integral of (2x+1)/(x^2-6x+13)

解

\int \frac{2 x + 1}{x ^{2} - 6 x + 13} d x

ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} arctan (\frac{x - 3}{2}) + C

解答ステップ

\int \frac{2 x + 1}{x ^{2} - 6 x + 13} d x

簡素化

x^{2} : xx

= \int \frac{2 x + 1}{xx - 6 x + 13} d x

拡張

\frac{2 x + 1}{xx - 6 x + 13} : \frac{2 x}{xx - 6 x + 13} + \frac{1}{xx - 6 x + 13}

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}{xx - 6 x + 13} d x + \int \frac{1}{xx - 6 x + 13} d x

\int \frac{2 x}{xx - 6 x + 13} d x = ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3))

\int \frac{1}{xx - 6 x + 13} d x = \frac{1}{2} arctan (\frac{x - 3}{2})

= ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} arctan (\frac{x - 3}{2})

定数を解答に追加する

= ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} arctan (\frac{x - 3}{2}) + C