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

解

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

x^{2} + 2 + 2 ln \frac{1}{2} x + \frac{1}{2} (2 + x^{2}) + C

解答ステップ

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

拡張

\frac{x + 2}{x ^{2} + 2} : \frac{x}{x ^{2} + 2} + \frac{2}{x ^{2} + 2}

= \int \frac{x}{x ^{2} + 2} + \frac{2}{x ^{2} + 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}{x ^{2} + 2} d x + \int \frac{2}{x ^{2} + 2} d x

\int \frac{x}{x ^{2} + 2} d x = x^{2} + 2

\int \frac{2}{x ^{2} + 2} d x = 2 ln \frac{1}{2} x + \frac{1}{2} (2 + x^{2})

= x^{2} + 2 + 2 ln \frac{1}{2} x + \frac{1}{2} (2 + x^{2})

定数を解答に追加する

= x^{2} + 2 + 2 ln \frac{1}{2} x + \frac{1}{2} (2 + x^{2}) + C