integral of (2x^3-x^2+4x+8)/(x^4+4x^2)

解

\int \frac{2 x ^{3} - x ^{2} + 4 x + 8}{x ^{4} + 4 x ^{2}} d x

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

解答ステップ

\int \frac{2 x ^{3} - x ^{2} + 4 x + 8}{x ^{4} + 4 x ^{2}} d x

拡張

\frac{2 x ^{3} - x ^{2} + 4 x + 8}{x ^{4} + 4 x ^{2}} : \frac{2 x ^{3}}{x ^{4} + 4 x ^{2}} - \frac{x ^{2}}{x ^{4} + 4 x ^{2}} + \frac{4 x}{x ^{4} + 4 x ^{2}} + \frac{8}{x ^{4} + 4 x ^{2}}

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{2 x ^{3}}{x ^{4} + 4 x ^{2}} d x - \int \frac{x ^{2}}{x ^{4} + 4 x ^{2}} d x + \int \frac{4 x}{x ^{4} + 4 x ^{2}} d x + \int \frac{8}{x ^{4} + 4 x ^{2}} d x

\int \frac{2 x ^{3}}{x ^{4} + 4 x ^{2}} d x = ln x^{2} + 4

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

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

\int \frac{8}{x ^{4} + 4 x ^{2}} d x = - \frac{2}{x} - arctan (\frac{x}{2})

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

簡素化

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

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

定数を解答に追加する

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