integral of (2x^3+2x^2+6x+9)/(x^3+3x)

解

\int \frac{2 x ^{3} + 2 x ^{2} + 6 x + 9}{x ^{3} + 3 x} d x

2 x - \frac{1}{2} ln x^{2} + 3 + 3 ln ∣ x ∣ + C

解答ステップ

\int \frac{2 x ^{3} + 2 x ^{2} + 6 x + 9}{x ^{3} + 3 x} d x

拡張

\frac{2 x ^{3} + 2 x ^{2} + 6 x + 9}{x ^{3} + 3 x} : \frac{2 x ^{3}}{x ^{3} + 3 x} + \frac{2 x ^{2}}{x ^{3} + 3 x} + \frac{6 x}{x ^{3} + 3 x} + \frac{9}{x ^{3} + 3 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{2 x ^{3}}{x ^{3} + 3 x} d x + \int \frac{2 x ^{2}}{x ^{3} + 3 x} d x + \int \frac{6 x}{x ^{3} + 3 x} d x + \int \frac{9}{x ^{3} + 3 x} d x

\int \frac{2 x ^{3}}{x ^{3} + 3 x} d x = - 23 arctan (\frac{x}{3}) + 2 x

\int \frac{2 x ^{2}}{x ^{3} + 3 x} d x = ln x^{2} + 3

\int \frac{6 x}{x ^{3} + 3 x} d x = 23 arctan (\frac{x}{3})

\int \frac{9}{x ^{3} + 3 x} d x = 9 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3)

= - 23 arctan (\frac{x}{3}) + 2 x + ln x^{2} + 3 + 23 arctan (\frac{x}{3}) + 9 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3)

簡素化

- 23 arctan (\frac{x}{3}) + 2 x + ln x^{2} + 3 + 23 arctan (\frac{x}{3}) + 9 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3) : 2 x - \frac{1}{2} ln x^{2} + 3 + 3 ln ∣ x ∣

= 2 x - \frac{1}{2} ln x^{2} + 3 + 3 ln ∣ x ∣

定数を解答に追加する

= 2 x - \frac{1}{2} ln x^{2} + 3 + 3 ln ∣ x ∣ + C