integral from 1 to 2 of ((2x+1)(x-2))/x

解

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

- 2 ln (2)

十進法表記

- 1.38629 \dots

解答ステップ

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

拡張

\frac{( 2 x + 1 ) ( x - 2 )}{x} : 2 x - 3 - \frac{2}{x}

= \int_{1}^{2} 2 x - 3 - \frac{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_{1}^{2} 2 x d x - \int_{1}^{2} 3 d x - \int_{1}^{2} \frac{2}{x} d x

\int_{1}^{2} 2 x d x = 3

\int_{1}^{2} 3 d x = 3

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

= 3 - 3 - 2 ln (2)

簡素化

= - 2 ln (2)