integral of (5-2x)/(sqrt(5+4x-x^2))

解

\int \frac{5 - 2 x}{5 + 4 x - x ^{2}} d x

arcsin (\frac{1}{3} (x - 2)) + 2 - x^{2} + 4 x + 5 + C

解答ステップ

\int \frac{5 - 2 x}{5 + 4 x - x ^{2}} d x

拡張

\frac{5 - 2 x}{5 + 4 x - x ^{2}} : \frac{5}{5 + 4 x - x ^{2}} - \frac{2 x}{5 + 4 x - x ^{2}}

= \int \frac{5}{5 + 4 x - x ^{2}} - \frac{2 x}{5 + 4 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{5}{5 + 4 x - x ^{2}} d x - \int \frac{2 x}{5 + 4 x - x ^{2}} d x

\int \frac{5}{5 + 4 x - x ^{2}} d x = 5 arcsin (\frac{1}{3} (x - 2))

\int \frac{2 x}{5 + 4 x - x ^{2}} d x = 2 (- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2)))

= 5 arcsin (\frac{1}{3} (x - 2)) - 2 (- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2)))

簡素化

5 arcsin (\frac{1}{3} (x - 2)) - 2 (- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2))) : arcsin (\frac{1}{3} (x - 2)) + 2 - x^{2} + 4 x + 5

= arcsin (\frac{1}{3} (x - 2)) + 2 - x^{2} + 4 x + 5

定数を解答に追加する

= arcsin (\frac{1}{3} (x - 2)) + 2 - x^{2} + 4 x + 5 + C