integral of (x+1)/(sqrt(x^2-9))

解

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

x^{2} - 9 + ln (\frac{1}{3} x^{2} - 9 + x) + C

解答ステップ

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

拡張

\frac{x + 1}{x ^{2} - 9} : \frac{x}{x ^{2} - 9} + \frac{1}{x ^{2} - 9}

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

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

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

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

定数を解答に追加する

= x^{2} - 9 + ln (\frac{1}{3} x^{2} - 9 + x) + C