integral from 1 to e of (x^5-2x+1)/x

解答

\int_{1}^{e} \frac{x ^{5} - 2 x + 1}{x} d x

3 + \frac{e ^{5} - 1}{5} - 2 e

十进制

27.04606 \dots

求解步骤

\int_{1}^{e} \frac{x ^{5} - 2 x + 1}{x} d x

乘开

\frac{x ^{5} - 2 x + 1}{x} : x^{4} - 2 + \frac{1}{x}

= \int_{1}^{e} x^{4} - 2 + \frac{1}{x} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{e} x^{4} d x - \int_{1}^{e} 2 d x + \int_{1}^{e} \frac{1}{x} d x

\int_{1}^{e} x^{4} d x = \frac{e ^{5} - 1}{5}

\int_{1}^{e} 2 d x = 2 e - 2

\int_{1}^{e} \frac{1}{x} d x = 1

= \frac{e ^{5} - 1}{5} - (2 e - 2) + 1

化简

= 3 + \frac{e ^{5} - 1}{5} - 2 e

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