integral of (2x+6)/(sqrt(16-6x-x^2))

解

\int \frac{2 x + 6}{16 - 6 x - x ^{2}} d x

- 2 - x^{2} - 6 x + 16 + C

解答ステップ

\int \frac{2 x + 6}{16 - 6 x - x ^{2}} d x

拡張

\frac{2 x + 6}{16 - 6 x - x ^{2}} : \frac{2 x}{16 - 6 x - x ^{2}} + \frac{6}{16 - 6 x - x ^{2}}

= \int \frac{2 x}{16 - 6 x - x ^{2}} + \frac{6}{16 - 6 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{2 x}{16 - 6 x - x ^{2}} d x + \int \frac{6}{16 - 6 x - x ^{2}} d x

\int \frac{2 x}{16 - 6 x - x ^{2}} d x = 2 (- - x^{2} - 6 x + 16 - 3 arcsin (\frac{1}{5} (x + 3)))

\int \frac{6}{16 - 6 x - x ^{2}} d x = 6 arcsin (\frac{1}{5} (x + 3))

= 2 (- - x^{2} - 6 x + 16 - 3 arcsin (\frac{1}{5} (x + 3))) + 6 arcsin (\frac{1}{5} (x + 3))

簡素化

2 (- - x^{2} - 6 x + 16 - 3 arcsin (\frac{1}{5} (x + 3))) + 6 arcsin (\frac{1}{5} (x + 3)) : - 2 - x^{2} - 6 x + 16

= - 2 - x^{2} - 6 x + 16

定数を解答に追加する

= - 2 - x^{2} - 6 x + 16 + C