integral of (x^2-5x-6)/(x^2-4)

解

\int \frac{x ^{2} - 5 x - 6}{x ^{2} - 4} d x

x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 + C

解答ステップ

\int \frac{x ^{2} - 5 x - 6}{x ^{2} - 4} d x

拡張

\frac{x ^{2} - 5 x - 6}{x ^{2} - 4} : \frac{x ^{2}}{x ^{2} - 4} - \frac{5 x}{x ^{2} - 4} - \frac{6}{x ^{2} - 4}

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} - 4} d x - \int \frac{5 x}{x ^{2} - 4} d x - \int \frac{6}{x ^{2} - 4} d x

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

\int \frac{5 x}{x ^{2} - 4} d x = \frac{5}{2} ln x^{2} - 4

\int \frac{6}{x ^{2} - 4} d x = - 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)

= - ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x - \frac{5}{2} ln x^{2} - 4 - (- 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1))

簡素化

- ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x - \frac{5}{2} ln x^{2} - 4 - (- 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)) : x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1

= x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1

定数を解答に追加する

= x - \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 + C