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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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