integral of (2x^2+2x-7)/(x^2-4)

解

\int \frac{2 x ^{2} + 2 x - 7}{x ^{2} - 4} d x

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

解答ステップ

\int \frac{2 x ^{2} + 2 x - 7}{x ^{2} - 4} d x

拡張

\frac{2 x ^{2} + 2 x - 7}{x ^{2} - 4} : \frac{2 x ^{2}}{x ^{2} - 4} + \frac{2 x}{x ^{2} - 4} - \frac{7}{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{2 x ^{2}}{x ^{2} - 4} d x + \int \frac{2 x}{x ^{2} - 4} d x - \int \frac{7}{x ^{2} - 4} d x

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

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

\int \frac{7}{x ^{2} - 4} d x = - \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

= - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + ln x^{2} - 4 - (- \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1))

簡素化

- 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + ln x^{2} - 4 - (- \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)) : 2 x - \frac{1}{4} ln \frac{x}{2} + 1 + \frac{1}{4} ln \frac{x}{2} - 1 + ln x^{2} - 4

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

定数を解答に追加する

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