integral of (2x^2+x+1)/((x-1)(x^2+1))

解

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

2 ln ∣ x - 1 ∣ + arctan (x) + C

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

= 2 ln ∣ x - 1 ∣ + arctan (x)

定数を解答に追加する

= 2 ln ∣ x - 1 ∣ + arctan (x) + C