integral from 0 to 1 of (1+2^x)^22^{-x}

解

\int_{0}^{1} (1 + 2^{x})^{2} 2^{- x} d x

\frac{3}{2 ln ( 2 )} + 2

十進法表記

4.16404 \dots

解答ステップ

\int_{0}^{1} (1 + 2^{x})^{2} \cdot 2^{- x} d x

拡張

(1 + 2^{x})^{2} \cdot 2^{- x} : 2^{- x} + 2 + 2^{x}

= \int_{0}^{1} 2^{- x} + 2 + 2^{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_{0}^{1} 2^{- x} d x + \int_{0}^{1} 2 d x + \int_{0}^{1} 2^{x} d x

\int_{0}^{1} 2^{- x} d x = \frac{1}{2 ln ( 2 )}

\int_{0}^{1} 2 d x = 2

\int_{0}^{1} 2^{x} d x = \frac{1}{ln ( 2 )}

= \frac{1}{2 ln ( 2 )} + 2 + \frac{1}{ln ( 2 )}

簡素化

\frac{1}{2 ln ( 2 )} + 2 + \frac{1}{ln ( 2 )} : \frac{3}{2 ln ( 2 )} + 2

= \frac{3}{2 ln ( 2 )} + 2