Answer: $\ln|x^3 - 8| + C$

Check the f′/f pattern: Let $f(x) = x^3 - 8$. Then $f'(x) = 3x^2$. The integrand is $\frac{3x^2}{x^3 - 8} = \frac{f'(x)}{f(x)}$. Why the correct answer works: Since the numerator equals $f'(x)$, we get $\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C = \ln|x^3 - 8| + C$. Why distractors fail: Option A incorrectly includes a $\frac{1}{3}$ factor, as if the numerator were $x^2$ instead of $3x^2$. Option B multiplies by $3$ instead of recognizing the numerator already matches $f'(x)$. Option C applies an irrelevant arctan form.

Answer: $\frac{1}{6}\arctan\left(\frac{3x}{2}\right) + C$

Rewrite in standard form: Factor: $4 + 9x^2 = 4\left(1 + \frac{9x^2}{4}\right) = 4\left(1 + \left(\frac{3x}{2}\right)^2\right)$. So the integral becomes $\frac{1}{4}\int \frac{1}{1 + \left(\frac{3x}{2}\right)^2}\, dx$. Apply substitution: Let $u = \frac{3x}{2}$, so $du = \frac{3}{2}\, dx$ and $dx = \frac{2}{3}\, du$. The integral becomes $\frac{1}{4} \cdot \frac{2}{3} \int \frac{1}{1 + u^2}\, du = \frac{1}{6}\arctan(u) + C = \frac{1}{6}\arctan\left(\frac{3x}{2}\right) + C$. Why distractors fail: Option B uses incorrect values for both the coefficient and the argument. Option C incorrectly places $\frac{9x}{4}$ in the argument. Option D has coefficient $\frac{1}{3}$ instead of $\frac{1}{6}$, missing the factor from the substitution.

Answer: $\arcsin\left(\frac{x}{3}\right) + C$

Recognize the standard form: The integral $\int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \arcsin\left(\frac{x}{a}\right) + C$. Here $a^2 = 9$, so $a = 3$. Why the correct answer works: Substituting $a = 3$ directly into the formula gives $\arcsin\left(\frac{x}{3}\right) + C$. Why distractors fail: Option A uses the arctan pattern, which applies to $\int \frac{1}{a^2 + x^2}\, dx$, not the square root form. Option C incorrectly treats this as a logarithmic integral. Option D misapplies the formula by placing the $\frac{1}{3}$ outside instead of inside the argument of arcsin.

Answer: It is valid because $\frac{d}{dx}(x^2 + 6x + 13) = 2x + 6$, matching the $f'/f$ pattern exactly

Verify the f′/f condition: Let $f(x) = x^2 + 6x + 13$. Then $f'(x) = 2x + 6$, which matches the numerator exactly. Why the correct answer works: Since the integrand is $\frac{f'(x)}{f(x)}$, the antiderivative is $\ln|f(x)| + C$. The student's work is correct. Why distractors fail: Option A is wrong — the $f'/f$ logarithmic form does not require real roots in the denominator. Option B is wrong — completing the square is needed only when the numerator does not match $f'(x)$. Option D raises a valid technical point (the denominator is always positive, so absolute values are optional), but this does not invalidate the student's answer — the result with absolute values is still correct.

Answer: Because the derivative of $\arcsin(u)$ is $\frac{1}{\sqrt{1 - u^2}}$, which matches the structure of the integrand after substitution

Connect antiderivatives to derivatives: Integration reverses differentiation. We know $\frac{d}{du}[\arcsin(u)] = \frac{1}{\sqrt{1 - u^2}}$. With the substitution $u = x/a$, the integrand $\frac{1}{\sqrt{a^2 - x^2}}$ takes exactly this form. Why the correct answer works: Option A correctly identifies that the structure of $\frac{1}{\sqrt{1 - u^2}}$ is the derivative of arcsin, so the antiderivative must involve arcsin. Why distractors fail: Option B is false — arctan is well-defined for all real arguments, including expressions with square roots. Option C is incorrect — the result does not contain a square root; the square root is in the integrand, not the antiderivative. Option D reverses the domains: arctan is defined on all reals, while arcsin has domain $[-1, 1]$.

Answer: $\frac{1}{2}\arctan\left(\frac{x + 3}{2}\right) + C$

Complete the square: $x^2 + 6x + 13 = (x + 3)^2 + 4 = (x + 3)^2 + 2^2$. Apply the arctan formula: With $u = x + 3$ and $a = 2$: $\int \frac{1}{u^2 + 4}\, du = \frac{1}{2}\arctan\left(\frac{u}{2}\right) + C = \frac{1}{2}\arctan\left(\frac{x+3}{2}\right) + C$. Why distractors fail: Option B is missing the $\frac{1}{a} = \frac{1}{2}$ coefficient. Option C uses $x - 3$ instead of $x + 3$, reflecting an error in completing the square. Option D treats the integral as a logarithmic form, but the numerator is $1$, not the derivative of the denominator.

Answer: $\arcsin\left(\frac{x + 3}{4}\right) + C$

Complete the square under the radical: $7 - 6x - x^2 = -(x^2 + 6x - 7) = -((x+3)^2 - 16) = 16 - (x+3)^2$. So $a = 4$. Apply the arcsin formula: With $u = x + 3$, we get $\int \frac{1}{\sqrt{16 - u^2}}\, du = \arcsin\left(\frac{u}{4}\right) + C = \arcsin\left(\frac{x+3}{4}\right) + C$. Why distractors fail: Option B uses $x - 3$ instead of $x + 3$, an error in completing the square. Option C includes an unnecessary $\frac{1}{4}$ coefficient. Option D uses arctan instead of arcsin; the square root in the denominator signals the arcsin pattern.

Answer: The logarithmic form $\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C$

Check the numerator against the derivative of the denominator: Let $f(x) = x^2 + 5$. Then $f'(x) = 2x$. The integrand is exactly $\frac{f'(x)}{f(x)}$. Why the correct answer works: Since the numerator is the derivative of the denominator, we apply $\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C = \ln|x^2 + 5| + C$. Why distractors fail: Option A is wrong because the arcsin pattern requires $\sqrt{a^2 - x^2}$ in the denominator. Option B would apply if the numerator were a constant (not $2x$). Option D is unnecessary since the denominator does not factor over the reals and the integral has a simpler logarithmic form.

Answer: The logarithmic form requires the numerator to be the derivative of the denominator, but $\frac{d}{dx}(x^2 + 4x + 5) = 2x + 4 \neq 1$

Identify the error in reasoning: The $\ln|f(x)| + C$ result applies when the integrand is $\frac{f'(x)}{f(x)}$. Here $f(x) = x^2 + 4x + 5$ and $f'(x) = 2x + 4$, but the numerator is $1$, not $2x + 4$. What the correct approach is: Completing the square: $x^2 + 4x + 5 = (x + 2)^2 + 1$. The integral becomes $\int \frac{1}{(x+2)^2 + 1}\, dx = \arctan(x + 2) + C$. Why distractors fail: Option A is wrong because having no real roots does not prevent integration. Option C is incorrect because partial fractions apply when the denominator factors over the reals. Option D is too restrictive — the logarithmic form works for any $\frac{f'(x)}{f(x)}$, not only when $f$ is linear.

Answer: $\frac{1}{5}\arcsin\left(\frac{5x}{4}\right) + C$

Rewrite to match the arcsin pattern: Write $16 - 25x^2 = 16\left(1 - \frac{25x^2}{16}\right) = 16\left(1 - \left(\frac{5x}{4}\right)^2\right)$. Then $\sqrt{16 - 25x^2} = 4\sqrt{1 - \left(\frac{5x}{4}\right)^2}$. Apply substitution: Let $u = \frac{5x}{4}$, so $du = \frac{5}{4}\, dx$. The integral becomes $\frac{1}{4} \cdot \frac{4}{5}\int \frac{1}{\sqrt{1 - u^2}}\, du = \frac{1}{5}\arcsin(u) + C = \frac{1}{5}\arcsin\left(\frac{5x}{4}\right) + C$. Why distractors fail: Option A uses coefficient $\frac{1}{4}$ instead of $\frac{1}{5}$. Option B omits the coefficient entirely. Option D has an incorrect argument $\frac{x}{4}$ and wrong coefficient $5$.

Answer: $\frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$

Recall the arctan integral pattern: The standard result is $\int \frac{1}{a^2 + x^2}\, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$. Why the correct answer works: Option C matches the standard formula exactly, with the $\frac{1}{a}$ coefficient and $\frac{x}{a}$ as the argument of arctan. Why distractors fail: Option A uses arcsin, which applies to the $\frac{1}{\sqrt{a^2 - x^2}}$ form. Option B has the coefficient as $a$ instead of $\frac{1}{a}$. Option D is a logarithmic form, which would only be correct for $\int \frac{f'(x)}{f(x)}\, dx$; the derivative of $a^2 + x^2$ is $2x$, not $1$.

Answer: Write $2x + 8 = (2x + 4) + 4$, then integrate $\frac{2x+4}{x^2+4x+8}$ as a logarithm and $\frac{4}{x^2+4x+8}$ using completing the square with arctan

Split the numerator strategically: The derivative of $x^2 + 4x + 8$ is $2x + 4$. Write $2x + 8 = (2x + 4) + 4$ to separate the integral into $\int \frac{2x+4}{x^2+4x+8}\, dx + \int \frac{4}{x^2+4x+8}\, dx$. Evaluate each part: The first integral is $\ln|x^2+4x+8| + C_1$ by the $f'/f$ rule. For the second, complete the square: $x^2+4x+8 = (x+2)^2+4$, giving $4 \cdot \frac{1}{2}\arctan\left(\frac{x+2}{2}\right) + C_2 = 2\arctan\left(\frac{x+2}{2}\right) + C_2$. Why distractors fail: Option B fails because partial fractions require factoring over the reals, and $x^2+4x+8$ is irreducible. Option C ignores that the numerator $2x+8$ has a linear term, so arctan alone does not suffice. Option D fails because $du = (2x+4)\, dx \neq (2x+8)\, dx$, so the substitution does not account for the extra constant.