integral of [(2x-3)/(x^2+2x+5)]

解

\int [\frac{2 x - 3}{x ^{2} + 2 x + 5}] d x

ln x^{2} + 2 x + 5 - \frac{5}{2} arctan (\frac{1}{2} (x + 1)) + C

解答ステップ

\int \frac{2 x - 3}{x ^{2} + 2 x + 5} d x

拡張

\frac{2 x - 3}{x ^{2} + 2 x + 5} : \frac{2 x}{x ^{2} + 2 x + 5} - \frac{3}{x ^{2} + 2 x + 5}

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 + 5} d x - \int \frac{3}{x ^{2} + 2 x + 5} d x

\int \frac{2 x}{x ^{2} + 2 x + 5} d x = ln x^{2} + 2 x + 5 - arctan (\frac{1}{2} (x + 1))

\int \frac{3}{x ^{2} + 2 x + 5} d x = \frac{3}{2} arctan (\frac{1}{2} (x + 1))

= ln x^{2} + 2 x + 5 - arctan (\frac{1}{2} (x + 1)) - \frac{3}{2} arctan (\frac{1}{2} (x + 1))

類似した元を足す:

- arctan (\frac{1}{2} (x + 1)) - \frac{3}{2} arctan (\frac{1}{2} (x + 1)) = - \frac{5}{2} arctan (\frac{1}{2} (x + 1))

= ln x^{2} + 2 x + 5 - \frac{5}{2} arctan (\frac{1}{2} (x + 1))

定数を解答に追加する

= ln x^{2} + 2 x + 5 - \frac{5}{2} arctan (\frac{1}{2} (x + 1)) + C