integral of (-6x^3+8x^2-30x+10)/(x^3+5x)

解

\int \frac{- 6 x ^{3} + 8 x ^{2} - 30 x + 10}{x ^{3} + 5 x} d x

- 6 x + 3 ln x^{2} + 5 + 2 ln ∣ x ∣ + C

解答ステップ

\int \frac{- 6 x ^{3} + 8 x ^{2} - 30 x + 10}{x ^{3} + 5 x} d x

拡張

\frac{- 6 x ^{3} + 8 x ^{2} - 30 x + 10}{x ^{3} + 5 x} : - \frac{6 x ^{3}}{x ^{3} + 5 x} + \frac{8 x ^{2}}{x ^{3} + 5 x} - \frac{30 x}{x ^{3} + 5 x} + \frac{10}{x ^{3} + 5 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{6 x ^{3}}{x ^{3} + 5 x} d x + \int \frac{8 x ^{2}}{x ^{3} + 5 x} d x - \int \frac{30 x}{x ^{3} + 5 x} d x + \int \frac{10}{x ^{3} + 5 x} d x

\int \frac{6 x ^{3}}{x ^{3} + 5 x} d x = - 65 arctan (\frac{x}{5}) + 6 x

\int \frac{8 x ^{2}}{x ^{3} + 5 x} d x = 4 ln x^{2} + 5

\int \frac{30 x}{x ^{3} + 5 x} d x = 65 arctan (\frac{x}{5})

\int \frac{10}{x ^{3} + 5 x} d x = 2 ln ∣ x ∣ - ln x^{2} + 5

= - (- 65 arctan (\frac{x}{5}) + 6 x) + 4 ln x^{2} + 5 - 65 arctan (\frac{x}{5}) + 2 ln ∣ x ∣ - ln x^{2} + 5

簡素化

- (- 65 arctan (\frac{x}{5}) + 6 x) + 4 ln x^{2} + 5 - 65 arctan (\frac{x}{5}) + 2 ln ∣ x ∣ - ln x^{2} + 5 : - 6 x + 3 ln x^{2} + 5 + 2 ln ∣ x ∣

= - 6 x + 3 ln x^{2} + 5 + 2 ln ∣ x ∣

定数を解答に追加する

= - 6 x + 3 ln x^{2} + 5 + 2 ln ∣ x ∣ + C