integral from 0 to 1 of 1/(x^2-8x+15)

解

\int_{0}^{1} \frac{1}{x ^{2} - 8 x + 15} d x

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

十進法表記

0.09116 \dots

解答ステップ

\int_{0}^{1} \frac{1}{x ^{2} - 8 x + 15} d x

以下の部分分数を得る:

\frac{1}{x ^{2} - 8 x + 15} : - \frac{1}{2 ( x - 3 )} + \frac{1}{2 ( x - 5 )}

= \int_{0}^{1} - \frac{1}{2 ( x - 3 )} + \frac{1}{2 ( x - 5 )} 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}{2 ( x - 3 )} d x + \int_{0}^{1} \frac{1}{2 ( x - 5 )} d x

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

\int_{0}^{1} \frac{1}{2 ( x - 5 )} d x = ln (2) - \frac{1}{2} ln (5)

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

簡素化

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

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