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If $\int x^{3} e^{- x} d x = - e^{- x} [a x^{3} + b x^{2} + c x + d] + K$ , then $(a, b, c, d) =$

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$(- 1, - 3, 6, 6)$

$(2, 4, 1, 3)$

$(1, - 3, 6, - 6)$

$(1, 3, 6, 6)$

answer is B.

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Detailed Solution

$\int x^{3} e^{- x} d x$

$= - x^{3} (e^{- x}) - 3 x^{2} e^{- x} - 6 x e^{- x} - 6 e^{- x} + c$

$= - e^{- x} (x^{3} + 3 x^{2} + 6 x + 6) + c$

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