integral from 1 to 4 of e^{-x}+(ln(x))/x

解

\int_{1}^{4} e^{- x} + \frac{ln ( x )}{x} d x

\frac{1}{e} - \frac{1}{e ^{4}} + 2 ln^{2} (2)

十進法表記

1.31046 \dots

解答ステップ

\int_{1}^{4} e^{- x} + \frac{ln ( x )}{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}^{4} e^{- x} d x + \int_{1}^{4} \frac{ln ( x )}{x} d x

\int_{1}^{4} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{4}}

\int_{1}^{4} \frac{ln ( x )}{x} d x = 2 ln^{2} (2)

= \frac{1}{e} - \frac{1}{e ^{4}} + 2 ln^{2} (2)