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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

定数を解答に追加する

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