integral of (-8x+5)/(x^3+4x)

解

\int \frac{- 8 x + 5}{x ^{3} + 4 x} d x

\frac{5}{4} ln ∣ x ∣ - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2})) + C

解答ステップ

\int \frac{- 8 x + 5}{x ^{3} + 4 x} d x

以下の部分分数を得る:

\frac{- 8 x + 5}{x ^{3} + 4 x} : \frac{5}{4 x} + \frac{- 5 x - 32}{4 ( x ^{2} + 4 )}

= \int \frac{5}{4 x} + \frac{- 5 x - 32}{4 ( x ^{2} + 4 )} 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{5}{4 x} d x + \int \frac{- 5 x - 32}{4 ( x ^{2} + 4 )} d x

\int \frac{5}{4 x} d x = \frac{5}{4} ln ∣ x ∣

\int \frac{- 5 x - 32}{4 ( x ^{2} + 4 )} d x = - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2}))

= \frac{5}{4} ln ∣ x ∣ - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2}))

定数を解答に追加する

= \frac{5}{4} ln ∣ x ∣ - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2})) + C