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

解

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

\frac{4 x ^{3}}{x ^{2} + 1} + \frac{4 x}{x ^{2} + 1} + 4 ln ∣ x ∣ - 2 ln x^{2} + 1 + C

解答ステップ

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

拡張

\frac{4 x ^{3} + 4 x + 4}{x ^{3} + x} : \frac{4 x ^{3}}{x ^{3} + x} + \frac{4 x}{x ^{3} + x} + \frac{4}{x ^{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{4 x ^{3}}{x ^{3} + x} d x + \int \frac{4 x}{x ^{3} + x} d x + \int \frac{4}{x ^{3} + x} d x

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

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

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

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

簡素化

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

= \frac{4 x ^{3}}{x ^{2} + 1} + \frac{4 x}{x ^{2} + 1} + 4 ln ∣ x ∣ - 2 ln x^{2} + 1

定数を解答に追加する

= \frac{4 x ^{3}}{x ^{2} + 1} + \frac{4 x}{x ^{2} + 1} + 4 ln ∣ x ∣ - 2 ln x^{2} + 1 + C