Integration by parts is the key skill for evaluating ∫ $x^2$ sin x dx, based on the formula ∫ u dv = u v - ∫ v du, often applied repeatedly. Choose u = $x^2$ so du = 2x dx, and dv = sin x dx so v = -cos x for the first step. This leads to $-x^2$ cos x + 2 ∫ x cos x dx, requiring another parts application where u = x and dv = cos x dx. The full computation yields $-x^2$ cos x + 2x sin x + 2 cos x + C after integrating twice. A tempting distractor like choice C fails by having a negative 2 cos x term, possibly from a sign error in the second integration step. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

This integral requires integration by parts to handle the product of x and $e^x$. For integration by parts, we use ∫u dv = uv - ∫v du, choosing u = x (so du = dx) and dv = $e^x$ dx (so v = $e^x$). Applying the formula: $∫xe^x$ dx = $xe^x$ - $∫e^x$ dx = $xe^x$ - $e^x$ + C = $e^x$(x-1) + C. Choice C shows the incomplete form before finishing the integration, which is a common error when students forget to complete the final integral. When using integration by parts with polynomial times exponential, always let u be the polynomial part to reduce its degree.

This problem requires integration by parts to find the antiderivative of $\sqrt{x}$ ln x for x > 0. To apply integration by parts, choose u = ln x because it differentiates to 1/x, and dv = $x^{1/2}$ dx because it integrates to (2/3) $x^{3/2}$. Then du = (1/x) dx and v = (2/3) $x^{3/2}$. Substituting into the formula gives (2/3) $x^{3/2}$ ln x - ∫ (2/3) $x^{3/2}$ (1/x) dx, which simplifies to (2/3) $x^{3/2}$ ln x - (4/9) $x^{3/2}$ + C. A tempting distractor might be (2/3) $x^{3/2}$ ln x + (4/9) $x^{3/2}$ + C, from neglecting the negative sign in parts. A transferable strategy for integration by parts is to select u as the log for rational derivative, and dv as the power function, adjusting exponents carefully in the process.

This problem requires integration by parts to find the antiderivative of x $e^{2x}$. To apply integration by parts, choose u = x because its derivative simplifies to 1, and dv = $e^{2x}$ dx because exponentials are straightforward to integrate. Then du = dx and v = (1/2) $e^{2x}$. Substituting into the formula gives (x/2) $e^{2x}$ - ∫ (1/2) $e^{2x}$ dx, which evaluates to (x/2) $e^{2x}$ - (1/4) $e^{2x}$ + C. A tempting distractor might be the expression with a plus sign instead of minus, resulting from omitting the negative sign in the parts formula. A transferable strategy for integration by parts is to let u be the polynomial term that decreases in degree when differentiated, and dv the exponential or transcendental function that integrates easily.

This integral requires integration by parts for xe^(-x). Choose u = x (so du = dx) and dv = e^(-x) dx (so v = -e^(-x)), noting the negative from integrating e^(-x). Applying the formula: ∫xe^(-x) dx = x(-e^(-x)) - ∫(-e^(-x)) dx = -xe^(-x) + ∫e^(-x) dx = -xe^(-x) - e^(-x) + C = -e^(-x)(x + 1) + C. Choice C has the wrong overall sign, missing the negative that comes from integrating e^(-x). When integrating products with e^(-x), remember that ∫e^(-x) dx = -e^(-x), introducing a crucial negative sign.

Integration by parts is the key skill for evaluating ∫ x ln x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = x dx so v = (1/2) $x^2$. This choice works well as ln x simplifies upon differentiation, pairing with the polynomial growth in dv. The result is (1/2) $x^2$ ln x - ∫ (1/2) $x^2$ (1/x) dx, simplifying to (1/2) $x^2$ ln x - (1/4) $x^2$ + C. A tempting distractor like choice B fails with a positive (1/4) $x^2$ term, likely from neglecting the subtraction in the parts formula. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Integration by parts is the key skill for evaluating ∫ (ln $x)/x^2$ dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = $x^{-2}$ dx so v = $-x^{-1}$. This selection simplifies the logarithmic part, leaving a rational integral. The computation yields - (ln x)/x - ∫ - (1/x) (1/x) dx, simplifying to - (ln x + 1)/x + C. A tempting distractor like choice C fails by having a positive overall sign, possibly from forgetting the negative v in the initial product. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Integration by parts is the key skill for evaluating ∫ ln x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = dx so v = x. This selection is ideal because ln x differentiates to a simple rational function, while dv is basic. The result is x ln x - ∫ x (1/x) dx, simplifying to x ln x - x + C. A tempting distractor like choice B fails with a positive x term, which would result from forgetting the negative sign in the parts formula. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Integration by parts is the key skill for evaluating ∫ $e^x$ sin x dx, based on the formula ∫ u dv = u v - ∫ v du, applied twice to cycle back. Choose u = sin x so du = cos x dx, and dv = $e^x$ dx so v = $e^x$ for the first step. This produces $e^x$ sin x - ∫ $e^x$ cos x dx, then repeat with u = cos x and dv = $e^x$ dx. Solving the system yields (1/2) $e^x$ (sin x - cos x) + C. A tempting distractor like choice B fails by having a positive cos x term, which might come from a sign error in resolving the cyclic integrals. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

This integral requires integration by parts twice for $x^2 e^x$. First, let $u = x^2$ (so $du = 2x , dx$) and $dv = e^x , dx$ (so $v = e^x$), giving $∫ x^2 e^x , dx = x^2 e^x - ∫ 2x e^x , dx$. For the remaining integral $∫ 2x e^x , dx$, apply parts again with $u = 2x$ and $dv = e^x , dx$, yielding $2x e^x - 2 e^x$. Combining: $∫ x^2 e^x , dx = x^2 e^x - 2x e^x + 2 e^x + C = e^x (x^2 - 2x + 2) + C$. Choice D shows only the first application of parts without completing the process. For polynomial times exponential, repeated integration by parts reduces the polynomial degree to zero.