integral of (3x^2-2x+19)/((x-1)(x^2+9))

解

\int \frac{3 x ^{2} - 2 x + 19}{( x - 1 ) ( x ^{2} + 9 )} d x

2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}) + C

解答ステップ

\int \frac{3 x ^{2} - 2 x + 19}{( x - 1 ) ( x ^{2} + 9 )} d x

以下の部分分数を得る:

\frac{3 x ^{2} - 2 x + 19}{( x - 1 ) ( x ^{2} + 9 )} : \frac{2}{x - 1} + \frac{x - 1}{x ^{2} + 9}

= \int \frac{2}{x - 1} + \frac{x - 1}{x ^{2} + 9} 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{2}{x - 1} d x + \int \frac{x - 1}{x ^{2} + 9} d x

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

\int \frac{x - 1}{x ^{2} + 9} d x = \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})

= 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})

定数を解答に追加する

= 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}) + C