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

解

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

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

十進法表記

53.21528 \dots

解答ステップ

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

拡張

\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

総和規則を適用する:

\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

簡素化

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