integral of (3x-1)/(x^2(x+1))

解

\int \frac{3 x - 1}{x ^{2} ( x + 1 )} d x

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

解答ステップ

\int \frac{3 x - 1}{x ^{2} ( x + 1 )} d x

以下の部分分数を得る:

\frac{3 x - 1}{x ^{2} ( x + 1 )} : \frac{4}{x} - \frac{1}{x ^{2}} - \frac{4}{x + 1}

= \int \frac{4}{x} - \frac{1}{x ^{2}} - \frac{4}{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{4}{x} d x - \int \frac{1}{x ^{2}} d x - \int \frac{4}{x + 1} d x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

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

簡素化

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

定数を解答に追加する

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