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

解

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

\frac{1}{5} (- ln x^{2} + 1 + arctan (x)) + \frac{2}{5} ln ∣ x - 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{x}{( x ^{2} + 1 ) ( x - 2 )} : \frac{- 2 x + 1}{5 ( x ^{2} + 1 )} + \frac{2}{5 ( x - 2 )}

= \int \frac{- 2 x + 1}{5 ( x ^{2} + 1 )} + \frac{2}{5 ( x - 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{- 2 x + 1}{5 ( x ^{2} + 1 )} d x + \int \frac{2}{5 ( x - 2 )} d x

\int \frac{- 2 x + 1}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (- ln x^{2} + 1 + arctan (x))

\int \frac{2}{5 ( x - 2 )} d x = \frac{2}{5} ln ∣ x - 2 ∣

= \frac{1}{5} (- ln x^{2} + 1 + arctan (x)) + \frac{2}{5} ln ∣ x - 2 ∣

定数を解答に追加する

= \frac{1}{5} (- ln x^{2} + 1 + arctan (x)) + \frac{2}{5} ln ∣ x - 2 ∣ + C