integral of (x+3)/(sqrt(x^2-4))

解

\int \frac{x + 3}{x ^{2} - 4} d x

x^{2} - 4 + 3 ln (\frac{1}{2} x^{2} - 4 + x) + C

解答ステップ

\int \frac{x + 3}{x ^{2} - 4} d x

拡張

\frac{x + 3}{x ^{2} - 4} : \frac{x}{x ^{2} - 4} + \frac{3}{x ^{2} - 4}

= \int \frac{x}{x ^{2} - 4} + \frac{3}{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{3}{x ^{2} - 4} d x

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

\int \frac{3}{x ^{2} - 4} d x = 3 ln (\frac{1}{2} x^{2} - 4 + x)

= x^{2} - 4 + 3 ln (\frac{1}{2} x^{2} - 4 + x)

定数を解答に追加する

= x^{2} - 4 + 3 ln (\frac{1}{2} x^{2} - 4 + x) + C