integral of (x^6-x^3+16)/(x^4+4x^2)

解

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

\frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2}) + C

解答ステップ

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

以下の部分分数を得る:

\frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} : x^{2} - 4 + \frac{4}{x ^{2}} + \frac{- x + 12}{x ^{2} + 4}

= \int x^{2} - 4 + \frac{4}{x ^{2}} + \frac{- x + 12}{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 x^{2} d x - \int 4 d x + \int \frac{4}{x ^{2}} d x + \int \frac{- x + 12}{x ^{2} + 4} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 4 d x = 4 x

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

\int \frac{- x + 12}{x ^{2} + 4} d x = - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2})

= \frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2})

定数を解答に追加する

= \frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2}) + C