This integral requires u-substitution to handle the compound logarithmic function. Let u = ln x, so du = (1/x) dx, making the integral ∫(4/u) du = 4 ln|u| + C. Substituting back gives 4 ln|ln x| + C, which correctly handles the natural logarithm of a logarithm. The absolute value is necessary since ln x can be negative for 0 < x < 1. Choice B omits the constant 4, while choice C would result from incorrectly substituting just x instead of ln x. When the integrand has the form f'(x)/f(x), use u-substitution with u = f(x) to get ln|f(x)| + C.

This integral uses u-substitution for an exponential function with a linear exponent. Let u = 2x - 1, so du = 2 dx, which means 8 dx = 4 du. The integral becomes ∫4eᵘ du = 4eᵘ = 4e^(2x-1). The chain rule adjustment divides 8 by 2 to give the coefficient 4. Choice B incorrectly keeps the factor 8, while choice C has the wrong coefficient. For exponential functions with linear exponents ax + b, substitute u = ax + b and divide by a.

This integral requires u-substitution for a rational function with a quadratic denominator raised to a power. Let u = x² + 1, so du = 2x dx, which means 10x dx = 5 du. The integral becomes ∫(5/u²) du = 5∫u^(-2) du = 5(-1/u) = -5/u = -5/(x² + 1). The antiderivative of u^(-2) is -1/u. Choice A has the wrong sign, while choice E represents the derivative rather than the antiderivative. When the numerator is a multiple of the derivative of the base of a power in the denominator, use u-substitution.

This integral uses u-substitution for a rational function with a linear denominator. Let u = 3x + 2, so du = 3 dx, which means 6 dx = 2 du. The integral becomes ∫(2/u) du = 2 ln|u| + C = 2 ln|3x + 2| + C. The coefficient 6 divided by the derivative factor 3 gives 2. Choice A incorrectly multiplies by 6, while choice C omits the adjustment factor. When integrating a constant over a linear expression ax + b, divide the constant by a.

This integral uses u-substitution for a rational function where the numerator is a multiple of the derivative of the denominator. Let u = x² + 1, so du = 2x dx, which means 5(2x) dx = 10x dx becomes ∫(5/u) du = 5 ln|u| + C = 5 ln(x² + 1) + C. Since x² + 1 > 0, we can omit absolute value bars. Choice B uses 5/2 incorrectly, while choice D would result from an arctangent substitution. When the numerator is a multiple of the derivative of the denominator, use logarithmic integration.

This integral uses u-substitution for a function with a linear expression raised to a negative power. Let $u = 2x + 1$, so $du = 2 , dx$, which means $4 , dx = 2 , du$. The integral becomes $\int \frac{2}{u^3} , du = 2 \int u^{-3} , du = 2 \left( -\frac{1}{2} u^{-2} \right) = - \frac{1}{u^2} = -\frac{1}{(2x + 1)^2}$. The antiderivative of $u^{-3}$ is $-\frac{1}{2} u^{-2}$. Choice A doubles the coefficient incorrectly. When integrating powers of linear expressions, substitute and use the power rule with chain rule adjustment.

This integral requires u-substitution for a rational trigonometric function. The integrand 6sin(3x)/cos(3x) = 6tan(3x) can be rewritten, but it's better to use u = cos(3x), so du = -3sin(3x) dx. This gives us ∫6sin(3x)/cos(3x) dx = ∫(-2/u) du = -2ln|u| = -2ln|cos(3x)|. The factor 6 divided by -3 gives -2. Choice B has the wrong sign, while choice E represents the derivative form. When integrating f'(x)/f(x), use u-substitution to get ln|f(x)| + C.

This integral requires u-substitution to handle the rational function with a quadratic denominator. Let u = t² + 4, so du = 2t dt, which means 6t dt = 3 du. The integral becomes ∫(3/u) du = 3 ln|u| + C. Substituting back gives 3 ln(t² + 4) + C (absolute value bars can be dropped since t² + 4 > 0). Choice D might tempt students who think of arctangent formulas, but that applies to ∫1/(t² + 4) dt, not our integral with t in the numerator. When the numerator contains the derivative of the denominator (up to a constant), use u-substitution with the denominator as u.

This integral requires u-substitution for a trigonometric function with a linear argument. Let u = 5x, so du = 5 dx, making the integral ∫sin(u) du = -cos(u) + C = -cos(5x) + C. The antiderivative of sin(u) is -cos(u), and we don't need to adjust for the chain rule here since we already have the factor 5 in front. Choice B has the wrong sign, while choice C incorrectly keeps the factor 5 with the cosine. When integrating sin(ax), the result is -(1/a)cos(ax) + C, but here we have the full derivative factor already present.

This integral uses u-substitution to handle a rational function where the numerator is the derivative of the denominator. Let u = x² + 9, so du = 2x dx, which means 8x dx = 4 du. The integral becomes ∫(4/u) du = 4 ln|u| + C = 4 ln(x² + 9) + C. Since x² + 9 > 0 for all real x, we can drop the absolute value bars. Choice C might tempt students thinking of arctangent formulas, but those apply when the numerator is constant, not when it contains x. When the numerator is a constant multiple of the derivative of the denominator, use ln of the denominator.