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

解

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

- 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + x + C

解答ステップ

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

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

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

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

\int 1 d x = x

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

定数を解答に追加する

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