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

解

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

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

解答ステップ

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

以下の部分分数を得る:

\frac{3 x ^{2} + x + 4}{x ^{4} + 3 x ^{2} + 2} : \frac{x + 1}{x ^{2} + 1} + \frac{- x + 2}{x ^{2} + 2}

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

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

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

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

定数を解答に追加する

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