integral of (x-1)/(x^2+4x+13)

解

\int \frac{x - 1}{x ^{2} + 4 x + 13} d x

\frac{1}{2} ln \frac{( x + 2 ) ^{2}}{9} + 1 - \frac{2}{3} arctan (\frac{1}{3} (x + 2)) - \frac{1}{3} arctan (\frac{x + 2}{3}) + C

解答ステップ

\int \frac{x - 1}{x ^{2} + 4 x + 13} d x

拡張

\frac{x - 1}{x ^{2} + 4 x + 13} : \frac{x}{x ^{2} + 4 x + 13} - \frac{1}{x ^{2} + 4 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{x}{x ^{2} + 4 x + 13} d x - \int \frac{1}{x ^{2} + 4 x + 13} d x

\int \frac{x}{x ^{2} + 4 x + 13} d x = \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{9} + 1 - \frac{2}{3} arctan (\frac{1}{3} (x + 2))

\int \frac{1}{x ^{2} + 4 x + 13} d x = \frac{1}{3} arctan (\frac{x + 2}{3})

= \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{9} + 1 - \frac{2}{3} arctan (\frac{1}{3} (x + 2)) - \frac{1}{3} arctan (\frac{x + 2}{3})

定数を解答に追加する

= \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{9} + 1 - \frac{2}{3} arctan (\frac{1}{3} (x + 2)) - \frac{1}{3} arctan (\frac{x + 2}{3}) + C