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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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