integral from 1 to e of (9x^2+9x+9)/x

解答

\int_{1}^{e} \frac{9 x ^{2} + 9 x + 9}{x} d x

9 e + \frac{9 ( e ^{2} - 1 )}{2}

十进制

53.21528 \dots

求解步骤

\int_{1}^{e} \frac{9 x ^{2} + 9 x + 9}{x} d x

乘开

\frac{9 x ^{2} + 9 x + 9}{x} : 9 x + 9 + \frac{9}{x}

= \int_{1}^{e} 9 x + 9 + \frac{9}{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} 9 x d x + \int_{1}^{e} 9 d x + \int_{1}^{e} \frac{9}{x} d x

\int_{1}^{e} 9 x d x = \frac{9 ( e ^{2} - 1 )}{2}

\int_{1}^{e} 9 d x = 9 e - 9

\int_{1}^{e} \frac{9}{x} d x = 9

= \frac{9 ( e ^{2} - 1 )}{2} + 9 e - 9 + 9

化简

= 9 e + \frac{9 ( e ^{2} - 1 )}{2}

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