integral of (3x^2+1)/(x^5-x)

解

\int \frac{3 x ^{2} + 1}{x ^{5} - x} d x

- \frac{1}{2} ln x^{2} + 1 + \frac{3}{4} ln x^{2} - 1 + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - ln ∣ x ∣ + C

解答ステップ

\int \frac{3 x ^{2} + 1}{x ^{5} - x} d x

拡張

\frac{3 x ^{2} + 1}{x ^{5} - x} : \frac{3 x ^{2}}{x ^{5} - x} + \frac{1}{x ^{5} - 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{3 x ^{2}}{x ^{5} - x} d x + \int \frac{1}{x ^{5} - x} d x

\int \frac{3 x ^{2}}{x ^{5} - x} d x = - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1)

\int \frac{1}{x ^{5} - x} d x = - ln ∣ x ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣

= - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1) - ln ∣ x ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣

簡素化

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

= - \frac{1}{2} ln x^{2} + 1 + \frac{3}{4} ln x^{2} - 1 + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - ln ∣ x ∣

定数を解答に追加する

= - \frac{1}{2} ln x^{2} + 1 + \frac{3}{4} ln x^{2} - 1 + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - ln ∣ x ∣ + C