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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

定数を解答に追加する

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