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

解

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

- \frac{1}{2} ln (3)

十進法表記

- 0.54930 \dots

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ^{2} - 6 x + 5} : - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 5 )}

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

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

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

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

簡素化

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

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