integral from 0 to 1 of (x-4)/(x^2+5x+6)

解

\int_{0}^{1} \frac{x - 4}{x ^{2} + 5 x + 6} d x

20 ln (2) - 13 ln (3)

十進法表記

- 0.41901 \dots

解答ステップ

\int_{0}^{1} \frac{x - 4}{x ^{2} + 5 x + 6} d x

以下の部分分数を得る:

\frac{x - 4}{x ^{2} + 5 x + 6} : - \frac{6}{x + 2} + \frac{7}{x + 3}

= \int_{0}^{1} - \frac{6}{x + 2} + \frac{7}{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{6}{x + 2} d x + \int_{0}^{1} \frac{7}{x + 3} d x

\int_{0}^{1} \frac{6}{x + 2} d x = 6 (ln (3) - ln (2))

\int_{0}^{1} \frac{7}{x + 3} d x = 7 (2 ln (2) - ln (3))

= - 6 (ln (3) - ln (2)) + 7 (2 ln (2) - ln (3))

簡素化

- 6 (ln (3) - ln (2)) + 7 (2 ln (2) - ln (3)) : 20 ln (2) - 13 ln (3)

= 20 ln (2) - 13 ln (3)