Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of $F(x) = 12x^{1/2} - 5$, integrate each term by increasing the power by one and dividing by the new power. The integral of $12x^{1/2}$ is $8x^{3/2}$, and the integral of -5 is $-5x$, yielding $8x^{3/2} - 5x + C$. This reverses differentiation as the derivative of $8x^{3/2} - 5x + C$ is $12x^{1/2} - 5$. A tempting distractor like choice D fails because it uses 12 instead of 8 for the coefficient, forgetting to divide by $3/2$ properly. Always remember to add the constant of integration (C) and verify by differentiating the result.

This problem tests basic antiderivative reasoning skills. The antiderivative of $R'(x) = 3x^{2/3} - 12x^{-1/3}$ involves reversing differentiation with fractional powers. For $3x^{2/3}$, add 1 to get $5/3$, divide by $5/3$, resulting in $\dfrac{9}{5}x^{5/3}$. For $-12x^{-1/3}$, add 1 to get $2/3$, divide by $2/3$, yielding $-18x^{2/3}$. Choice C fails by doubling to $-36x^{2/3}$, mishandling the coefficient. Convert all terms to exponents and carefully compute new coefficients as a reliable strategy, adding $+C$.

This question requires finding an antiderivative by applying the power rule to each term separately. For C'(x) = 9x² + 8x⁻¹/², we integrate term by term: ∫9x² dx = 9x³/3 = 3x³, and ∫8x⁻¹/² dx = 8x¹/²/(1/2) = 16x¹/². Combining these results with the constant of integration gives 3x³ + 16x¹/² + C. The key insight is that x⁻¹/² integrates to x¹/² multiplied by 2, not divided by 2, because we divide by the new exponent (1/2). Choice B incorrectly computes the second term as 4x¹/², likely from misapplying the power rule by multiplying rather than dividing by 1/2. When integrating negative or fractional powers, always remember to add 1 to the exponent and divide by the new exponent.

This problem asks for an antiderivative of f'(x) = sec²x + 10. Finding antiderivatives reverses differentiation: the antiderivative of sec²x is tan x (since d/dx(tan x) = sec²x), and the antiderivative of 10 is 10x. Including the constant of integration gives tan x + 10x + C. Choice E shows -tan x + 10x + C, incorrectly using a negative sign with tan x. Remember: the derivative of tan x is sec²x, so the antiderivative of sec²x is tan x, not -tan x.

This problem tests your ability to find antiderivatives using basic integration rules. To find an antiderivative of v(t) = 6t² - 4t + 9, we reverse the differentiation process by applying the power rule for integration: ∫tⁿ dt = tⁿ⁺¹/(n+1) + C. For 6t², we get 6t³/3 = 2t³; for -4t, we get -4t²/2 = -2t²; and for 9, we get 9t. Since antiderivatives are not unique, we must include the constant of integration C, giving us 2t³ - 2t² + 9t + C. Choice A is tempting because it shows the correct integration of each term, but it's missing the crucial constant C that represents the family of all antiderivatives. Remember that when finding antiderivatives, always add the constant of integration unless you're evaluating a definite integral.

This problem tests basic antiderivative reasoning skills. The antiderivative of r(t) = 5t⁴ - $3t^{-2}$ is found by reversing differentiation using the power rule on each term. For 5t⁴, increase the exponent to 5 and divide by 5, giving t⁵. For $-3t^{-2}$, increase to -1 and divide by -1, resulting in $3t^{-1}$. Choice B might tempt if one forgets to change the sign when integrating the negative term, but it incorrectly has $-3t^{-1}$. A key strategy for integrating power functions is to handle negative exponents carefully and always add +C.

This problem involves integrating T'(t) = 7t⁶ + 2/t³, which we rewrite as T'(t) = 7t⁶ + 2t⁻³. Applying the power rule: ∫7t⁶ dt = 7t⁷/7 = t⁷, and ∫2t⁻³ dt = 2t⁻²/(-2) = -t⁻² = -1/t². The complete antiderivative is t⁷ - 1/t² + C. The positive term 2/t³ becomes negative after integration due to dividing by the negative new exponent -2. Choice C incorrectly shows +1/t², missing the sign change that occurs when integrating negative powers. When integrating t⁻ⁿ where n > 0, the result will have the opposite sign of the original coefficient due to division by a negative number.

This problem tests integration of polynomial terms with various powers. To find an antiderivative of g(x) = 4 - 9x² + (1/2)x⁴, we integrate each term separately using the power rule. For the constant 4, we get 4x; for -9x², we get -9x³/3 = -3x³; for (1/2)x⁴, we get (1/2)x⁵/5 = x⁵/10. The complete antiderivative is 4x - 3x³ + x⁵/10 + C. Choice B (4x - 9x³ + (1/2)x⁵ + C) incorrectly integrates -9x² as -9x³ instead of -3x³, forgetting to divide by the new exponent 3. Always remember to divide by (n+1) when integrating xⁿ, as this step is essential for reversing the differentiation process.

This problem asks for an antiderivative of the marginal cost function $M(x) = 9x^2 + 2x^{-1}$. Finding antiderivatives means reversing differentiation: for each term, add 1 to the exponent and divide by the new exponent. For $9x^2$, we get $9x^3 / 3 = 3x^3$. For $2x^{-1}$, we add 1 to get $x^0 = 1$, but the antiderivative of $1/x$ is $\ln|x|$, so we get $2\ln|x|$. Including the constant of integration, the antiderivative is $3x^3 + 2\ln|x| + C$. Choice D incorrectly shows $3x^2 + \dfrac{2}{x} + C$, which appears to be a partial differentiation rather than integration. When integrating, remember that $\int(1/x) , dx = \ln|x| + C$, not a power rule application.

This acceleration problem requires finding an antiderivative using the power rule for integration. To integrate $a(t) = 12t^3 + 6t$, we apply $\int t^n , dt = \frac{t^{n+1}}{n+1} + C$ to each term: $12t^3$ becomes $\frac{12t^4}{4} = 3t^4$, and $6t$ becomes $\frac{6t^2}{2} = 3t^2$. The complete antiderivative is $3t^4 + 3t^2 + C$. Choice E shows $3t^4 - 3t^2 + C$, which has the wrong sign on the second term—this error occurs when students mistakenly change signs during integration, perhaps confusing it with differentiation rules. Remember that integration preserves the signs of the original terms unless the division step introduces a sign change (which only happens with negative exponents).