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

解

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

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

\int_{1}^{2} \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}