integral from 1 to 2 of 1/(x^3+4x)

解

\int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

\frac{- ln ( 2 ) + ln ( 5 )}{8}

十進法表記

0.11453 \dots

解答ステップ

\int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

以下の部分分数を得る:

\frac{1}{x ^{3} + 4 x} : \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

= \int_{1}^{2} \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )} 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}{4 x} d x - \int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x

\int_{1}^{2} \frac{1}{4 x} d x = \frac{1}{4} ln (2)

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= \frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5))

簡素化

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)) : \frac{- ln ( 2 ) + ln ( 5 )}{8}

= \frac{- ln ( 2 ) + ln ( 5 )}{8}