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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

定数を解答に追加する

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