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

解

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

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

十進法表記

4.19452 \dots

解答ステップ

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

拡張

\frac{x ^{2} + 1}{x} : x + \frac{1}{x}

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

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

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

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