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

解

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

\frac{7}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{1}{2 ( x ^{2} - 1 )} - \frac{3}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )} + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - (- \frac{1}{2 ( x ^{2} - 1 )}) + 3 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) : \frac{7}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{1}{2 ( x ^{2} - 1 )} - \frac{3}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )}

= \frac{7}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{1}{2 ( x ^{2} - 1 )} - \frac{3}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )}

定数を解答に追加する

= \frac{7}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{1}{2 ( x ^{2} - 1 )} - \frac{3}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )} + C