integral of (x^3+4)/(x^5+14x^3+40x)

解

\int \frac{x ^{3} + 4}{x ^{5} + 14 x ^{3} + 40 x} d x

\frac{1}{10} ln ∣ x ∣ + \frac{1}{6} (- \frac{1}{2} ln x^{2} + 4 - 2 arctan (\frac{x}{2})) + \frac{1}{15} (\frac{1}{2} ln x^{2} + 10 + \frac{5 5}{2} arctan (\frac{x}{10})) + C

解答ステップ

\int \frac{x ^{3} + 4}{x ^{5} + 14 x ^{3} + 40 x} d x

以下の部分分数を得る:

\frac{x ^{3} + 4}{x ^{5} + 14 x ^{3} + 40 x} : \frac{1}{10 x} + \frac{- x - 4}{6 ( x ^{2} + 4 )} + \frac{x + 25}{15 ( x ^{2} + 10 )}

= \int \frac{1}{10 x} + \frac{- x - 4}{6 ( x ^{2} + 4 )} + \frac{x + 25}{15 ( x ^{2} + 10 )} d 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{1}{10 x} d x + \int \frac{- x - 4}{6 ( x ^{2} + 4 )} d x + \int \frac{x + 25}{15 ( x ^{2} + 10 )} d x

\int \frac{1}{10 x} d x = \frac{1}{10} ln ∣ x ∣

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

\int \frac{x + 25}{15 ( x ^{2} + 10 )} d x = \frac{1}{15} (\frac{1}{2} ln x^{2} + 10 + \frac{5 5}{2} arctan (\frac{x}{10}))

= \frac{1}{10} ln ∣ x ∣ + \frac{1}{6} (- \frac{1}{2} ln x^{2} + 4 - 2 arctan (\frac{x}{2})) + \frac{1}{15} (\frac{1}{2} ln x^{2} + 10 + \frac{5 5}{2} arctan (\frac{x}{10}))

定数を解答に追加する

= \frac{1}{10} ln ∣ x ∣ + \frac{1}{6} (- \frac{1}{2} ln x^{2} + 4 - 2 arctan (\frac{x}{2})) + \frac{1}{15} (\frac{1}{2} ln x^{2} + 10 + \frac{5 5}{2} arctan (\frac{x}{10})) + C