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Question at position 12 Solve ∫x2e2x+1dx\int x^2 e^{2x+1} \, dx.e2x+1(x2−x+12)+Ce^{2x+1}\left( x^2 - x + \frac{1}{2} \right) + Cxe2x+12−e2x+14+C\frac{x e^{2x+1}}{2} - \frac{e^{2x+1}}{4}+ Ce2x+12(x2−x)+C\frac{e^{2x+1}}{2} \left( x^2 - x \right) + Ce2x+12(x2−x+12)+C\frac{e^{2x+1}}{2} \left( x^2 - x + \frac{1}{2} \right) + C

Options
A.𝑒 2 𝑥 + 1 ( 𝑥 2 − 𝑥 + 1 2 ) + 𝐶
B.𝑥 𝑒 2 𝑥 + 1 2 − 𝑒 2 𝑥 + 1 4 + 𝐶
C.𝑒 2 𝑥 + 1 2 ( 𝑥 2 − 𝑥 ) + 𝐶
D.𝑒 2 𝑥 + 1 2 ( 𝑥 2 − 𝑥 + 1 2 ) + 𝐶
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Step-by-Step Analysis
To tackle the integral ∫ x^2 e^{2x+1} dx, I first note that e^{2x+1} = e · e^{2x}, so the integral is e · ∫ x^2 e^{2x} dx. The standard approach for ∫ x^n e^{ax} dx yields a polynomial in x multiplied by e^{ax}; here with n = 2 and a = 2, we expect a result of the form e^{2x} times a quadratic in x, up to a constant factor, and then multiplied by the extra e from e^{2x+1}. Option 1: e^{2x+1} · ( x^2 − x + 1/2 ) + C - This expression corresponds to a polynomial with c......Login to view full explanation

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