integral of (x-3)/(x^2+4x+8)

解

\int \frac{x - 3}{x ^{2} + 4 x + 8} d x

\frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1) + C

解答ステップ

\int \frac{x - 3}{x ^{2} + 4 x + 8} d x

拡張

\frac{x - 3}{x ^{2} + 4 x + 8} : \frac{x}{x ^{2} + 4 x + 8} - \frac{3}{x ^{2} + 4 x + 8}

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 + 8} d x - \int \frac{3}{x ^{2} + 4 x + 8} d x

\int \frac{x}{x ^{2} + 4 x + 8} d x = \frac{1}{2} ln \frac{1}{4} (x + 2)^{2} + 1 - arctan (\frac{1}{2} (x + 2))

\int \frac{3}{x ^{2} + 4 x + 8} d x = \frac{3}{2} arctan (\frac{1}{2} (x + 2))

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

簡素化

\frac{1}{2} ln \frac{1}{4} (x + 2)^{2} + 1 - arctan (\frac{1}{2} (x + 2)) - \frac{3}{2} arctan (\frac{1}{2} (x + 2)) : \frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1)

= \frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1)

定数を解答に追加する

= \frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1) + C