integral of (x+4)/(x^2+4x+6)

解

\int \frac{x + 4}{x ^{2} + 4 x + 6} d x

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

解答ステップ

\int \frac{x + 4}{x ^{2} + 4 x + 6} d x

拡張

\frac{x + 4}{x ^{2} + 4 x + 6} : \frac{x}{x ^{2} + 4 x + 6} + \frac{4}{x ^{2} + 4 x + 6}

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 + 6} d x + \int \frac{4}{x ^{2} + 4 x + 6} d x

\int \frac{x}{x ^{2} + 4 x + 6} d x = \frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2))

\int \frac{4}{x ^{2} + 4 x + 6} d x = 22 arctan (\frac{1}{2} (x + 2))

= \frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2)) + 22 arctan (\frac{1}{2} (x + 2))

簡素化

\frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2)) + 22 arctan (\frac{1}{2} (x + 2)) : \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 + 2 arctan (\frac{1}{2} (x + 2))

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

定数を解答に追加する

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