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

解

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

\frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{4} sin (2 arctan (x - 1)) + \frac{1}{2} arctan (x - 1) - \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{x ^{2} - 2 x + 2} + C

解答ステップ

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

拡張

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

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

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

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

簡素化

\frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{2} sin (2 arctan (x - 1)) - \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{x ^{2} - 2 x + 2} + \frac{1}{2} (arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1))) : \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{4} sin (2 arctan (x - 1)) + \frac{1}{2} arctan (x - 1) - \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{x ^{2} - 2 x + 2}

= \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{4} sin (2 arctan (x - 1)) + \frac{1}{2} arctan (x - 1) - \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{x ^{2} - 2 x + 2}

定数を解答に追加する

= \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{4} sin (2 arctan (x - 1)) + \frac{1}{2} arctan (x - 1) - \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{x ^{2} - 2 x + 2} + C