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

解

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

\frac{4}{5} arctan (x) + \frac{3}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln x^{2} + 1 + C

解答ステップ

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

拡張

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

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

\int \frac{x ^{2}}{( x ^{2} + 1 ) ( x - 2 )} d x = \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + \frac{4}{5} ln ∣ x - 2 ∣

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

= \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + \frac{4}{5} ln ∣ x - 2 ∣ - (\frac{1}{5} (- \frac{1}{2} ln x^{2} + 1 - 2 arctan (x)) + \frac{1}{5} ln ∣ x - 2 ∣)

簡素化

\frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + \frac{4}{5} ln ∣ x - 2 ∣ - (\frac{1}{5} (- \frac{1}{2} ln x^{2} + 1 - 2 arctan (x)) + \frac{1}{5} ln ∣ x - 2 ∣) : \frac{4}{5} arctan (x) + \frac{3}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln x^{2} + 1

= \frac{4}{5} arctan (x) + \frac{3}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln x^{2} + 1

定数を解答に追加する

= \frac{4}{5} arctan (x) + \frac{3}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln x^{2} + 1 + C