The skill here is integrating using u-substitution. The denominator is (x² + 7x + $1)^4$, and numerator 2x + 7 is its derivative. Set u = x² + 7x + 1 with du = (2x + 7) dx, so ∫ $u^{-4}$ du = $-1/(3u^3$) + C. This is -1/[3(x² + 7x + $1)^3$] + C. Choice D omits the 3 in the denominator, forgetting the power rule division. Identify negative powers in denominators with matching numerators for substitution.

This integral requires the skill of integration by substitution, often called u-substitution. The denominator is 4x + 5, with derivative 4, and the numerator is a constant 9. Set u = 4x + 5, du = 4 dx, dx = du/4, so the integral is (9/4) ∫ du/u. This integrates to (9/4) ln|u| + C, or (9/4) ln|4x + 5| + C. A tempting distractor is choice B, using 9 without dividing by 4, likely from ignoring the coefficient in du. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

This integral requires the skill of integration by substitution, often called u-substitution. The integrand is sin(3t + π), where 3t + π is the inner function, and its derivative is 3, suggesting a need to account for the chain rule. Setting u = 3t + π gives du = 3 dt, so dt = du/3, transforming the integral to (1/3) ∫ sin u du. Integrating yields - (1/3) cos u + C, or - (1/3) cos(3t + π) + C. A tempting distractor is choice A, which omits the 1/3 factor, likely from forgetting to include the adjustment for du. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

This integral requires the skill of integration by substitution, often called u-substitution. The denominator is 1 + tan(3x), and sec²(3x) is the derivative of tan(3x) times 3. Set u = 1 + tan(3x), du = 3 sec²(3x) dx, so sec²(3x) dx = du/3, transforming to (1/3) ∫ du/u. This integrates to (1/3) ln|u| + C, or (1/3) ln|1 + tan(3x)| + C. A tempting distractor is choice A, omitting the 1/3, perhaps from ignoring the chain rule factor of 3. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

This integral requires the skill of integration by substitution, often called u-substitution. The numerator 12x³ is the derivative of the inner function 3x⁴ + 1 in the denominator, making substitution straightforward. Let u = 3x⁴ + 1, so du = 12x³ dx, turning the integral into ∫ du/u. This integrates to ln|u| + C, or ln|3x⁴ + 1| + C. A tempting distractor is choice B, which multiplies by 4 unnecessarily, perhaps from miscounting the coefficient in du. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.

This integral requires u-substitution to handle the composite function. Notice that the integrand contains $(3t^2+5)^4$ multiplied by $6t$, and the derivative of the inner function $3t^2+5$ is exactly $6t$. Setting $u = 3t^2 + 5$ gives $du = 6t,dt$, transforming the integral to $\int u^4,du = \frac{u^5}{5} + C$. Substituting back yields $\frac{(3t^2+5)^5}{5} + C$. Choice C incorrectly divides by an extra factor of 6, forgetting that $6t,dt$ is already accounted for in the substitution. When you see a function raised to a power multiplied by its derivative, u-substitution directly integrates to increase the power by 1 and divide by the new power.

This integral requires u-substitution for the exponential function with a linear exponent. The inner function is $3x + 7$, whose derivative is 3, matching the coefficient in front of the exponential. Setting $u = 3x + 7$ gives $du = 3,dx$, so the integral becomes $\int e^u,du = e^u + C$. Substituting back yields $e^{3x+7} + C$. Choice C incorrectly divides by 3, which would be necessary if the original integral lacked the factor of 3 in front. When integrating $ae^{ax+b}$ where $a$ is the coefficient of $x$ in the exponent, the result is simply $e^{ax+b} + C$.

This integral requires u-substitution where we recognize that $8x$ is the derivative of $4x^2 - 3$. Setting $u = 4x^2 - 3$ gives $du = 8x,dx$, which perfectly matches the $8x,dx$ in our integral. The integral becomes $\int u^6,du = \frac{u^7}{7} + C$. Substituting back yields $\frac{(4x^2-3)^7}{7} + C$. Choice B incorrectly divides by an additional factor of 8, forgetting that the $8x$ is already accounted for in the substitution. When the integrand is of the form $f'(x)[f(x)]^n$, the antiderivative is $\frac{[f(x)]^{n+1}}{n+1} + C$.

This integral features a rational function where u-substitution is key because the numerator is the derivative of a function in the denominator. Setting u = 1 + x⁴, we get du = 4x³ dx, which exactly matches the 4x³ dx in our integrand. The integral becomes ∫ du/u = ln|u| + C = ln(1+x⁴) + C. Choice B (¼ln(1+x⁴) + C) incorrectly assumes we need to compensate with a factor of ¼, but the 4x³ is completely absorbed by the substitution. When the numerator of a fraction is exactly the derivative of the denominator, the integral is simply the natural logarithm of the denominator.

This integral features an exponential function with a squared linear expression in the exponent, perfect for u-substitution. Let u = (x+4)², so du = 2(x+4) dx, which means (x+4) dx = ½ du. The integral becomes ∫ eᵘ · ½ du = ½ ∫ eᵘ du = ½eᵘ + C = ½e^((x+4)²) + C. Choice A (e^((x+4)²) + C) forgets the factor of ½ that comes from the chain rule working backwards. When integrating e^(f(x)) where f'(x) appears as a factor, divide by any constant in f'(x).