integral of (x-3)/(x^2+5x+10)

解

\int \frac{x - 3}{x ^{2} + 5 x + 10} d x

\frac{1}{2} ln 4 x^{2} + 20 x + 40 - \frac{11}{15} arctan (\frac{2}{15} x + \frac{5}{3}) + C

解答ステップ

\int \frac{x - 3}{x ^{2} + 5 x + 10} d x

拡張

\frac{x - 3}{x ^{2} + 5 x + 10} : \frac{x}{x ^{2} + 5 x + 10} - \frac{3}{x ^{2} + 5 x + 10}

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} + 5 x + 10} d x - \int \frac{3}{x ^{2} + 5 x + 10} d x

\int \frac{x}{x ^{2} + 5 x + 10} d x = \frac{1}{2} ln 4 x^{2} + 20 x + 40 - \frac{5}{3} arctan (\frac{2}{15} x + \frac{5}{3})

\int \frac{3}{x ^{2} + 5 x + 10} d x = \frac{2 3}{5} arctan (\frac{2}{15} x + \frac{5}{3})

= \frac{1}{2} ln 4 x^{2} + 20 x + 40 - \frac{5}{3} arctan (\frac{2}{15} x + \frac{5}{3}) - \frac{2 3}{5} arctan (\frac{2}{15} x + \frac{5}{3})

類似した元を足す:

- \frac{5}{3} arctan (\frac{2}{15} x + \frac{5}{3}) - \frac{2 3}{5} arctan (\frac{2}{15} x + \frac{5}{3}) = - \frac{11}{15} arctan (\frac{2}{15} x + \frac{5}{3})

= \frac{1}{2} ln 4 x^{2} + 20 x + 40 - \frac{11}{15} arctan (\frac{2}{15} x + \frac{5}{3})

定数を解答に追加する

= \frac{1}{2} ln 4 x^{2} + 20 x + 40 - \frac{11}{15} arctan (\frac{2}{15} x + \frac{5}{3}) + C