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

解

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

- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣ + C

解答ステップ

\int \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 - \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 \frac{1}{4 ( x - 1 )} d x + \int \frac{1}{4 ( x - 5 )} d x

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{1}{4 ( x - 5 )} d x = \frac{1}{4} ln ∣ x - 5 ∣

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣

定数を解答に追加する

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣ + C