integral from 1 to e of (t^6+1)/t

解

\int_{1}^{e} \frac{t ^{6} + 1}{t} d t

\frac{e ^{6} - 1}{6} + 1

十進法表記

68.07146 \dots

解答ステップ

\int_{1}^{e} \frac{t ^{6} + 1}{t} d t

拡張

\frac{t ^{6} + 1}{t} : t^{5} + \frac{1}{t}

= \int_{1}^{e} t^{5} + \frac{1}{t} d t

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} t^{5} d t + \int_{1}^{e} \frac{1}{t} d t

\int_{1}^{e} t^{5} d t = \frac{e ^{6} - 1}{6}

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

= \frac{e ^{6} - 1}{6} + 1