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

解

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

x^{2} + 4 + 2 ln (\frac{1}{2} x + 4 + x^{2}) + C

解答ステップ

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

拡張

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

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

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

\int \frac{2}{x ^{2} + 4} d x = 2 ln (\frac{1}{2} x + 4 + x^{2})

= x^{2} + 4 + 2 ln (\frac{1}{2} x + 4 + x^{2})

定数を解答に追加する

= x^{2} + 4 + 2 ln (\frac{1}{2} x + 4 + x^{2}) + C