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$\int e^{x} (x^{4} + 3 x^{3} + 6 x^{2} - 5 x + 1) d x = e^{x} g (x) + C t h e n g (0) =___$

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answer is C.

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Detailed Solution

$\int e^{x} (x^{4} + 3 x^{3} + 6 x^{2} - 5 x + 1) d x$

$= \int e^{x} (x^{4} + 4 x^{3} - x^{3} - 3 x^{2} + 9 x^{2} + 18 x - 23 x - 23 + 24) d x$

$= \int e^{x} ((x^{4} - x^{3} + 9 x^{2} - 23 x) + (4 x^{3} - 3 x^{2} + 18 x - 23)) d x + \int 24 e^{x} d x$

$= e^{x} (x^{4} - x^{3} + 9 x^{2} - 23 x) + 24 e^{x} + C$ $∵ \int e^{x} (f (x) + f^{1} (x)) d x = e^{x} f (x) + C$

$= e^{x} (x^{4} - x^{3} + 9 x^{2} - 23 x + 24) + C$

$g (x) = e^{x} (x^{4} - x^{3} + 9 x^{2} - 23 x + 24)$

Now $g (0) = 24$

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