integral from 0 to 1 of 1/(x^2-5x+6)

解

\int_{0}^{1} \frac{1}{x ^{2} - 5 x + 6} d x

2 ln (2) - ln (3)

十進法表記

0.28768 \dots

解答ステップ

\int_{0}^{1} \frac{1}{x ^{2} - 5 x + 6} d x

以下の部分分数を得る:

\frac{1}{x ^{2} - 5 x + 6} : - \frac{1}{x - 2} + \frac{1}{x - 3}

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

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

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

= - (- ln (2)) + ln (2) - ln (3)

簡素化

= 2 ln (2) - ln (3)