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

解

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

6 ln ∣ x ∣ + \frac{5}{x} - 6 ln ∣ x + 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

= 6 ln ∣ x ∣ - (- \frac{5}{x}) - 6 ln ∣ x + 1 ∣

簡素化

= 6 ln ∣ x ∣ + \frac{5}{x} - 6 ln ∣ x + 1 ∣

定数を解答に追加する

= 6 ln ∣ x ∣ + \frac{5}{x} - 6 ln ∣ x + 1 ∣ + C