integral from-2 to-1 of 1/(x^3+x)

解

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

\frac{ln ( 5 ) - 3 ln ( 2 )}{2}

十進法表記

- 0.23500 \dots

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ^{3} + x} : \frac{1}{x} - \frac{x}{x ^{2} + 1}

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

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

\int_{- 2}^{- 1} \frac{x}{x ^{2} + 1} d x = - \frac{1}{2} (ln (5) - ln (2))

= - ln (2) - (- \frac{1}{2} (ln (5) - ln (2)))

簡素化

- ln (2) - (- \frac{1}{2} (ln (5) - ln (2))) : \frac{ln ( 5 ) - 3 ln ( 2 )}{2}

= \frac{ln ( 5 ) - 3 ln ( 2 )}{2}