int e^{x} cdot x^{5} , dx is | vedclass

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Question

(C) We use the formula for integration by parts: $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.Let $I = \int e^{x} \cdot x^{5

Vedclass · Accepted Answer

$e^{x}[x^{5}-5 x^{4}+20 x^{3}-60 x^{2}+120 x-120]+C$

Vedclass · Answer

(C) We use the formula for integration by parts: $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.Let $I = \int e^{x} \cdot x^{5} \, dx$.Using the generalized integration by parts formula $\int e^{x} f(x) \, dx = e^{x} [f(x) - f'(x) + f''(x) - f'''(x) + f^{(4)}(x) - f^{(5)}(x)] + C$.Here,$f(x) = x^{5}$.Then,$f'(x) = 5x^{4}$,$f''(x) = 20x^{3}$,$f'''(x) = 60x^{2}$,$f^{(4)}(x) = 120x$,and $f^{(5)}(x) = 120$.Substituting these into the formula:$I = e^{x} [x^{5} - 5x^{4} + 20x^{3} - 60x^{2} + 120x - 120] + C$.

$\int e^{x} \cdot x^{5} \, dx$ is

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