integral of (x+1)/(sqrt(3-2x-x^2))

解

\int \frac{x + 1}{3 - 2 x - x ^{2}} d x

- - x^{2} - 2 x + 3 + C

解答ステップ

\int \frac{x + 1}{3 - 2 x - x ^{2}} d x

拡張

\frac{x + 1}{3 - 2 x - x ^{2}} : \frac{x}{3 - 2 x - x ^{2}} + \frac{1}{3 - 2 x - x ^{2}}

= \int \frac{x}{3 - 2 x - x ^{2}} + \frac{1}{3 - 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{x}{3 - 2 x - x ^{2}} d x + \int \frac{1}{3 - 2 x - x ^{2}} d x

\int \frac{x}{3 - 2 x - x ^{2}} d x = - - x^{2} - 2 x + 3 - arcsin (\frac{1}{2} (x + 1))

\int \frac{1}{3 - 2 x - x ^{2}} d x = arcsin (\frac{1}{2} (x + 1))

= - - x^{2} - 2 x + 3 - arcsin (\frac{1}{2} (x + 1)) + arcsin (\frac{1}{2} (x + 1))

簡素化

= - - x^{2} - 2 x + 3

定数を解答に追加する

= - - x^{2} - 2 x + 3 + C