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

解

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

\frac{5}{6}

十進法表記

0.83333 \dots

解答ステップ

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

指数の規則を適用する:

a^{- b} = \frac{1}{a ^{b}}

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

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

\int_{1}^{2} \frac{1}{( 1 + x ) ^{2}} d x = \frac{1}{6}

= 1 - \frac{1}{6}

簡素化

1 - \frac{1}{6} : \frac{5}{6}

= \frac{5}{6}