integral from 0 to 1 of 1/((x+1)(x-2))

解

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

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

十進法表記

- 0.46209 \dots

解答ステップ

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

以下の部分分数を得る:

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

= \int_{0}^{1} - \frac{1}{3 ( x + 1 )} + \frac{1}{3 ( 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_{0}^{1} \frac{1}{3 ( x + 1 )} d x + \int_{0}^{1} \frac{1}{3 ( x - 2 )} d x

\int_{0}^{1} \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln (2)

\int_{0}^{1} \frac{1}{3 ( x - 2 )} d x = - \frac{1}{3} ln (2)

= - \frac{1}{3} ln (2) - \frac{1}{3} ln (2)

簡素化

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

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