integral of (24-x)/(x^2+2x-24)

解

\int \frac{24 - x}{x ^{2} + 2 x - 24} d x

- \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1) + C

解答ステップ

\int \frac{24 - x}{x ^{2} + 2 x - 24} d x

拡張

\frac{24 - x}{x ^{2} + 2 x - 24} : \frac{24}{x ^{2} + 2 x - 24} - \frac{x}{x ^{2} + 2 x - 24}

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{24}{x ^{2} + 2 x - 24} d x - \int \frac{x}{x ^{2} + 2 x - 24} d x

\int \frac{24}{x ^{2} + 2 x - 24} d x = - \frac{12}{5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1)

\int \frac{x}{x ^{2} + 2 x - 24} d x = \frac{1}{2} ln x^{2} + 2 x - 24 + \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)

= - \frac{12}{5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1) - (\frac{1}{2} ln x^{2} + 2 x - 24 + \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1))

簡素化

- \frac{12}{5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1) - (\frac{1}{2} ln x^{2} + 2 x - 24 + \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)) : - \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)

= - \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1)

定数を解答に追加する

= - \frac{12}{5} ln \frac{1}{5} (x + 1) + 1 + \frac{12}{5} ln \frac{1}{5} (x + 1) - 1 - \frac{1}{2} ln x^{2} + 2 x - 24 - \frac{1}{10} (ln \frac{1}{5} (x + 1) + 1 - ln \frac{1}{5} (x + 1) - 1) + C