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

解

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

- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x + C

解答ステップ

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

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

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

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

\int 1 d x = x

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x

定数を解答に追加する

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x + C