Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of $v(t) = 6t^2 - 4t + 9$, integrate each term separately by increasing the exponent by one and dividing by the new exponent. The integral of $6t^2$ is $2t^3$, the integral of $-4t$ is $-2t^2$, and the integral of 9 is $9t$, resulting in $2t^3 - 2t^2 + 9t + C$. This reverses differentiation because differentiating $2t^3 - 2t^2 + 9t + C$ yields back $6t^2 - 4t + 9$. A tempting distractor like choice E fails because it uses $-4t^2$ instead of $-2t^2$, which comes from incorrectly integrating $-4t$ without dividing by 2. Always remember to add the constant of integration (C) and verify by differentiating the result.

Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of (f'(x) = 2cos x + 5), recall that the integral of (cos x) is (sin x) and of a constant is the variable times that constant. The integral is (2sin x + 5x + C). This reverses differentiation since the derivative of (2sin x + 5x + C) is (2cos x + 5). A tempting distractor like choice D fails because it uses (-2sin x) instead of (2sin x), confusing the integral of cosine with negative sine. Always remember to add the constant of integration (C) and verify by differentiating the result.

Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of (T'(t) = $7t^{-2}$ + 4), integrate term by term by adding one to each exponent and dividing by the new exponent. The integral of $(7t^{-2}$) is $(-7t^{-1}$), and the integral of 4 is (4t), giving $(-7t^{-1}$ + 4t + C). This reverses differentiation since differentiating $(-7t^{-1}$ + 4t + C) returns $(7t^{-2}$ + 4). A tempting distractor like choice B fails because it omits the negative sign from integrating $(t^{-2}$), which should produce a -1 in the denominator. Always remember to add the constant of integration (C) and verify by differentiating the result.

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.

Finding antiderivatives involves applying basic integration rules to reverse the process of differentiation. To find the antiderivative of (r(t) = 8 - $2t^{-3}$), integrate each term by adding one to the exponent and dividing by the new exponent. The integral of 8 is (8t), and the integral of $(-2t^{-3}$) is $(t^{-2}$), resulting in (8t + $t^{-2}$ + C). This reverses differentiation because differentiating (8t + $t^{-2}$ + C) gives back (8 - $2t^{-3}$). A tempting distractor like choice B fails because it incorrectly assigns a negative sign to $(t^{-2}$), mishandling the integration of the negative coefficient. 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 tests your ability to find antiderivatives using basic integration rules. To find an antiderivative of v(t) = 6t² - 4t + 5, 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 5, we get 5t. Since antiderivatives are not unique, we must include the constant of integration C, giving us 2t³ - 2t² + 5t + C. Choice A incorrectly omits the constant C, which is a common error when students forget that indefinite integrals represent families of functions. Remember that every antiderivative must include '+ C' to account for all possible vertical shifts of the function.

This problem tests basic antiderivative skills with linear functions. To find an antiderivative of F(x) = 4 - 10x, we integrate each term: the constant 4 becomes 4x (since ∫c dx = cx), and -10x becomes -10x²/2 = -5x². Adding the constant of integration gives us 4x - 5x² + C. Choice B incorrectly shows 4x - 10x² + C, which results from forgetting to divide by the new exponent when integrating -10x—a common mistake where students only increase the exponent without completing the power rule formula. The key strategy for polynomial integration is to systematically apply xⁿ⁺¹/(n+1) to each term, treating constants as x⁰ terms.

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.