int e^{-x}(x^3-2x^2+3x-4) dx= | vedclass

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(A) We use the formula for the integral of the form $\int e^{ax} f(x) dx = \frac{e^{ax}}{a} [f(x) - \frac{f'(x)}{a} + \frac{f''(x)}{a^2} - \frac{f'''(

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$-e^{-x}(x^3+x^2+5x+1)+c$

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(A) We use the formula for the integral of the form $\int e^{ax} f(x) dx = \frac{e^{ax}}{a} [f(x) - \frac{f'(x)}{a} + \frac{f''(x)}{a^2} - \frac{f'''(x)}{a^3} + \dots]$.Here, $a = -1$ and $f(x) = x^3 - 2x^2 + 3x - 4$.Calculating the derivatives:$f'(x) = 3x^2 - 4x + 3$$f''(x) = 6x - 4$$f'''(x) = 6$$f^{(4)}(x) = 0$Substituting these into the formula:$\int e^{-x} f(x) dx = \frac{e^{-x}}{-1} [f(x) - \frac{f'(x)}{-1} + \frac{f''(x)}{(-1)^2} - \frac{f'''(x)}{(-1)^3}] + c$$= -e^{-x} [f(x) + f'(x) + f''(x) + f'''(x)] + c$$= -e^{-x} [(x^3 - 2x^2 + 3x - 4) + (3x^2 - 4x + 3) + (6x - 4) + 6] + c$$= -e^{-x} [x^3 + (-2+3)x^2 + (3-4+6)x + (-4+3-4+6)] + c$$= -e^{-x} [x^3 + x^2 + 5x + 1] + c$.

$\int e^{-x}(x^3-2x^2+3x-4) dx=$

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