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

解

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

e^{2}

十進法表記

7.38905 \dots

解答ステップ

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

拡張

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

= \int_{1}^{e} 2 t + \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} 2 t d t + \int_{1}^{e} \frac{1}{t} d t

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

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

= e^{2} - 1 + 1

簡素化

= e^{2}