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

解

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

- 2 ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 1 + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= - 2 ln ∣ x ∣ - (- \frac{1}{x}) + ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 1

簡素化

= - 2 ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 1

定数を解答に追加する

= - 2 ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 1 + C