integral from-5 to-1 of 1/(x^2-8x+7)

解

\int_{- 5}^{- 1} \frac{1}{x ^{2} - 8 x + 7} d x

\frac{ln ( 3 ) + ln ( \frac{2}{3} )}{6}

十進法表記

0.11552 \dots

解答ステップ

\int_{- 5}^{- 1} \frac{1}{x ^{2} - 8 x + 7} d x

以下の部分分数を得る:

\frac{1}{x ^{2} - 8 x + 7} : - \frac{1}{6 ( x - 1 )} + \frac{1}{6 ( x - 7 )}

= \int_{- 5}^{- 1} - \frac{1}{6 ( x - 1 )} + \frac{1}{6 ( x - 7 )} 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_{- 5}^{- 1} \frac{1}{6 ( x - 1 )} d x + \int_{- 5}^{- 1} \frac{1}{6 ( x - 7 )} d x

\int_{- 5}^{- 1} \frac{1}{6 ( x - 1 )} d x = - \frac{1}{6} ln (3)

\int_{- 5}^{- 1} \frac{1}{6 ( x - 7 )} d x = \frac{1}{6} ln (\frac{2}{3})

= - (- \frac{1}{6} ln (3)) + \frac{1}{6} ln (\frac{2}{3})

簡素化

- (- \frac{1}{6} ln (3)) + \frac{1}{6} ln (\frac{2}{3}) : \frac{ln ( 3 ) + ln ( \frac{2}{3} )}{6}

= \frac{ln ( 3 ) + ln ( \frac{2}{3} )}{6}