integral of (x^3+1)/(x(x^2+1)^2)

解

\int \frac{x ^{3} + 1}{x ( x ^{2} + 1 ) ^{2}} d x

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

解答ステップ

\int \frac{x ^{3} + 1}{x ( x ^{2} + 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{x ^{3} + 1}{x ( x ^{2} + 1 ) ^{2}} : \frac{1}{x} + \frac{- x + 1}{x ^{2} + 1} + \frac{- x - 1}{( x ^{2} + 1 ) ^{2}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

簡素化

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

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

定数を解答に追加する

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