integral of (x^2+x-6)/(x^2-9)

解

\int \frac{x ^{2} + x - 6}{x ^{2} - 9} d x

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

解答ステップ

\int \frac{x ^{2} + x - 6}{x ^{2} - 9} d x

拡張

\frac{x ^{2} + x - 6}{x ^{2} - 9} : \frac{x ^{2}}{x ^{2} - 9} + \frac{x}{x ^{2} - 9} - \frac{6}{x ^{2} - 9}

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 ^{2}}{x ^{2} - 9} d x + \int \frac{x}{x ^{2} - 9} d x - \int \frac{6}{x ^{2} - 9} d x

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

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

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

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

簡素化

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

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

定数を解答に追加する

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