integral of ((x+1))/(x^2-2x+5)

解

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

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

解答ステップ

\int \frac{x + 1}{x ^{2} - 2 x + 5} d x

拡張

\frac{x + 1}{x ^{2} - 2 x + 5} : \frac{x}{x ^{2} - 2 x + 5} + \frac{1}{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{x}{x ^{2} - 2 x + 5} d x + \int \frac{1}{x ^{2} - 2 x + 5} d x

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

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

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

定数を解答に追加する

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