integral of (1-x)/(sqrt(5-2x-x^2))

解

\int \frac{1 - x}{5 - 2 x - x ^{2}} d x

2 arcsin (\frac{1}{6} (x + 1)) + - x^{2} - 2 x + 5 + C

解答ステップ

\int \frac{1 - x}{5 - 2 x - x ^{2}} d x

拡張

\frac{1 - x}{5 - 2 x - x ^{2}} : \frac{1}{5 - 2 x - x ^{2}} - \frac{x}{5 - 2 x - x ^{2}}

= \int \frac{1}{5 - 2 x - x ^{2}} - \frac{x}{5 - 2 x - x ^{2}} d x

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{1}{5 - 2 x - x ^{2}} d x - \int \frac{x}{5 - 2 x - x ^{2}} d x

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

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

= arcsin (\frac{1}{6} (x + 1)) - (- - x^{2} - 2 x + 5 - arcsin (\frac{1}{6} (x + 1)))

簡素化

= 2 arcsin (\frac{1}{6} (x + 1)) + - x^{2} - 2 x + 5

定数を解答に追加する

= 2 arcsin (\frac{1}{6} (x + 1)) + - x^{2} - 2 x + 5 + C