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

解

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

ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ - 3 ln ∣ x ∣ + \frac{1}{x} + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

3 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) + \frac{1}{x} - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ : ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ - 3 ln ∣ x ∣ + \frac{1}{x}

= ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ - 3 ln ∣ x ∣ + \frac{1}{x}

定数を解答に追加する

= ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ - 3 ln ∣ x ∣ + \frac{1}{x} + C