integral of (x+1)/(x^2+x-6)

解

\int \frac{x + 1}{x ^{2} + x - 6} d x

2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x) + C

解答ステップ

\int \frac{x + 1}{x ^{2} + x - 6} d x

拡張

\frac{x + 1}{x ^{2} + x - 6} : \frac{x}{x ^{2} + x - 6} + \frac{1}{x ^{2} + x - 6}

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} + x - 6} d x + \int \frac{1}{x ^{2} + x - 6} d x

\int \frac{x}{x ^{2} + x - 6} d x = 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

\int \frac{1}{x ^{2} + x - 6} d x = - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x) + C