2} = \frac{5x^2}{2}$. $\int -7\,dx = -7x$. Why the correct answer works: Combining the results gives $x^3 + \frac{5x^2}{2} - 7x + C$, which is Option A. Why distractors fail: Option B is the derivative, not the antiderivative. Option C fails to divide the $5x$ term by 2. Option D fails to divide the $3x^2$ coefficient by 3.

Answer: $\frac{x^3}{3} + 2\ln|x| + C$

Simplify the integrand first: $\frac{x^3+2}{x} = \frac{x^3}{x} + \frac{2}{x} = x^2 + \frac{2}{x}$. Integrate term by term: $\int x^2\,dx = \frac{x^3}{3}$ and $\int \frac{2}{x}\,dx = 2\ln|x|$. Why distractors fail: Option B integrates $x^3 + 2$ without dividing by $x$ first. Option C omits the absolute value. Option D treats the integrand as a single composite function suitable for a logarithmic substitution, which is incorrect here.

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

Integrate term by term: $\int e^x\,dx = e^x$. $\int \frac{1}{x}\,dx = \ln|x|$. $\int x^{-3}\,dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$. Note the equivalence trap: Options A and B are algebraically identical: $-\frac{1}{2x^2}$ and $\frac{x^{-2}}{-2}$ are the same expression. However, as presented, Option A is in simplified form which is the conventional final answer. Why distractors fail: Option B, while algebraically equal to A, is marked as the unsimplified form — only one answer can be correct and A is the simplified standard. Option C omits the absolute value. Option D divides by $-3$ instead of $-2$ when integrating $x^{-3}$.

Answer: $4\ln|x| - x^3 + \arcsin(x) + C$

Integrate each term: $\int \frac{4}{x}\,dx = 4\ln|x|$. $\int -3x^2\,dx = -x^3$. $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x)$. Why the correct answer works: Option A correctly applies the logarithmic, power, and inverse trig rules with proper coefficients. Why distractors fail: Option B omits the absolute value on $\ln$ and uses $\arctan$ instead of $\arcsin$. Option C fails to divide the $-3$ coefficient by 3. Option D differentiates $4/x$ instead of integrating.

Answer: $6\ln|x| + \arcsin(x) - 6\ln(1/2) - \pi/6$

Find the general antiderivative: $\int \left(\frac{6}{x} + \frac{1}{\sqrt{1-x^2}}\right)dx = 6\ln|x| + \arcsin(x) + C$. Apply the initial condition: $F(1/2) = 6\ln(1/2) + \arcsin(1/2) + C = 6\ln(1/2) + \frac{\pi}{6} + C = 0$, so $C = -6\ln(1/2) - \frac{\pi}{6}$. Why the correct answer works: Option A substitutes this specific value of $C$ into the general antiderivative, satisfying both the derivative condition and the initial condition. Why distractors fail: Option B uses $\arctan$ instead of $\arcsin$. Option C omits the absolute value and does not apply the initial condition. Option D leaves $C$ undetermined, violating the initial condition requirement.

Answer: $C = -3$

Substitute the point into F(x): $F(3) = \frac{27}{3} - 6 + C = 9 - 6 + C = 3 + C$. Setting $F(3) = 0$ gives $3 + C = 0$. Solve for C: $C = -3$. This selects the unique member of the antiderivative family passing through $(3,0)$. Why distractors fail: Option A gives $F(3) = 6 \neq 0$. Option C gives $F(3) = 3 \neq 0$. Option D gives $F(3) = -6 \neq 0$.

Answer: Correct, because rewriting $\sqrt{4-x^2}$ as $2\sqrt{1-(x/2)^2}$ yields the arcsin pattern with $a=2$.

Rewrite the integrand: $\frac{1}{\sqrt{4-x^2}} = \frac{1}{\sqrt{4(1-(x/2)^2)}} = \frac{1}{2\sqrt{1-(x/2)^2}}$. Apply the generalized arcsin formula: Using $\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\left(\frac{x}{a}\right) + C$ with $a = 2$, the result is $\arcsin\left(\frac{x}{2}\right) + C$. Why distractors fail: Option A is wrong because it does match the arcsin pattern after factoring. Option C confuses the arcsin and arctan forms. Option D has the wrong domain restriction; the domain is $|x| < 2$, not $|x| < 4$.

Answer: $\ln|x| + C$

Identify the special case n = −1: The power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ is undefined when $n = -1$ because it produces division by zero. The logarithmic antiderivative handles this case. Why the correct answer works: Option C, $\ln|x| + C$, is correct because the absolute value ensures the expression is defined for all $x \neq 0$, including negative values. Why distractors fail: Option A produces division by zero. Option B omits the absolute value, making it undefined for $x < 0$. Option D is the derivative of $\frac{1}{x}$, not its antiderivative.

Answer: Because when $n = -1$, the formula produces division by zero, so a different antiderivative ($\ln|x|$) is needed.

Identify the algebraic issue: Substituting $n = -1$ into $\frac{x^{n+1}}{n+1}$ gives $\frac{x^0}{0} = \frac{1}{0}$, which is undefined. Why the correct answer works: Option B correctly identifies division by zero as the reason and notes the logarithmic antiderivative as the appropriate replacement. Why distractors fail: Option A is incorrect because $x^{-1}$ is continuous on $(0, \infty)$ and $(-\infty, 0)$. Option C is wrong because the power rule works for all real $n \neq -1$, including negative values. Option D incorrectly claims $x^0/(0)$ equals 1.

Answer: $3\ln|x| + 2e^x + C$

Integrate each term: $\int \frac{3}{x}\,dx = 3\ln|x|$ and $\int 2e^x\,dx = 2e^x$. Why the correct answer works: Option B correctly uses $\ln|x|$ (with absolute value) and keeps the coefficient 2 with $e^x$. Why distractors fail: Option A omits the absolute value on $\ln$. Option C incorrectly applies the power rule to $\frac{3}{x}$ by treating it as a derivative. Option D incorrectly moves the coefficient 2 into the exponent of $e$.

Answer: The student applied the power rule at $n = -1$, where the formula is undefined due to division by zero.

Identify the error: The power rule formula $\frac{x^{n+1}}{n+1}$ requires $n \neq -1$. When $n = -1$, the denominator becomes $-1 + 1 = 0$, making the expression undefined. Why the correct answer works: Option B accurately describes the error: applying the power rule at its excluded value, resulting in division by zero. Why distractors fail: Option A is incorrect because the student did include $C$. Option C is wrong because integration by parts is unnecessary; the correct antiderivative is simply $\ln|x| + C$. Option D is wrong because $-1+1 = 0$ is correct arithmetic; the issue is that 0 appears in the denominator.

Answer: $\arctan(3x) + C$

Rewrite the integrand to match the arctan form: $\frac{3}{1+9x^2} = \frac{3}{1+(3x)^2}$. Using the formula $\int \frac{a}{a^2+u^2}\,du$ or recognizing $\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$ with suitable rearrangement. Detailed computation: Factor: $\frac{3}{1+(3x)^2}$. Let $u = 3x$, $du = 3\,dx$, so $dx = du/3$. Then $\int \frac{3}{1+u^2} \cdot \frac{du}{3} = \int \frac{1}{1+u^2}\,du = \arctan(u) + C = \arctan(3x) + C$. Why distractors fail: Option A uses $9x$ inside arctan instead of $3x$. Option C includes an extra factor of $1/3$. Option D includes an extra factor of 3.

Answer: When evaluating a definite integral using the Fundamental Theorem of Calculus

Understand when C cancels: In a definite integral $\int_a^b f(x)\,dx = F(b) - F(a)$, the constant $C$ appears in both $F(b) + C$ and $F(a) + C$, so it cancels in the subtraction. Why the correct answer works: Option B is correct because definite integration involves a subtraction $F(b) - F(a)$ that eliminates any additive constant. Why distractors fail: Option A is too broad; indefinite integrals require $C$. Option C is wrong because initial value problems use $C$ to find a specific solution. Option D is irrelevant; the type of integrand does not determine whether $C$ matters.

Answer: The general formula $\int a^x\,dx = \frac{a^x}{\ln a} + C$ reduces to $\int e^x\,dx = e^x + C$ because $\ln e = 1$.

State the general formula: For any base $a > 0$, $a \neq 1$: $\int a^x\,dx = \frac{a^x}{\ln a} + C$. Show the special case: When $a = e$, $\ln e = 1$, so $\frac{e^x}{\ln e} = \frac{e^x}{1} = e^x$. Thus the $e^x$ rule is a special case of the general rule. Why distractors fail: Option A ignores the $\frac{1}{\ln a}$ factor. Option C falsely claims the rules are unrelated. Option D multiplies by $\ln a$ instead of dividing.

Answer: The student applied the power rule for $\int x^n\,dx$ to an exponential function $\int a^x\,dx$.

Identify the error pattern: The expression $\frac{2^{x+1}}{x+1}$ mimics $\frac{x^{n+1}}{n+1}$, revealing that the student treated the base 2 as though $x$ were in the base (power rule) rather than in the exponent (exponential rule). State the correct result: The correct antiderivative is $\int 2^x\,dx = \frac{2^x}{\ln 2} + C$. Why distractors fail: Option A is inaccurate because the logarithmic rule involves $\int 1/x\,dx$. Option C is wrong because the student did include $C$. Option D is irrelevant to the error shown.

Answer: It is incorrect; differentiating gives $3^x (\ln 3)^2$, not $3^x$.

Differentiate the proposed answer: $\frac{d}{dx}[3^x \ln 3 + C] = (\ln 3) \cdot \frac{d}{dx}[3^x] = (\ln 3)(3^x \ln 3) = 3^x(\ln 3)^2$. Compare with the original integrand: The derivative is $3^x (\ln 3)^2 \neq 3^x$. The correct antiderivative should be $\frac{3^x}{\ln 3} + C$. Why distractors fail: Option A falsely claims the differentiation yields $3^x$. Option C incorrectly computes the derivative. Option D acknowledges the role of $\ln 3$ but arrives at the wrong conclusion about correctness.

Answer: $e^x + C$

Recall the exponential antiderivative rule: The function $e^x$ is its own antiderivative: $\int e^x\,dx = e^x + C$. Why the correct answer works: Option B is the standard result. Differentiating $e^x + C$ gives $e^x$, confirming correctness. Why distractors fail: Option A incorrectly applies the power rule to the exponential. Option C also misapplies the power rule structure. Option D simplifies to $x + C$, whose derivative is 1, not $e^x$.

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

Apply the power rule: $\int x^{-3/2}\,dx = \frac{x^{-3/2+1}}{-3/2+1} + C = \frac{x^{-1/2}}{-1/2} + C = -2x^{-1/2} + C = -\frac{2}{\sqrt{x}} + C$. Why the correct answer works: Option A correctly computes $n+1 = -1/2$ and divides by $-1/2$, yielding multiplication by $-2$. Why distractors fail: Option B differentiates instead of integrating. Option C has the wrong sign (positive instead of negative). Option D uses the original exponent rather than the new exponent.

Answer: $f(x) = x^3 + 3$

Differentiate F(x): $F'(x) = \frac{d}{dx}\left[\frac{x^4}{4}\right] + \frac{d}{dx}[3x] + \frac{d}{dx}[C] = x^3 + 3 + 0 = x^3 + 3$. Why the correct answer works: Option B matches the result of differentiating $F(x)$ term by term. Why distractors fail: Option A fails to apply the power rule for differentiation (drops the factor of $\frac{1}{4}$ cancellation). Option C multiplies by 4 without cancelling the denominator 4, giving $4x^3$. Option D integrates $F(x)$ instead of differentiating.

Answer: $\frac{2}{3}x^{3/2} + 4\ln|x| + C$

Rewrite and integrate term by term: $\sqrt{x} = x^{1/2}$. By the power rule: $\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$. Also, $\int \frac{4}{x}\,dx = 4\ln|x|$. Why the correct answer works: Option A correctly combines both terms with the absolute value on the logarithm. Why distractors fail: Option B differentiates $\sqrt{x}$ instead of integrating. Option C omits the absolute value on $\ln$. Option D incorrectly uses $\frac{3}{2}$ as the coefficient instead of $\frac{2}{3}$ (it multiplied by $\frac{3}{2}$ rather than dividing).

Answer: $\arctan(x) + C$

Recognize the inverse trig form: The integrand $\frac{1}{1+x^2}$ matches the derivative of $\arctan(x)$. Why the correct answer works: Option C is correct: $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$, so $\int \frac{1}{1+x^2}\,dx = \arctan(x) + C$. Why distractors fail: Option A ($\arcsin$) corresponds to $\frac{1}{\sqrt{1-x^2}}$. Option B would be the antiderivative of $\frac{2x}{1+x^2}$, not $\frac{1}{1+x^2}$. Option D ($\text{arcsec}$) corresponds to $\frac{1}{|x|\sqrt{x^2-1}}$.

Answer: $2\arcsin(x) - 5\arctan(x) + 7e^x + C$

Integrate term by term: $\int \frac{2}{\sqrt{1-x^2}}\,dx = 2\arcsin(x)$. $\int \frac{-5}{1+x^2}\,dx = -5\arctan(x)$. $\int 7e^x\,dx = 7e^x$. Why the correct answer works: Option B correctly combines all three terms with proper coefficients and signs. Why distractors fail: Option A reverses arcsin and arctan. Option C moves the coefficient 7 into the exponent of $e$. Option D reverses the signs on the inverse trig terms.

Answer: $\frac{5^x}{\ln 5} + C$

Apply the general exponential antiderivative: For $a > 0$, $a \neq 1$: $\int a^x\,dx = \frac{a^x}{\ln a} + C$. Here $a = 5$. Why the correct answer works: Option C correctly divides $5^x$ by $\ln 5$. Why distractors fail: Option A multiplies by $\ln 5$ instead of dividing (this is the derivative of $5^x$). Option B misapplies the power rule for polynomials to an exponential. Option D omits the $\frac{1}{\ln 5}$ factor.

Answer: $x^{-1}$

Determine when the power rule fails: The power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ works for all real $n$ except $n = -1$, where the denominator is zero. Why the correct answer works: Option C has $n = -1$, the only case requiring $\int x^{-1}\,dx = \ln|x| + C$. Why distractors fail: Option A ($n=-2$), Option B ($n=1/3$), and Option D ($n=0$) all have $n \neq -1$, so the standard power rule applies to each.

Answer: $\frac{x^5}{5} + C$

Apply the Power Rule for Integrals: The power rule states $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. Here $n = 4$, so the result is $\frac{x^5}{5} + C$. Why the correct answer works: Option A correctly increases the exponent by 1 (to 5) and divides by the new exponent (5). Why distractors fail: Option B gives the derivative, not the antiderivative. Option C divides by the original exponent 4 instead of the new exponent 5. Option D increases the exponent correctly but divides by 4 instead of 5.

Answer: $\int \frac{1}{\sqrt{1-x^2}}\,dx$

Recall the inverse trig antiderivative formulas: $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x) + C$. The other standard forms yield $\arctan(x)$ and $\text{arcsec}(x)$. Why the correct answer works: Option B matches the standard form whose derivative is $\frac{1}{\sqrt{1-x^2}}$, which is the derivative of $\arcsin(x)$. Why distractors fail: Option A yields $\arctan(x) + C$. Option C yields $\text{arcsec}|x| + C$. Option D can be decomposed via partial fractions and does not yield an inverse trig function directly.