integral of 1/(1+sqrt(x+2))

解

\int \frac{1}{1 + x + 2} d x

- 2 ln x + 2 + 1 + 2 x + 2 + 1 + C

解答ステップ

\int \frac{1}{1 + x + 2} d x

\frac{1}{1 + x + 2} = - \frac{x + 2}{1 + x + 2} + 1

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

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

\int 1 d x = x

= - (x + 2 ln x + 2 + 1 - 2 x + 2 - 1) + x

簡素化

= - 2 ln x + 2 + 1 + 2 x + 2 + 1

定数を解答に追加する

= - 2 ln x + 2 + 1 + 2 x + 2 + 1 + C