integral of (2x+6)/(sqrt(9-x^2))

解

\int \frac{2 x + 6}{9 - x ^{2}} d x

- 2 9 - x^{2} + 6 arcsin (\frac{1}{3} x) + C

解答ステップ

\int \frac{2 x + 6}{9 - x ^{2}} d x

拡張

\frac{2 x + 6}{9 - x ^{2}} : \frac{2 x}{9 - x ^{2}} + \frac{6}{9 - x ^{2}}

= \int \frac{2 x}{9 - x ^{2}} + \frac{6}{9 - 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}{9 - x ^{2}} d x + \int \frac{6}{9 - x ^{2}} d x

\int \frac{2 x}{9 - x ^{2}} d x = - 2 9 - x^{2}

\int \frac{6}{9 - x ^{2}} d x = 6 arcsin (\frac{1}{3} x)

= - 2 9 - x^{2} + 6 arcsin (\frac{1}{3} x)

定数を解答に追加する

= - 2 9 - x^{2} + 6 arcsin (\frac{1}{3} x) + C