integral of (x^2+9)/(x^2-9)

解

\int \frac{x ^{2} + 9}{x ^{2} - 9} d x

- 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1) + x + C

解答ステップ

\int \frac{x ^{2} + 9}{x ^{2} - 9} d x

\frac{x ^{2} + 9}{x ^{2} - 9} = \frac{18}{x ^{2} - 9} + 1

= \int \frac{18}{x ^{2} - 9} + 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{18}{x ^{2} - 9} d x + \int 1 d x

\int \frac{18}{x ^{2} - 9} d x = - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1)

\int 1 d x = x

= - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1) + x

定数を解答に追加する

= - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1) + x + C