integral of (9x^2-8x+24)/(x^3+4x)

解

\int \frac{9 x ^{2} - 8 x + 24}{x ^{3} + 4 x} d x

\frac{3}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 6 ln ∣ x ∣ + C

解答ステップ

\int \frac{9 x ^{2} - 8 x + 24}{x ^{3} + 4 x} d x

拡張

\frac{9 x ^{2} - 8 x + 24}{x ^{3} + 4 x} : \frac{9 x ^{2}}{x ^{3} + 4 x} - \frac{8 x}{x ^{3} + 4 x} + \frac{24}{x ^{3} + 4 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{9 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{8 x}{x ^{3} + 4 x} d x + \int \frac{24}{x ^{3} + 4 x} d x

\int \frac{9 x ^{2}}{x ^{3} + 4 x} d x = \frac{9}{2} ln x^{2} + 4

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

\int \frac{24}{x ^{3} + 4 x} d x = 24 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= \frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 24 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

簡素化

\frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 24 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : \frac{3}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 6 ln ∣ x ∣

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

定数を解答に追加する

= \frac{3}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 6 ln ∣ x ∣ + C