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

解

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

3 ln ∣ x ∣ - \frac{2}{x} - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2}) + C

解答ステップ

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

以下の部分分数を得る:

\frac{3 x ( 3 x + 2 ) + 4}{x ^{4} + 2 x ^{2}} : \frac{3}{x} + \frac{2}{x ^{2}} + \frac{- 3 x + 7}{x ^{2} + 2}

= \int \frac{3}{x} + \frac{2}{x ^{2}} + \frac{- 3 x + 7}{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{3}{x} d x + \int \frac{2}{x ^{2}} d x + \int \frac{- 3 x + 7}{x ^{2} + 2} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{- 3 x + 7}{x ^{2} + 2} d x = - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2})

= 3 ln ∣ x ∣ - \frac{2}{x} - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2})

定数を解答に追加する

= 3 ln ∣ x ∣ - \frac{2}{x} - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2}) + C