integral of ((x^2+x+6))/(x^3+4x^2+6x)

解

\int \frac{( x ^{2} + x + 6 )}{x ^{3} + 4 x ^{2} + 6 x} d x

\frac{1}{2} arctan (\frac{1}{2} (x + 2)) - 22 arctan (\frac{1}{2} (x + 2)) + ln ∣ x ∣ + C

解答ステップ

\int \frac{x ^{2} + x + 6}{x ^{3} + 4 x ^{2} + 6 x} d x

拡張

\frac{x ^{2} + x + 6}{x ^{3} + 4 x ^{2} + 6 x} : \frac{x ^{2}}{x ^{3} + 4 x ^{2} + 6 x} + \frac{x}{x ^{3} + 4 x ^{2} + 6 x} + \frac{6}{x ^{3} + 4 x ^{2} + 6 x}

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 ^{2}}{x ^{3} + 4 x ^{2} + 6 x} d x + \int \frac{x}{x ^{3} + 4 x ^{2} + 6 x} d x + \int \frac{6}{x ^{3} + 4 x ^{2} + 6 x} d x

\int \frac{x ^{2}}{x ^{3} + 4 x ^{2} + 6 x} d x = \frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2))

\int \frac{x}{x ^{3} + 4 x ^{2} + 6 x} d x = \frac{1}{2} arctan (\frac{x + 2}{2})

\int \frac{6}{x ^{3} + 4 x ^{2} + 6 x} d x = ln ∣ x ∣ - \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} + 1 - 2 arctan (\frac{1}{2} (x + 2)) + \frac{1}{2} arctan (\frac{x + 2}{2}) + ln ∣ x ∣ - \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 - 2 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)) + \frac{1}{2} arctan (\frac{x + 2}{2}) + ln ∣ x ∣ - \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 - 2 arctan (\frac{1}{2} (x + 2)) : \frac{1}{2} arctan (\frac{1}{2} (x + 2)) - 22 arctan (\frac{1}{2} (x + 2)) + ln ∣ x ∣

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

定数を解答に追加する

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