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

解

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

\frac{1}{2} ln x^{2} + 6 x + 13 - \frac{5}{2} arctan (\frac{1}{2} (x + 3)) + C

解答ステップ

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

拡張

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

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

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

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

類似した元を足す:

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

= \frac{1}{2} ln x^{2} + 6 x + 13 - \frac{5}{2} arctan (\frac{1}{2} (x + 3))

定数を解答に追加する

= \frac{1}{2} ln x^{2} + 6 x + 13 - \frac{5}{2} arctan (\frac{1}{2} (x + 3)) + C